PHYSICS FOR YOU
MARCH ’18
3
4
PHYSICS FOR YOU
MARCH ’18
An Institute For Excellence In Science ...
An Insti tute ForE xc el lenc e I n S ci ence
PHYSICS FOR YOU
MARCH ’18
5
6
PHYSICS FOR YOU
MARCH ’18
Volume 26 Managing Editor Mahabir Singh Editor Anil Ahlawat
No. 3
March 2018
Corporate Office: Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (HR). Tel : 0124-6601200 e-mail :
[email protected] website : www.mtg.in Regd. Office: 406, Taj Apartment, Near Safdarjung Hospital, New Delhi - 110029.
Competition Edge
S T N E T N O C
Physics Musing Problem Set 56
8
JEE Main Practice Paper
11
NEET Practice Paper
18
Physics Musing Solution Set 55
26
JEE Advanced Practice Paper
27
Gear Up for AIIMS
40
BITSAT Full Length Practice Paper
53
Class 11
Brain Map
46
MPP
69
Class 12
Brain Map
47
CBSE Board Practice Paper
73
MPP
82
Subscribe online at www.mtg.in Individual Subscription Rates
Combined Subscription Rates
1 yr.
2 yrs.
3 yrs.
Mathematics Today
330
600
775
PCM
1 yr.
2 yrs.
3 yrs.
900
1500
1900
Chemistry Today
330
600
775
PCB
900
1500
1900
Physics For You
330
600
775
PCMB
1000
1800
2300
Biology Today
330
600
775
Send D.D/M.O in favour of MTG Learning Media (P) Ltd. Payments should be made directly to : MTG Learning Media (P) Ltd, Plot No. 99, Sector 44, Gurgaon - 122003 (Haryana) We have not appointed any subscription agent.
Printed and Published by Mahabir Singh on behalf of MTG Learning Media Pvt. Ltd. Printed at HT Media Ltd., B-2, Sector-63, Noida, UP-201307 and published at 406, Taj Apartment, Ring Road, Near Safdarjung Hospital, New Delhi - 110029. Editor : Anil Ahlawat Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives. MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only. Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited.
PHYSICS FOR YOU
|
MARCH ‘18
7
PHYSICS MUSING P
hysics Musing was started in August 2013 issue of Physics For You. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / NEET / AIIMS / JIPMER with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / NEET. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved ve or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast ve correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.
SUBJECTIVE TYPE 1.
Te final image I o the object O shown in the figure is ormed at point 20 cm below a thin equalconcave lens, which is at a depth o 65 cm rom principal axis o a convex lens. From the given geometry, calculate the radius o curvature (in cm) o lens kept at A.
2.
wo persons A and B wear glasses o optical powers (in air) P 1 = + 2 D and P 2 = + 1 D respectively. Te glasses have reractive index 1.5. Now they jump into a swimming pool and look at each other. B appears to be present at distance 2 m (rom A) to A. A appears to be present at distance 1 m (rom B) to B. Find out the reractive index o water in the swimming pool.
3.
A block is hanged by means o two identical wires having cross section area 1 mm2 as shown in the diagram. I temperature is lowered by 10°C, ind the mass (in kg) to be added to hanging mass such that junction remains at initial position. Given that co-efficient o linear expansion a = 2 × 10 –5 °C–1 and Young's modulus Y = 5 × 1011 N m–2 or the wire.
4.
In the figure shown, a conducting rod AB o length l , resistance R and mass m can move vertically downward due to gravity. Other parts are kept fixed. B = constant = B0. MN and PQ are vertical, smooth, conducting rails. Te capacitance o the capacitor is C . Te rod is released rom rest. Find the maximum current in the circuit.
5.
In the figure, a conducting rod o length l = 1 m and mass m = 1 kg moves with initial velocity u = 5 m s–1 on a fixed horizontal rame containing inductor L = 2 H and Q P resistance R = 1 W. PQ B and MN are smooth, L R u conducting wires. Tere is a uniorm magnetic field N M o strength B = 1 . Initially there is no current in the inductor. Find the total charge flown through the inductor by the time velocity o rod becomes v f = 1 m s–1 and the rod has travelled a distance x = 3 m.
6.
A uniorm rod o mass m is hinged at a point L/4 rom one end o the rod and is at rest. An impulse I is imparted on the other end o the rod perpendicular to it. Find the angular speed o the rod just afer the application o the impulse.
By Akhil Tewari, Author Foundation o f Physics for JEE Main & Advanced, Professor, IITians PACE, Mumbai. 8
PHYSICS FOR YOU
|
MARCH ‘18
PHYSICS FOR YOU
|
MARCH ‘18
9
7.
In the figure shown an observer O1 floats (static) on water surface with ears in air while another observer O2 is moving upwards with constant velocity v 1 = v /5 in water. Te source moves down with constant velocity v S = v /5 and emits sound of frequency u. Te velocity of sound in air is v and that in water is 4v . Find the frequency of sound received by O2.
when one foot is exerting 100 N force. Neglect friction within cycle parts and the rolling friction.
8.
Te tension in the upper portion of the chain is equal to (a) 100 N (b) 120 N (c) 160 N (d) 240 N
9.
Te power delivered by the cyclist is equal to (a) 28 p W (b) 10 p W (c) 64 p W (d) 32 p W
10.
Te net force of the friction on the rear wheel due to the road is (a) 100 N (b) 62 N (c) 32.6 N (d) 18.3 N
COMPREHENSION TYPE
A bicycle has pedal rods of length 16 cm connected to a sprocketed disc of radius 10 cm. Te bicycle wheels are 70 cm in diameter and the chain runs over a gear of radius 4 cm. Te speed of the cycle is constant and the cyclist applies 100 N force that is always perpendicular to the pedal rod, as shown. Assume tension in the lower part of chain negligible. Te cyclist is peddling at a constant rate of 2 rps. Assume that the force applied by other foot is zero
10
PHYSICS FOR YOU
|
MARCH ‘18
2018
Exam on 8th April (Offline) 15th & 16th April (Online)
1.
(a) (c) 2.
magnitudes of 2.50 × 104 N C–1 and 7.00 × 10 4 N C–1, respectively. Assuming that no other electric field lines cross the surfaces of the parallelopipe, the net charge contained within is (a) –67.5e0 C (b) 37.5e0 C (c) 105e0 C (d) –105e0 C
A man is in a satellite which is at a distance of 2000 km from the surface of the earth. Te man has to be thrown out so that he can escape the earth completely. With what velocity should he be thrown out? 2GM
R
2GM
(b)
GM R + 2000
(d)
R
+ 2000
1
GM
20
5
One end of a light A Ring spring of natural length l d and spring constant k h is fixed on a rigid wall and the other end is 37° attached to a smooth P d B Rod ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of 37° with the horizontal as shown in figure. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. (sin 37° = 3/5) 3 g k k 4 g + (a) d (b) d + 2d 16m 3d 4m (c)
3 gd −
kl 2 4m
(d) l
3 g 2d
−
4.
5.
k 16m (a) 4 × 10–3 J (c) 8 × 10 –4 J
3.
Te electric field E1 E 2 at one face of a parallelopiped is 6 cm uniform over the entire face and is m c E 1 directed out of the 0 30° 0 . 5 face. At the opposite face, the electric field E2 is also uniform over the entire face and is directed into that face as shown in figure. Te two faces are inclined at 30° from the horizontal. E1 and E2 (both horizontal) have
(b) 6 × 10–4 J (d) 3 × 10–3 J
6.
Given that ln (a/P b) = az /kBq where P is pressure, z is distance, kB is Boltzmann constant and q is temperature. Te dimensions of b are (a) [M0L0 0] (b) [M–1L1–2] (c) [M0L2 0] (d) [M1L–1–2]
7.
In the given figure, a mass 5 kg slides without friction on an inclined plane making an angle 30° with the horizontal. If this mass is connected to another mass of 10 kg through massless and frictionless
A thick rubber rope of density r and length L is suspended from a rigid support. If Y is the Young’s modulus of elasticity of the material of the rope, then the increase in length of the rope due to its own weight is 2 1 ρ gL ρ gL ρ gL (a) r gL2Y (b) (c) (d) 2 6Y Y 2Y A part of a circuit in steady state along with the current flowing in the branches. Value of each resistance is shown in figure. Calculate the energy stored in the 4 mF capacitor.
PHYSICS FOR YOU
|
MARCH ‘18
11
pulleys, then what is the acceleration of this mass when it is moving upwards is (ake g = 10 m s–2.) (a) 0.33 m s–2 (b) 3.3 m s–2 (c) 33 m s–2 (d) 3.03 m s–2 8.
Tree closed vessels A, B and C are at the same temperature T and contain gases which obey the Maxwell distribution of speed. Vessel A contains only O2, B only N2 and C a mixture of equal quantities of O2 and N2. If the average speed of O2 molecules in vessel A is v 1 and that of the N2 molecules in vessels B is v 2, then average speed of the O2 molecules in vessel C will be (a) (v 1 + v 2)/2 (b) v 1 v (c) (v 1v 2)1/2 (d) 1
has been graduated at a temperature of 20°C? (a = 1.1 × 10–5 °C–1). Assume that the length of the bar does not change.) (a) 1.98 × 10–1 mm (b) 1.98 × 10–2 mm (c) 1.98 × 10–3 mm (d) 1.98 × 10–4 mm 12.
(c) 13.
2
9.
10.
wo long parallel wires carry currents of equal magnitude but in opposite P directions. Tese wires are L Q suspended from rod PQ by I four chords of same I length L as shown in the 2L sin figure. Te mass per unit length of the wire is l. Determine the value of q assuming it to be small. (a) I
µ0 gL 4 πλ
(b) I
(c) I
µ0 2 πλ g
(d)
12
4 πε0 rh
ce 2
1 1 − (b) λ λ 0
4 πε0 h
rce2
1 1 − λ λ 0
2 πε0 rhc 1 1 1 1 − − (d) 2 λ λ λ λ e 0 e2 0 A bar measured with a Vernier caliper is found to be 180 mm long. Te temperature during the measurement is 10 °C. What will be the error in measurement if the scale of the Vernier caliper
(c) 11.
µ0 I g
4 πε0 rhc
PHYSICS FOR YOU
|
MARCH ‘18
3( π − 1)
1 3
ρ gH
2
w
(d)
a 3(π + 1)
(d)
1 2
O
2
ρ gwH
Consider two deuterons moving towards each other with equal speeds in a deutron gas. Te closest separation between them becomes 2 fm? Assume that the nuclear force is not effective for separations greater than 2 fm. At what temperature will the deuterons have this speed on an average? (a) 7.3 × 10 8 K (c) 2.8 × 10 9 K
µ0
A monochromatic light of wavelength l is incident on an isolated sphere of radius r . Te threshold wavelength of the metal sphere is l0 (> l). Te number of photoelectrons emitted before the emission of photoelectrons stop is (a)
14.
6
a
Water is filled to a height H behind a dam of width w as shown in figure. Determine the H resultant force exerted by the water on the dam. Density of w water is r. (a) r gHw2 (b) r gwH 2 (c)
4 πλ gL
2 π λ
From a disc of radius a, an isosceles right angled triangle with the hypotenuse as the diameter of the disc is removed. Te distance of the centre of mass of the remaining portion from the centre of the disc is (π − 1)a (a) 3(p – 1)a (b)
(b) 3.4 × 109 K (d) 5.9 × 108 K
15.
Te circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit? (a) 2.31 A (b) 1.33 A (c) 1.71 A (d) 2.00 A
16.
A rod leans against a stationary cylindrical a w body as shown in the R R figure. Now, its right v end slides to the right x on the floor with a constant speed v . Choose the correct option. (a) Te angular acceleration a is –2Rv 2(2x 2 – R2)/[x 2(x 2 – R2)3/2]. (b) Te angular speed w is 2Rv /[x (x 2 – R2)1/2]. (c) Te angular speed w is Rv /[x (x 2 – R2)1/2]. (d) Te angular acceleration a is –Rv 2/[(x 2 – R2)3/2].
17.
wo cars are moving along two straight tracks on the ground with the same speed 20 m s –1. Te angle between two tracks is
π 3
22.
rad and their intersection
point is O. At a certain moment, the cars are at distance 300 m and 400 m rom O. I they are moving towards O, then find minimum distance between them. (ake, 3 = 1.73)
A beam o singly ionised atoms o carbon (each charge +e) all have the 5.00 cm same speed and enter a mass spectrometer, as 15.0cm shown in figure. Te ions strike the photographic plate in two different locations 5.00 cm apart. Te 12C6 isotope traces a path o smaller radius, 15.0 cm. What is the mass number o other isotope? (a) 12 (b) 13 (c) 15 (d) 14 +
(a) 73.4 m (b) 86.5 m (c) 48.6 m (d) 32.9 m 18.
19.
20.
I a battery o em 8 V and negligible internal resistance is connected between terminals P and Q o the circuit shown in figure. Ten mark the correct statement. (a) Te equivalent resistance o the circuit is 3 W. (b) Te current through 2.5 W resistance is 1A. (c) Te current drawn rom the battery is 2A. (d) Te current drawn rom 10 W resistance is zero.
Te de Broglie wavelength o a neutron corresponding to root mean square speed at 927 °C is l. What will be the de Broglie wavelength o the neutron corresponding to root mean square speed at 27 °C? λ 2
(b)
l
(c) 2l
dt
=
AB L dQ
=
I 2
L
R2
24.
An unpolarised beam o light is incident on a group o our polarising sheets which are arranged in such a way that the characteristic direction o each polarising sheet makes an angle o 30° with o the preceding sheet. Te percentage o incident light transmitted by third polariser will be (a) 100% (b) 37.5% (c) 28% (d) 12.5%
25.
Consider telecommunication through optical fibres. Which o the ollowing statements is not true? (a) Optical ibres may have homogeneous core with a suitable cladding. (b) Optical fibres can be o graded reractive index. (c) Optical fibres are subject to electromagnetic intererence rom outside. (d) Optical fibres have extremely low transmission loss.
26.
A uniorm rope o length 12 m and mass 6 kg, is swinging vertically rom rigid base. From its ree end, one 2 kg mass is attached. At its bottom end
(T1 − T 2 ) A
(T1 + T 2 ) dt 2L A dQ 2 2 (T1 + T 2 ) (c) I B = 0, = 2L dt dQ AB 2 2 (T1 − T 2 ) (d) I C = B ≠ 0, = dt L
(b) I C = 0, B = 0,
R3
R1
(a) immediately afer closing o switch S are 3.33 A and 2.73 A respectively. (b) a long time afer closing the switch are 4.55 A and 1.82 A. (c) immediately ater reopening the switch are 0 and 1.82 A. (d) a long time ater reopening the switch are 1.82 A and 3.33 A.
dQ
dQ
+
option regarding the rate o flow o heat along dt the wire. (a) I C = 0,
S
–
(d) 4l
A well lagged wire o length L and cross-sectional area A has its ends maintained at temperatures T 1 and T 2. Te thermal conductivity o the wire is given by K = B + CT , where T is the temperature and B and C are constants. Choose the correct
In the circuit shown in figure, e = 100 V, R1 = 10.0 W, R2 = 20.0 W, R3 = 30.0 W and L = 2.00 H. Te values o I 1 and I 2, I 1
A long narrow horizontal slit lies 1 mm above a plane mirror. Te intererence pattern produced by the slit and its image is viewed on a screen placed at a distance 1 m rom the slit. Te wavelength o light is 600 nm. Ten the distance o the first maxima above the mirror is equal to (a) 0.30 mm (b) 0.15 mm (c) 60 mm (d) 7.5 mm
(a) 21.
23.
+
PHYSICS FOR YOU
|
MARCH ‘18
13
one transverse wave is produced o wavelength 0.06 m. At upper end o rope, wavelength will be (a) 1.2 m (b) 0.12 m (c) 9.12 cm (d) 0.12 mm 27. An amount o heat Q is supplied to one mole o
a diatomic gas undergoing a process such that the volume V =
α
,
2
28. wo concentric shells o radii
R and 2R are shown in figure. 2R Initially a charge q is imparted R to the inner shell. Now key K 1 is closed and opened and then key K 2 is closed and opened. Afer the keys K 1 and K 2 are K 2 K 1 alternately closed n times each, find the potential difference between the shells. Note that finally the key K 2 remains closed. (a) –q/4pe02n+1 (b) q/2pe02nR (c) –q/4pe0R (d) zero 29. A ray o light strikes
a plane mirror at an 4° angle o incidence 45° as shown in the figure. 45° Afer reflection, the ray passes through a prism o reractive index 1.50, whose apex angle is 4°. Te angle through which the mirror should be rotated i the total deviation o the ray is to be 90° is (a) 1° clockwise (b) 1° anticlockwise (c) 2° clockwise (d) 2° anticlockwise. 30. wo nucleons are at a separation o 1 m. Te net
orce between them is F 1 i both are neutrons, F 2 i both are protons, and F 3 i one is a proton and the other is neutron. Choose the correct option. (a) F 1 > F 2 > F 3 (b) F 2 > F 1 > F 3 (c) F 1 = F 3 > F 2 (d) F 1 = F 2 > F 3 SOLUTIONS
1. (b) : Initial potential energy =
where R is the radius o earth. 1 2
−
GMm R
+
2000
,
mv 2,
where v is the velocity required to throw the man out. 14
PHYSICS FOR YOU
|
⇒
v =
where T is the absolute
T temperature and a is a positive constant. I the temperature increases rom T 0 to 3T 0, then find Q. (a) 2RT 0 (b) 4RT 0 (c) 5RT 0 (d) RT 0
Initial kinetic energy =
Applying the law o conservation o energy For a body to escape rom earth, the total energy (PE + KE) should be equal to zero 1 GMm = 0 Accordingly, mv 2 + − R + 2000 2
MARCH ‘18
2GM
R + 2000
2. (a) : I l is the stretched length o the spring, then
rom figure PB = d and PA = l 4 5 d , i.e., l = d = cos 37° = l 5 4 So, the stretch y and AB = h
=
l
=
l d −
5
=
sin 37° =
4 5 4
d
−
d
×
d
=
3 5
=
d 4 3 4
d
Now, taking point B as reerence level and applying law o conservation o mechanical energy between A and B, mgh or
+
1 2
ky 2
=
1 2
mv 2 2
d 1 mgd + k = mv 2 4 2 4 2 3
1
3 g
k 2d 16 m 3. (a) : o find the charge enclosed, we need to find the flux through the parallelopipe. f1 = AE1 cos 60° = (0.05 m) (0.06 m) (2.50 × 10 4 N C–1) cos 60° = 37.5 N m 2 C–1 f2 = AE2 cos 120° = (0.05 m) (0.06 m) (7.00 × 10 4 N C –1) cos 120° = –105 N m2 C–1 So, the total flux is 2 –1 2 –1 f = f1 + f2 = (37.5 – 105) N m C = –67.5 N m C Net charge enclosed, q = fe0 = –67.5 e0 C Tere must be a net charge (negative) in the parallelepiped since there is a net flux flowing into the surace. Also, there must be an external field, otherwise all lines would point toward the slab. L 4. (b) : Since the weight acts at centre o mass, Leq = 2 and F = mg I x is the extension produced, F . Leq F . (L/2) (ρLAg)L or x = = Y = A . Y A . Y . 2 Ax 2 ρ . gL ⇒ x = 2Y or v
=
d
+
5. (c) : Applying Kirchhoff's first law at junction M ,
we get the current I 1 = 3 A. Applying Kirchhoff's first law at junction P , we get the current I 2 = 1 A. For loop MNOP , we get V M – 5I 1 – I 1 – 2I 2 = V P or V M – V P = 6I 1 + 2I 2 = 20 V Energy stored in the capacitor is 1 2
CV 2
6. (c) : ln
=
1 2
×
4
10
×
−6
×
20
×
20
=
8
×
−4
10
J
α αz = P β kB θ
[az ] = [kBq] and [a] = [P b] [β] =
[kB θ] [Pz ]
2 −2
=
[ML T
][K
−1 −2
[ML T
−1
][K ]
][L]
= L2
7. (b) 8. (b) : v av
8 RT
, v ∝ T M 0 av For same temperature in vessel A, B and C , average speed of O2 molecule is same in vessel A and C and is equal to v 1. =
π
9. (a) : he force per
unit length between current carrying parallel wires is dF µo I1I 2 =
2 πd dL If two wires carry current in opposite directions the magnetic force is repulsive, due to which the parallel wires have moved out so that equilibrium is reached. Figure shows free body diagram of each wire. In equilibrium, ∑F y = 0, 2T cosq = (lL0) g (i) ∑F z = 0, 2T sinq =F B (ii) Now dividing eqn. (ii) by eqn. (i) we get F B
tan θ =
L0 λ g where, the magnetic force,
be stopped when the potential of sphere becomes V . Terefore, hc 1 1 hc hc or V = − eV = − e λ λ 0 λ λ 0 Charge on sphere, Q = (4pe0r )V Number of photoelectrons emitted, 1 Q (4 πε0r )V ( 4 πε0rhc) 1 = − n= = e e e2 λ λ 0 11. (b) : rue measurement = scale reading [1 + a(q – q0)] = 180 [1 – 10 × 1.1 × 10–5] Error = 180 – 180 [ 1 – 1.1 × 10 –4] = 1.98 × 10–2 mm 12. (c) 13. (d) : Let's consider a vertical y
F = ∫ P dA =
repulsion will slow them down. Te loss in kinetic energy will be equal to the gain in Coulombic potential energy. At the closest separation, the kinetic energy is zero and the potential energy is e2 4 πε0 r
For small q, tanq µ0 ∴ θ = I gL 4 πλ
sin q
q
10. (c) : As the sphere is isolated, therefore on emission
of photoelectrons from sphere, the potential of sphere is raised. Let the emission of photoelectrons
. If the initial kinetic energy of each deuteron
is K and the closet separation is 2 fm, we shall have 2 K =
dF × L = µ0 I L0 dL 0 4 π sin θ L
1
H 2 ∫ 0 ρg (H − y ) w dy = 2 ρgwH
14. (c) : As the deuterons move, the Coulombic
2
F B =
h
axis, starting from the bottom dy of the dam. Let's consider a y thin horizontal strip at a height w y above the bottom. We need to consider force due to the pressure of the water only as atmospheric pressure acts on both sides of the dam. Te pressure due to the water at the depth h P = r gh = r g (H – y ) Te force exerted on the shaded strip of area dA =wdy, dF = P dA = r g (H – y ) w dy Integrate to find the total force on the dam
e2 4 πε0 (2 fm) 19
2
(1.6 × 10− C) =
×
(9
×
9
2
2
10 N m C − )
15
2 ×10− m
or K = 5.7×10–14 J If the temperature of the gas is T , the average kinetic energy of random motion of each nucleus will be 1.5 kBT . Te temperature needed for the deuterons to have the average kinetic energy of 5.7 × 10–14 J will be given by PHYSICS FOR YOU
|
MARCH ‘18
15
1.5 kBT = 5.7 × 10 –14 J where kB = Botzmann’s constant or,
5 .7
T =
×
1.5 × 1.38
10
×
−14
10
19. (b) : Point O is a minima. Hence the first maxima
J
β will be at y = from O. 2 λ D ⇒ y = 2d 2(2d )
9
−23
JK
= 2.8 × 10 K.
−1
15. (d)
=
\
so Q
R sin θ
dx d R R(d θ / dt )( − cos θ) = = 2 dt dt sin θ sin θ d θ ω = − dt
ω =
v sin2 θ R cos θ
tan θ =
2
v ⋅ R
=
(R / x )
x
2
−R
/ x
2
= =
=
x( x 2 − R2 )1 / 2
= −
Rv (2x 2
x (x
2
2
−
v B sin120° v A + v B
100 × 50
R )
−
=
vA
−
through 2.5
=
1
20 − 20
l min
C
×
v B 3
=
2
B
4
v B
3 2
O
v B
v AB q
60°
27 + 273
3 2
λ ∝
1
T
kBT
= 2 λ
W
resistance is zero.
1 25 + 15
+
1 5+3
1 =
A
=
0 .8 A
10
10
1 +
40
1 +
8
1 =
4
or
Req = 4 W
\
Current drawn from battery is (8/4) A = 2 A. PHYSICS FOR YOU
|
MARCH ‘18
−
Tis gives, B(T2 − T 1 ) + C (T22 − T12 ) = − DL. dQ A C So, B (T1 − T 2 ) + T12 − T 22 = AD = 2 dt L dQ A C or = (T1 − T2 ) B + (T1 + T 2 ) . 2 dt L
(
)
.
22. (d) : Let R1 be the radius of trajectory of the isotope
Equivalent resistance is given by +
the wire is given by the definition of thermal conductivity as dQ dT KA . dt dx Since the wire is well lagged, we may assume that no heat enters or leaves it except at the ends, so dQ/dt must be constant. 1 dQ Let a constant D , A dt dT ( B + CT ) = − D. (As K = B + CT ) \ dx Tis differential equation can be solved by rearranging and integrating
2
60°
v A A
300 m
Current through 10 W resistance is
16
(927 + 273)
⇒
kBT
T L ∫ T 12 (B + CT ) dT = − D ∫ 0 dx .
D m 0 0 60°
8
10
2m ×
2
=
20 3 / 2
cos 120°
3 m
R )
2 3/2
18. (c) : Since the bridge is balanced, hence current
Req
2mE
=
2
= 86.5 m
1
h
=
3 =
21. (a) : he rate at which heat is flowing through
Rv
2
q =
=
h
=
D
m
2
60° By geometry, \ OD = OA = 300 m \ DB = 400 – 300 = 100 m \ l min = DB sin 60°
1
3
S
sin θ
\
h p
λ2 = λ ⋅
17. (b) : Here OA = 300 m, OB = 400 m, v AB
\
λ=
ωR cos θ
d ω d Rv α= = dt dt x x 2 − R 2
\
×1
β/2 O
20. (c) : Kinetic energy of neutron E
v =
v =
−9
4 × 1 × 10− = 0.15 mm.
16. (c) : From geometry,
x =
600 × 10
S
12
C6 and R2 that of the unknown isotope. he trajectory of the unknown isotope has a greater radius, and so the mass of the unknown isotope is greater than that of the 12C6 isotope. 2R2 – 2R1 = 5.00 × 10 –12 m R2 – R1 = 2.5 × 10 –12 m Since R1 = 15.0 × 10 –2 m R2 = 15.0 × 10 –2 m + 2.50 × 10 –2 m = 17.5 × 10 –2 m
m1v , R2 Bq
m2v Bq
For each isotope, R1 R2 m2 \ R1 m1 Since, m1 = a(12), m2 = a A where A is mass number o unknown isotope and a is a constant. −12 17.5 × 10 m R2 m2 A A So, \ = −12 R1 m1 12 12 15.0 × 10 m Solving or unknown atomic mass number, we obtain A = 14. Tus the unknown isotope is 14C6. =
=
I v 1 and v 2 are respective velocity at bottom and upper end, then v 1
=
µ
=
=
=
23. (c) : Immediately ater closing the switch, the
inductor opposes the ast build up o the current through it and hence current in the inductor is zero. Tis means I1
=
I 2
100
ε =
=
=
(10)(20) + (10)( 30)
+
( 20) ( 30)
=
4.55 A
=
2RT 0 (7 / 5) 1 −
=
R
+
R1R3
+
+
(10)( 30)
+
2α 3T −2αRT 3 W = ∫ dT = − dT T ∫ αT 3 T α T 2 T RT
0
= − 2R ∫ dT = − 2R(3T0 − T0 ) = − 4RT0 . ( 20)( 30)
=
2. 73A
25. (c)
26. (b) : ension at bottom end o rope T 1= 2 × 9.8 N
weight o rope acts on centre o gravity Tereore, tension at upper end o rope, T 2 = (6 + 2) × 9.8 = 8 × 9.8 N Tus, T 2 = 4T 1 Q
0
3T 0
When the switch is reopened, the lef loop is an open circuit. Te current immediately drops to zero when the switch is opened; hence I 1 = 0. Te current in R3 i.e., 4.55A – 2.73 A = 1.82 A which changes only slowly because there is an inductor in its branch. . Te current in R2 is the same as that in R3, 1.82 A. In the absence o any source o em all the currents drop to zero. 24. (c)
5RT 0
Now, work done by the gas, RT W = ∫ PdV = ∫ dV [Q PV = RT or 1 mole]. V α α Given, V = dT . or dV = − 2 2 T T 3 Substituting V and dV and integrating under the given limits, we get
R2R3
(100)(3) (10)(20)
=
3
ε 3
R1R2
...(i)
γ − 1
0
and I 2
µ
Q = ∆U + W = ∆U + ∫ PdV R (3T0 − T 0 ) Here, ∆U = nCV ∆T = 1⋅
3T 0
(100) (20 + 30)
T 2
27. (d) : From the first law o thermodynamics,
3.33 A
R1 + R2 10 + 20 A long time afer the current reaches its steady state value, the em across the inductor is zero, the inductor behaves as i it were replaced by a wire (i.e., short-circuited). ε(R2 + R3 ) I 1 = R1R2 + R1R3 + R2R3
and v 2 =
\ v 2 = 2v 1 (Q T 2 = 4T 1) Frequency o wave does not depend on medium, thereore v ∝ l I l1 and l2 are respective wavelength at bottom and upper end or rope, then l2 = 2l1 = 2 × 0.06 = 0.12 m
=
=
T 1
T 0
Substituting the values o DU and W in equation (i), we get Q = 5RT 0 – 4RT 0 = RT 0. 28. (a) 29. (b) : Deviation by prism = A (m – 1) = 4° (1.5 – 1) = 2° For 90° total deviation,
Deviation by mirror = 90 ° – 2° = 88° \ 180° – 2i = 88° 2i = 92° or i = 46° Mirror should be rotated 1° anticlockwise. 30. (c) : Nuclear orce o attraction between any two
nucleons (n - n, p - p; p - n) is same. Te difference comes up only due to electrostatic orce o repulsion between two protons. \ F 1 = F 3 ≠ F 2. As F 2 < F 3 or F 1 \ F 1 = F 3 > F 2
PHYSICS FOR YOU
|
MARCH ‘18
17
PRACTICE PAPER
th
Exam on
6 May 2018 1.
2.
3.
4.
One mole of helium gas, initially at SP (P 1 = 1 atm = 101.3 kPa, T 1 = 0 °C = 273.15 K), undergoes an isovolumetric process in which its pressure falls to half its initial value. Te helium gas then expands isobarically to twice its volume what is the work done by the gas? (a) 1135 J (b) 1535 J (c) 2335 J (d) 3335 J welve wires of equal resistance R are connected to form a cube. Te effective resistance between two opposite diagonal ends will be (a) (5/6) R (b) (6/5) R (c) 3 R (d) 12 R
6.
Large number of capacitors of rating 10 mF-200 V are available. Te minimum number of capacitors required to design a 10 mF-800 V capacitor is (a) 16 (b) 4 (c) 8 (d) 7
7.
A block of mass 4 kg hangs from a spring of force constant k = 400 N m –1. Te block is pulled down 15 cm below equilibrium and released. Find the kinetic energy when the block is 10 cm above equilibrium. (a) 1.5 J (b) 2.5 J (c) 3.5 J (d) 4.5 J
8.
In figure, two Zener diodes are connected in series in a voltage regulator circuit. Te maximum current through each of the diodes is 250 mA and the Zener voltage is 20 V. Calculate the resistance R for V = 50 V.
Te force between two charges situated in air is F . Te force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is (a) F /4 (b) 4 F (c) 16 F (d) F If r is mean density of earth, r its radius, g the acceleration due to gravity and G, the gravitational constant, on the basis of dimensional consistency, the correct expression is rG 3G (a) ρ = (b) ρ = 4 π r g 12 π g (c)
ρ=
3 g 4π r G
(d)
ρ=
3 r 4 π g G
R
(a) 10 W (b) 80 W V
(c) 20 W (d) 40 W 9.
One end of an infinitely long straight wire carrying a steady current I in the positive direction of the Z -axis is situated at the point P (0, a, 0), as shown in
18
A crate of mass 50 kg slides down a 30° incline. Te crate’s acceleration is 2.0 m s–2, and the incline is 10 m long. What is the magnitude of the frictional force that acts on the crate as it slides down the incline? (a) 245 N (b) 345 N (c) 145 N (d) 445 N PHYSICS FOR YOU
|
MARCH ‘18
Y P (0, a, 0)
R( 3 a, 0, 0) O
Q(a, 0, 0)
X
figure. Find ∫ B ⋅ dl along the line joining the points Q and R.
5.
RL
(a) (c)
µ 0 I 24
µ 0 I 12
(b)
(d)
µ 0 I 48
µ 0 I 21
10.
11.
12.
13.
14.
15.
16.
17.
wo boys start running straight toward each other rom two points that are 100 m apart. One runs with a speed o 5 m s–1, while the other moves as 7 m s–1. How close are they to the slower one’s starting point when they meet each other? (a) 41.7 m (b) 42.8 m (c) 45.9 m (d) 46.1 m A coil o inductance 0.50 H and resistance 100 W is connected to a 240 V, 50 Hz a.c. supply. Calculate time lag between current and voltage. (a) 3.2 ms (b) 5.2 ms (c) 7.2 ms (d) 9.2 ms Binding energy o nuclei P , Q and R are EP , EQ and ER respectively. In the usion processes 3P → Q + Energy (E1) 2Q → R + Energy (E2) Calculate, total energy (E3) released in the usion process 6P → R + Energy (E3). (a) E1 + E2 (b) E1 – E2 (c) E1 – 2E2 (d) 2E1 + E2 An object producing a pitch o 400 Hz flies past a stationary person. Te object was moving in a straight line with a velocity 200 m s–1. What is the change in requency noted by the person as the object flies past him? (Speed o sound in air = 300 m s –1) (a) 1200 Hz (b) 960 Hz (c) 240 Hz (d) 1440 Hz Te ratio o the KE required to be given to the satellite to escape earth’s gravitational field to the KE required to be given so that the satellite moves in a circular orbit just above earth’s atmosphere is (a) 1 (b) 1/2 (c) 2 (d) infinity Light o requency 7.21 × 1014 Hz is incident on a metal surace. Te maximum speed o the photoelectrons emitted is 6.0 × 105 m s –1. What is the threshold requency or the photoemission o electrons? (a) 4.73 × 1014 Hz (b) 7.31 × 1014 Hz (c) 4.73 × 1015 Hz (d) 7.34 × 1015 Hz
18.
(a)
K
4
(b)
K 2
(c)
K
6
(d)
K 8
19.
A satellite sent into space samples the density o matter within the solar system and gets a value o 2.5 hydrogen atoms per cubic centimeter. What is the mean ree path o the hydrogen atoms? ake the diameter o a hydrogen atoms as d = 0.24 nm. (a) 1.56 × 1012 m (b) 2.56 × 1012 m (c) 3.56 × 1012 m (d) 4.56 × 1012 m
20.
When two bar magnets have their like poles tied together, they make 12 oscillations per minute and when their unlike poles are tied together, they make 4 oscillations per minute. Find the ratio o their magnetic moments. (a)
2 3
(b)
3 2
(c)
5 4
(d)
4 5
21.
Te amplitude o a damped oscillator becomes hal in one minute. Te amplitude afer 3 minutes will be (1/x ) times the original, where x is (a) 2 × 3 (b) 23 (c) 32 (d) 3 × 22
22.
A 500 g wheel that has a moment o inertia o 0.015 kg m2 is initially turning at 30 rps. It comes to rest afer 163 revolution. How large is the torque that slowed it? (a) –0.26 N m (b) –0.45 N m (c) –0.55 N m (d) –0.95 N m
23.
For a series resonant LCR circuit with L = 2 H, C = 32 mF and R = 10 W, what is the Q-actor o this circuit? (a) 25 (b) 50 (c) 75 (d) 100
24.
A bullet o mass m, moving with a speed o u penetrates a block o wood o thickness x and emerges with a speed v . Te orce o resistance offered by the wood is given by
Find the mass o the block that a 40 hp engine can pull along a level road at 15 m s–1 i the coefficient o riction between block and road is 0.15. (a) 2532 kg (b) 1353 kg (c) 3553 kg (d) 4553 kg
(a)
m x
2
A hole o area 1 mm opens in the pipe near the lower end o a large water-storage tank, and a stream o water shoots rom it. I the top o the water in the tank is 20 m above the point o the leak, how much water escapes in 1s? (a) 19.8 mL s–1 (b) 28.8 mL s–1 (c) 38.8 mL s–1 (d) 48.8 mL s–1
In a Young’s double-slit experiment using monochromatic light o wavelength l, the intensity o light at a point on the screen where the path difference is l, is K units. What is the intensity o light at a point where the path difference is l/3?
(c) 25.
(u 2 − v 2 )
m
2 x
2
2
(v − u )
(b) (d)
m
2 x m x
(u 2 − v 2 )
(v 2 − u 2 )
A proton when accelerated through a potential difference o V volts has a wavelength l associated with it. An a particle in order to have the same wavelength l, must be accelerated through a potential difference (in volts) PHYSICS FOR YOU
|
MARCH ‘18
19
(a) 2V 26.
27.
(b)
(c)
V
V
4
(d)
V
(d)
A straight rod of length L extends from x = a to x = L + a with the mass per unit length A + B x 2. Te gravitational force exerted by the rod on a point mass m at x = 0 is
(2 p − 1)
29.
A metallic rod of length 1.0 m is rotated with an angular frequency of 400 rad s –1 about an axis normal to the rod and passing through one end. Te other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 and parallel to the axis exists around the rod and the ring. Calculate the emf developed between the centre and the ring. (a) 220 V (b) 120 V (c) 100 V (d) 200 V A boy stands on a freely rotating platform. With his arms extended, his rotational speed is 0.25 rps. But when he draws them in, his speed is 0.80 rps. Te ratio of his moment of inertia in the first case to that in the second case is (a) 2.2 (b) 3.2 (c) 4.2 (d) 5.2 Te dispersive powers of two thin prisms of refracting angles A and A′ are 0.03 and 0.05 respectively. Te refractive indices for yellow light for the prisms are 1.517 and 1.621 respectively. If the combination of prisms produces a deviation of 1° in the yellow light without producing any angular dispersion, then the values of A and A′ respectively are (a) 4.8°, 2.4° (b) 3.6°, 1.6° (c) 3.3°, 1.3° (d) 2.2°, 1.2° PHYSICS FOR YOU
|
1 1 − + BL a + L a 1 1 (b) Gm A − + BL a a + L 1 1 + − BL (c) Gm A (a + L) a 1 1 − BL (d) Gm A − a (a + L) (a) Gm A
2 p
Te images of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall, 3 m away, by means of a large convex lens. What is the maximum possible focal length of the lens? (a) 0.55 m (b) 0.65 m (c) 0.75 m (d) 0.85 m
20
33.
An open and closed organ pipe have the same length. Te ratio of pth mode of frequency of vibration of two pipes is (a) 1 (b) p
28.
31.
A parallel plate capacitor has circular plates of radius 6.0 cm each and a capacitance of 100 pF. It is connected to a 230 V a.c. supply with a frequency of 300 rad s–1. Determine the amplitude of magnetic field at a point 3.0 cm from the axis between the plates. (a) 5.6 × 10–10 (b) 6.5 × 10–11 (c) 7.6 × 10–11 (d) 7.6 × 10–10
8
A tank contains a pool of mercury 0.30 m deep, covered with a layer of water that is 1.2 m deep. Te density of water is 1.0 × 10 3 kg m–3 and that of mercury is 13.6 × 10 3 kg m–3. Find the pressure exerted by the double layer of liquids at t he bottom of the tank. Ignore the pressure of the atmosphere. (a) 52 kPa (b) 72 kPa (c) 82 kPa (d) 92 kPa
(c) p(2 p + 1)
30.
32.
MARCH ‘18
34.
A 400 g block originally moving at 120 cm s –1 travels 70 cm along a tabletop before coming to rest. What is the coefficient of friction between block and table? (a) 0.105 (b) 0.305 (c) 0.405 (d) 0.505
35.
A parallel beam of light of wavelength 550 nm is focused by a convex lens of diameter 10.0 cm on a screen at a distance of 25 cm from it. Find the diameter of the image of disc that is formed. (a) 2.24 × 10–4 cm (b) 3.34 × 10–4 cm (c) 4.44 × 10–4 cm (d) 5.44 ×10–4 cm
36.
What is the maximum amount of work that a Carnot engine can perform per kcal of heat input if it absorbs heat at 427 °C and exhausts heat at 177 °C? (a) 3.96 kJ (b) 1.49 kJ (c) 5.96 kJ (d) 7.96 kJ
37.
When a building is constructed at –10 °C, a steel beam (cross-sectional area 45 cm2) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force in the beam when the temperature is 25 °C? For steel, Y = 200 000 MPa, and asteel = 1.2 × 10–5 °C. (a) 380 kN (b) 480 kN (c) 580 kN (d) 780 kN
38.
39.
40.
41.
42.
43.
44.
Te hal-lie o 90 38Sr is 28 years. What is the disintegration rate o 15 mg o this isotope? (a) 2.12 Ci (b) 3.12 Ci (c) 1.22 Ci (d) 2.81 Ci An infinite number o electric charges each equal to 5 nC (magnitude) are placed along x -axis at x = 1 cm, x = 2 cm, x = 4 cm, x = 8 cm ... and so on. In the setup i the consecutive charges have opposite sign, then the electric field (in N C–1) at x = 0 is (a) 12 × 104 (b) 24 × 104 (c) 36 × 104 (d) 48 × 104 A ship is travelling due east at N –1 10 km h . What must be the 30° speed o a second ship heading E 30° east o north i it is always W due north rom the first ship? (a) 5 km h–1 S (b) 10 km h–1 (c) 15 km h–1 (d) 20 km h–1 A hydrogen sample in the ground state absorbs monochromatic radiation o wavelength l and emits radiation o six different wavelengths. Find the value o l. (a) 1000 Å (b) 680 Å (c) 1200 Å (d) 975 Å When 5 g o a certain type o coal is burned, it raises the temperature o 1000 mL o water rom 10 °C to 47 °C. Calculate the heat energy produced per gram o coal. Given, heat capacity o water is 1 cal g–1 °C –1. Neglect the small heat capacity o the coal. (a) 8400 cal g–1 (b) 9400 cal g–1 (c) 7400 cal g–1 (d) 6400 cal g–1 In a straight conductor o uniorm cross-section charge q is flowing or time t . Let s be the specific charge o an electron. Te momentum o all the ree electrons per unit length o the conductor, due to their drif velocity only is (a) q/ts (b) q/ts2 (c) (q / ts ) (d) qts A hose lying on the ground y Wall shoots a stream o water v 0 upward at an angle o 40° to the horizontal. Te speed 40° x o the water is 20 m s –1 as 8.0 m it leaves the hose. How high up will it strike a wall which is 8 m away? sin40° = 0.64 (a) 2.33 m (b) 5.33 m (c) 8.33 m (d) 11.33 m
45.
A parallel plate capacitor has two layers o dielectrics as shown in figure. Tis capacitor is connected across a battery, then the ratio o potential difference across the dielectric layers is K 1 = 2 K = 6 2 (a) 1/3 (b) 4/3 (c) 3/2 d 2d (d) 1/2 SOLUTIONS 2V 1
∫ V
1. (a) : W =
1
1
=
2
1 2
1
1
2
2
P1dV = P1V1 = RT1
(8.31)(273. 15)
=
1135 J
2. (a) : Te effective resistance between two diagonally
opposite ends = 5 R/6. 3. (d) : F =
F ′ = F ′ F
q1q 2 2
4 πε 0r
q1q 2 2
=
4 π ε(r / 2)
=4
4q1q 2 4 π εr 2
ε0 1 1 = 4 = 4 = 1 ε K 4
4. (c) :
3 g 4πr G
=
(LT
−2
−1
3
L(M
\
)
L T
F ′ = F −3
−2
)
= [ML
]= ρ
5. (c) 6. (a) : Number o capacitors to be connected in series
=
voltage rating required = voltage rating of given capacitor
C eq =
10 4
800 =
200
4
= 2.5 µ F
Number o rows required Capacity required 10 Capacity of each row 2 5
=
=
=
4
⋅
\
otal number o capacitors required = 4 × 4 = 16
7. (b) :
Given, A = 0.15 m; k = 400 N m–1
(From conservation o mechanical energy.) 1 2
1 2
1
mv 2 + kx 2 2
mv 2
=
1 2
=
1 2
kA2
k(A2 − x 2 ) =
1 2
×
2
400(0. 15
−
2
0. 1 ) = 2. 5 J
Since both the Zener diodes are in series, the voltage across RL = 2 × 20 V = 40 V. 8. (d) :
PHYSICS FOR YOU
|
MARCH ‘18
21
From Kirchhoff’s loop rule, the current in the Zener loop, I
50 V =
−
40 V
10 V =
R
\ 250 mA =
10 V
⇒ R=
2π
T
= ω ⇒ ∆t =
θ
π(57.5)
=
180 × 2 π × 50
ω
250 mA
12. (d) : As, 3P → Q + E1
10 V
=
250 × 10
−3
= 40 Ω A
9. (a) : From
figure, Y = p/4 and ∠PQO P (0, a, 0) ∠PRO = p/6. Consider a point S on the x -axis r lying between the 30° X 45° q S points Q and R. Let O Q(a, 0, 0) R( 3 a , 0, 0) PS = r . Te magnetic (90° – q) µ 0 I field B = is directed perpendicular to the line PS, 2 πr as shown in the figure. Te field B makes an angle (90° – q) with the positive direction o the x -axis. µ I ...(i) \ B ⋅ d x = Bdx ⋅ cos(90° − θ) = 0 ⋅ dx ⋅ sin θ 2 πr Let OS = x . Ten OP a = = tan θ ⇒ x = a cot q. OS x Differentiating, dx = –acosec2q d q.
Similarly,
∆t
=
= 3.2 × 10 –3 s = 3.2 ms.
R
10 V
R
θ
OP a = = sin θ PS r
⇒
E1 = EQ – 3EP Also 2Q → R + E2 ⇒ E2 = ER – 2EQ Now, 6P → R + E3 ⇒ ER – 6EP = E3 Now 2E1 + E2 = 2(EQ – 3EP ) + (ER – 2EQ) = E3 –1 –1 13. (b) : Here, u = 400 Hz, v s = 200 m s , v = 300 m s Beore crossing the person v × υ 300 × 400 υ = = = 1200 Hz 1 v − v s 300 − 200 Afer crossing the person, v × υ 300 × 400 υ = = = 240 Hz 2 v + v s 300 + 200 Change in requency = u1 – u2 = 1200 – 240 = 960 Hz ⇒
14. (c) : ve
1
\
KE1 KE2
1 = sin θ . r a
=
=
2 1 2
2
gR , v0
mv e2 = 2 0
mv
v e2 v 02
=
gR
=2
15. (a)
16. (b)
1 in eqn. (i) and integrating, 17. (a) : Using orricelli’s theorem or the escape speed, r we have or the volume flow R R R µ0 I sin θ ∫ B ⋅ dl = ∫ B ⋅ dx = ∫ 2π sin θ a (−acosec2θd θ) vA = 2 gh A Q Q Q
Substituting or dx and
3
3
2(9.8 × 10 )(20 × 10 ) × 1
=
µ0 I π/6 µ0 I π π µ0 I π µ0 I = − θ d = ∫ 2π 4 − 6 = 2 π × 12 = 24 . 2π π/ 4
18. (a)
10. (a) : Te two boys meet at the same place and time.
20. (c) : When the like poles are tied together, the net
Te time or the slower one to travel a distance x is x /5, while the other boy has to travel distance (100 – x ) in x 100 x time t = (100 – x )/7. Equating the times, 5 7 \ x = 41. 7 m
magnetic moment is (m1 + m2) and the moment o inertia is (I 1 + I 2). I1 + I 2 . \ Te time period T 1 = 2 π (m1 + m2 )B
= 19800 mm 3 s–1 = 19.8 mL s –1
−
=
11. (a) : I q is the phase difference between the current
and the voltage, −1
X ωL 2πυL 2(3.14)( 50 s )(0. 50 H) tan θ = L = = = = 1.57 100 Ω R R R ⇒ q =
tan–1(1.57) = 57.5° ×
π
180
rad
I Dt is the time lag between the current and voltage, 22
PHYSICS FOR YOU
|
MARCH ‘18
19.
(a)
When the unlike poles are tied together, the net magnetic moment is (m1 – m2), while the moment o inertia (being a scalar quantity) remains unchanged. I1 + I 2 . \ Te time period T 2 = 2 π (m1 − m2 )B T 22
(m1 + m2 )
m1 ⇒ Tus, 2 = m2 T 1 (m1 − m2 )
T22 + T 12
2 2 1 + υ2 = = T22 − T 12 υ12 − υ22 υ
.
Given, u1 = 12 per minute and u2 = 4 per minute. m1 \ m2
=
2
2
2
2
12 + 4 12 − 4
=
144 + 16 144 − 16
160 =
128
5 =
.
4
21. (b) : Amplitude o damped oscillation is
A = A0 e–bt As A = A0/2, when t = 1 minute, so A0 = A0 e–b × 1 or eb = 2 2
–6
23. (a) : Given, L = 2 H, C = 32 mF = 32 × 10 F and R = 10 W. I the resonant angular requency o the circuit is wr
then rom the condition o resonance, ω =
r
1
1
=
LC
(u
2
=
−
.
25. (d) : λ p = λ α ⇒
v
r L
ω
=
R 2
x =
) ( 2
m u =
h 2m p q pV
=
−
v
2
2 x h 2mα qαV ′
Squaring both sides, we get
m p q p V mα qα
2maqaV = 2m pq pV ; V ′ = As
m p mα
∴
V′ =
1 =
4
and
qp qα
vf v − f
)
2 m p q pV
′
υ=
p
Substituting or u in equation (i),
25.
2 x
\ Te orce o resistance, ma
2mα qαV ′ =
1
−
(2 H)(32 ×10 F)
24. (b) : Retardation a
or
125 rad s
= −6
Te Q-actor o the circuit, Q =
v 2l v For closed pipe, υ′ = (2 p − 1) 4l 2 p υ \ = 2 p − 1 υ′ 28. (c) : Let the distance between the object O and its real image I be x , as shown in figure. Ten, ...(i) x = u + v , where u and v are the numerical values o the distances o the object and Screen I the image rom the convex O lens. For the convex lens, f and u v x v will be positive, while u will be negative. From the lens ormula, v − f fv 1 1 1 1 1 1 − = ⇒ = − = ⇒u= ...(ii) +v −u +f u f v fv v − f 27. (d) : For open pipe,
When A = A0/x , t = 3 minute, then A0 = A0 e–b × 3 or x = e3b = (eb)3 = 23 x 2 2 22. (a) : w = w0 + 2aq, w0 = 2pu = (2p rad)(30 s–1) = 60 p rad s–1 (Q w = 0) \ q = 326p rad. 2 –2 a = – (60p) / 652p , or a = –17.3 rad s t = – 0.015 × 17.3 m s–2 = –0.26 N m
⇒
Te pressure P int exerted by the mercury column itsel is ound in the same manner P int = rmerc ghmerc = 13.6 × 10 × 9.8 × 0.30 m = 40 kPa Te total pressure at the bottom is thus 52 kPa.
⇒ v – vx + fx = 0
x 2 – 4 fx ≥ 0
v =
2
⇒
x ≥ 4 f ⇒
f ≤
x 4
.
Hence, or a fixed value o x (the distance between the object and the screen), the maximum value o the ocal length o a convex lens that can orm a real image at the screen is x /4. Since x = 3 m in this case,
2
29. (c)
=
=
pool. For a point below the surace o the mercury this may be regarded as a source o exerted pressure pext. Tus P ext = rwater ghwater = 1.0 × 9.8 × 1.2 = 12 kPa
⇒
x ± x 2 − 4 fx
For real values o v ,
f max
26. (a) : First find the pressure at the top o the mercury
v =
v 2 − vf v 2 = v − f v − f
+
2
1
1 1 V = V volts 4 2 8
+
vf
3 4
= 0. 75 m .
30. (b) : Because there is no net torque on the system
about the axis o rotation, the law o conservation o angular momentum tells us that angular momentum beore = angular momentum afer I 0w0 = I f w f I 0 ω f 0.80 = = = 3.2 I f ω 0 0.25 PHYSICS FOR YOU
|
MARCH ‘18
23
31. (a) : Since the net angular dispersion is zero.
35. (b) : Te diffraction pattern produced by a circular
q + q′ = 0 ⇒ (mv – mr ) A + (m′v – m′r ) A′ = 0
aperture consists of a central bright dis c surrounded by alternate dark and bright rings that are concentric with the central disc. Te angular radius of the central bright disc is given by
⇒ w(m y – 1) A + w′(m′ y – 1) A′ = 0 ⇒
ω(µ y − 1) A′ =− A ω′ (µ y ′ − 1)
Substituting the appropriate values, 0.03(1.517 − 1) A ′ ...(i) =− = −0.50 . 0.05(1.621 − 1) A Te negative sign shows that the prisms are oppositely placed. Te net deviation of the yellow light produced by the given combination, D = d – d′ = (m y – 1) A – (m′ y – 1) A′ ⇒ 1° = (1.517 – 1) A′ – (1.621 – 1) A′ 1° = (0.517) A – (0.621) A′ …(ii) From equations (i) and (ii), A = 4.8° and A′ = 2.4° 32. (c) : Let the magnetic field at a distance r = 3 cm = 3 × 10 –2 m from the axis be B. Te area of the circular section of radius r is A = pr 2. Te rms value of the V conduction current, I = = V rms wC X C rms
sin θ = 1.22
I d ′ =
A′ I rms A
2
2
=
r
R2
I rms
=
(3 cm)
2
(6 cm)
× (6.9 × 10
−6
sin θ =
µ0 I d ∫ B ⋅ dl = µ 0 Id ⇒ B × 2 πr = µ 0 I d ⇒ B = 2 πr
\ the amplitude of B,
B0 = 2 B =
2 (4 π × 10
−7
−1
H m )(1. 7 × 10
2 π(3 ×10 −2 m)
(1.22)(550 × 10 (10 × 10
−6
A)
−2
−9
m)
= 0.67 × 10
−5
.
m)
Radius of the bright disc, R = D tan q ≈ D sin q (Q q is small) where D = distance between the screen and the lens = 25 cm \ R = (25 cm)(0.67 × 10 –5) = 1.67 × 10–4 cm. Diameter of the disc image = 2R = 2 × (1.67 × 10 –4 cm) = 3.34 × 10–4 cm 36. (b) : Efficiency, h =
Q1 Q2 Q1 −
T1 T 2 T 1 −
=
Q1 = (1 kcal)(4.184 kJ/kcal) = 4.184 kJ h
Q1 Q2
700 K
−
=
4.184 kJ
=
−
450 K
700 K
Ten W = Q1 – Q2 = 1.49 kJ 37. (a) : As L = L0 (1 + a D T )
\
∆L
L0
= a DT = (1.2 × 10 –5 °C–1)(35 °C) = 4.2 × 10 –4
Ten F = YA
= 1.7 × 10–6 A From the Ampere’s circuital law,
A)
.
a Substituting l = 550 nm = 550 × 10 –9 m, and a = 10 cm
rms
= 230 × 300 × 10 –10 = 6.9 × 10 –6A As the total conduction current is linked with the total area A of the circular plates, the rms value of the displacement current linked with area A′.
λ
∆ L
= (2 × 10 11 N m–2)(45 × 10–4 m2) L0 (4.2 × 10–4) ≈ 380 kN
38. (a) : Te disintegration rate, R =
the decay constant,
λ=
ln 2
dN dt
= λ N , where
T 1/2 Te half-life, T 1/2 = 28 years = 28 × 3.156 × 10 7 s = 8.83 × 108 s Te mass of the given sample, M = 15 mg = 15 × 10 –3 g Te molar mass, M m = 90 g mol–1
= 7.6 × 10–11 33. (b) : Mass of the element of length dx at a distance x M N from the origin = ( A + B x 2)dx Te number of moles, = M m N A 2 \ dF = Gm( A + Bx )dx M (15 × 10 3 g) x 2 N A = × (6.022 × 1023 mol–1) ⇒ N = 1 M m (90 g mol ) a+L a+L ( A + B x 2 )dx A = 1 × 1020 = 1020. + F = Gm ∫ = Gm B dx x 2 x 2 ln 2 a a 0.693 20 10 −1 N = \ R = λ N = ×10 = 7.84 ×10 s 8 T 1/2 8.83×10 s 1 1 + = Gm A − BL 10 7.84 × 10 a a + L 10 = 7.84 × 10 Bq = Ci = 2.12 Ci 10 −
−
∫
3.7 × 10
34. (a) 24
PHYSICS FOR YOU
|
MARCH ‘18
9 5 × 10−9 5 × 10− 39. (c) : E = ⋅ − + 4 πε0 (1 × 10−2 )2 (2 × 10−2 )2
1
5 × 10−
9
(5 × 10− )
..... + 2 2 (8 × 10− ) 9
−2 2
(4 × 10 )
−
9 9 9 × 10 × 5 × 10− 1 1 1 1− ... + − + ⇒ E = 4 2 2 2 10 − (2) (4) (8)
⇒
1 1 ... + E = 45 × 10 4 . 1 + 2 + 2 (4) (16 ) 1 1 1 –45 × 104 ... + + + 2 2 2 (2) (8) (32)
4 1 45 × 10 1 4 1 45 × 10 ⇒ E = − + + + ... 1 2 2 2 (2) 4 (16) 1 − 16
Number of free electrons per unit length of conductor N = nA × 1 \ Momentum of all the free electrons is q / t q / t q p = N mv d = nA × m × n A e (e / m) t s –1 44. (b) : v 0 = 20 m s and q0 = 40°, we get v 0x = v 0 cos q0 = (20 m s –1)cos 40° = 15.3 m s –1 v 0 y = v 0 sin q0 = (20 m s –1) sin 40° = 12.8 m s–1 As, x = v 0x t 8 m = (15.3 m s –1)t , or, t = 0.52 s. =
Required height of the wall is given by, y = v 0 y t – –1
–2
2
1 2
45. (c) : Te capacitances of the two layers are given by
C 1
=
2ε0 A
; C 2
=
6ε0 A
2d d As q is same on each condenser, therefore,
V 1 ∴
q q and V 2 C 1 C 2 V 1 C 2 6ε0 A d = = × = 2d 2ε0 A V 2 C 1
=
=
3 2
EXAM CORNER 2018 Exam
Date
41. (d) : Since the total number of transitions is n(n − 1) =6⇒n= 4 2
VITEEE
4th to 15th April
JEE Main
8th April (Offline), 15 th & 16th April (Online)
Hence, the highest excitation state is n = 4. Te energy required to excite the atom from n = 1 to n = 4 is
SRMJEEE
16th to 30th April
Karnataka CET
18th & 19th April
WBJEE
22nd April
Kerala PET
23rd & 24th April
NEET
6th May
MHT CET
10th May
DE =
=
E4 – E1 =
−
13.6 4
2
eV
− −
13.6 2
eV
1
− 1 eV =12.75 eV 16
13.6 1
Tus, the energy of the incident photons, hc = 12.75 eV hu = 12.75 eV ⇒
COMEDK (Engg.) 13th May
λ
⇒
λ=
hc 12.75 eV
=
12420 eV
Å
12.75 eV
= 975 Å
42. (c) : Let Q′ be the heat per gram of coal. Ten –1
–1
(5 g)Q′ = mc Dt = (1000 g)(1.00 cal g °C )(37 °C) Q = 7400 cal g–1 q / t I 43. (a) : I = ne A v d or v d n Ae n Ae =
gt 2,
h = (12.8 m s ) (0.52 s) – (4.9 m s )(0.52 s) = 5.33 m.
1
E = 48 × 104 – 12 × 104 = 36 × 104 N C–1 N 40. (d) : v 1 = velocity of first v 2 ship relative to the earth ; 30° v 21 v 2 = velocity of second ship W E v 1 relative to the earth. Let v 21 = relative velocity of second ship with respect to first ship. S Ten v 2 = v 21 + v 1, where v 21 is due north. Tus, v 2sin30° = v 1 = 10 km h–1 ⇒ v 2 = 20 km h–1.
=
AMU (Engg.)
13th May (Revised)
BITSAT
16th to 31st May
JEE Advanced
20th May
AIIMS
27th May
JIPMER
3rd June
=
PHYSICS FOR YOU
|
MARCH ‘18
25
4. (d) : At equilibrium, pressure and temperature are
SOLUTION SET-55
1. (c) : Let v A and v B be velocities of shells A and B
respectively. Since, no external force acting. Hence, momentum is conserved. v A 2 \ mv A = 2mv B or v B 1 =
1
K Also A K B K A
=
2
mv A
2 1 2
=
2
2
2mv B
2 (where K A and K B are kinetic energies of A K B and B respectively) ...(i) From conservation of mechanical energy, U i + K i = U f + K f
or \
\
⇒
=
kQ2 10R
−
0 =
kQ2 2R
+
− k Q2
K A
+
k Q2
K B
3
1 2
K A =
2
kQ 2
5
R
2
mv A
⇒
4
kQ 2
15
R
=
K A = \
v A
2π − α
⇒
n1
kQ 2
15
R 8
kQ 2
15m
R
2. (b) :
α =
n2
(∵ V 1 = (2p – a) rA and V 2 = a rA m m ⇒ M 1(2p – a) = M 2a n1 = M and n2 = M 1 2 2 π M 1 16 π or α = = M1 + M 2 15 5. (c) : In process AB,
Q AB = DU AB + W AB As W AB = 0 and
k Q2
4
=
=
∆U AB =
1 − 1 K A + K B = + = 10 R 2R 2 R 5 K kQ2 4 K A + A = (using (i) 2 2R 5 2
or
+
−
same. \ P 1 = P 2 and T 1 = T 2 Also, let cross-sectional area of tube be A and radius of ring be r . V 1 V 2 \ n1 n2
∆U AB =
ƒ 2 5 2
f nR∆T = (∆PV ) 2
(∆PV )
Q AB = 2.5P 0V 0 In process BC, QBC = DU BC + W BC QBC = 0 + 2P 0V 0 ln 2 = 1.4P 0V 0 otal heat given to gas is Qnet = Q AB + QBC = 3.9P 0V 0 6. (b) : We analysed this problem
from the reference frame of elevator. otal buoyant force on the block, F B Length of light spot = AB = OB – OA = 2(L + x ) – 2x = 2L = constant. Hence the rate of change of length of light is zero. 3. (a) : Let v IM and v OM be the relative velocities of
2 3 ( g + a) = V ρ2 + V ρ1 5 5
For equilibrium, F B = T + Vd ( g + a) or T = F B – Vd ( g + a) 2 3 = ( g + a)V ρ2 + ρ1 − d 5 5
image and object w.r.t. mirror respectively. Terefore, v IM = –v OM (normal to plane mirror)
v I – v M = –(v O – v M ), where v M is the velocity of mirror w.r.t. ground. v I = 2v sinq ⇒ v I – v sinq = – (0 – v sinq) ⇒ ⇒
T =
10 + 10 × 10−3 2 × 1500 + 3 × 1000 − 800 5 5 2
=6N Contd. on Page No. 81
26
PHYSICS FOR YOU
|
MARCH ‘18
JEE
o n E x a t s m t a y h 1 M 2 0
PRACTICE PAPER 2018
ADVANCED PAPER - I
Section 1 (Maximum Marks : 28) • • • •
•
1.
2.
3.
This section contains SEVEN questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four options is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories : +4 If only the bubble(s) corresponding Full Marks : to all the correct option(s) is(are) darkened. +1 For darkening a bubble corresponding Partial Marks : to each correct option, provided NO incorrect option is darkened. 0 If none of the bubbles is darkened. Zero Marks : Negative Marks : –2 In all other cases. For example, if (a), (c) and (d) are all the correct options for a question, darkening all these three will get +4 marks; darkening only (a) and (d) will get +2 marks; and darkening (a) and (b) will get –2 marks, as a wrong option is also darkened.
A light bulb filament is constructed from 2 cm of tungsten wire of diameter 50 mm and is enclosed in an evacuated glass bulb. What temperature does the filament reach when it is operated at a power of 1 W? (Assume the emissivity of the tungsten surface to be 0.4.) (a) 1.94 × 10 3 K (b) 2.94 × 103 K (c) 3.88 × 10 3 K (d) 5.94 × 103 K
(d) tan b = tan a 4.
Te electric potential at a perpendicular distance r from a long charged straight wire of cross-sectional r radius a is given by V (r ) = – K ln where K is a a constant. ,
Now a second identical wire, carrying charge per unit length –q, is placed parallel to the first at a distance d from it. Both the wires have equal and opposite charges. V is the potential difference between the wires. Which of following expressions is/are correct? (a) |q| = 2 pK e0 (b) |q| = 4pK e0 d − a d − a (c) V = 2 K ln (d) V = 4K ln a a
A gun of mass M whose barrel makes an angle a with the horizontal fires a shell of mass m. Te gun is mounted on a frictionless track, so that recoil takes place with no resistive forces. Te velocity of the shell relative to the barrel is v . Te absolute velocity of the shell makes an angle b with the horizontal. Which of the following options is/are correct? mv cos α (a) Te recoil velocity of the gun is m + M mv sin α (b) Te recoil velocity of the gun is m + M m (c) tan b = 1 + tan a M
A circular parallel-plate capacitor of radius a and plate separation d is connected in series with a resistor R and a switch, initially open, to a constant voltage source V 0. Te switch is closed at time t = 0. Assuming that the charging time of the capacitor, t = CR, C is the capacitance, jd and B are the displacement current density and the magnetic flux density as a function of time t and r . (r is the radius of amperean loop between the capacitor plates). Ten, V V 0 2t / (e ) (a) jd = 20 (e t / ) (b) jd = 2 2πa R πa R −
(c) B = 5.
µ 0rV 0 2
4 πa R
(e
τ
−2t /τ
−
) (d) B =
µ 0rV 0 2
2πa R
(e
−t /τ
τ
)
A rod AC of length l and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving on the plane with velocity v strikes the rod at point B PHYSICS FOR YOU
|
MARCH ‘18
27
making angle 37° with l /4 A C the rod as shown in 37° B figure. Te collision is v elastic. (sin 37° = 3/5.) Afer collision, (a) the angular velocity o the rod will be 62v /55l (b) the centre o the rod will travel a distance pl /3 in the time in which it makes hal rotation (c) impulse o the impact orce is 24mv /55 (d) the angular velocity o the rod will be 72v /55l 6. A sphere o uniorm density r has a spherical cavity within it whose centre is at a distance a rom the centre o the sphere. Te gravitational field within the cavity is (a) uniorm (b) non-uniorm
(c)
4 3
πGρa
(d)
2 3
πGρa
7. Four rods P, Q, R and S o the same length and
material but o different radii r , r 2 , r 3 and 2r , respectively, are held between two rigid walls. Te temperature o all rods is increased through the same range. I the rods do not bend, then (a) the stress in the rods P, Q, R and S are in the ratio 1 : 2 : 3 : 4 (b) the orces on them exerted by the wall are in the ratio 1 : 2 : 3 : 4 (c) the energy stored in the rods due to elasticity are in the ratio 1 : 2 : 3 : 4 (d) the strains produced in the rods are in the ratio 1:2:3:4 Section 2 (Maximum Marks : 15) • • • •
This section contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases.
8. A thin wire o length l = 1 m is shaped into a semicircle with diameter. Te orce per unit length at the mid-point o the diameter when it carries a current I = 8 A is 1.03 × 10 –x N. Find x . (ake p = 3.14) 9. A sphere o mass M and radius r as slips on a rough horizontal plane. At some instant it has translational 28
PHYSICS FOR YOU
|
MARCH ‘18
velocity v 0 and rotational velocity about the centre v 0 . Te translational velocity afer the sphere 2r nv starts pure rolling is 0 . Find n. 7
10. A source emitting sound o requency 180 Hz is placed in ront o a wall at a distance o 2 m rom it. A detector is also placed in ront o the wall at the same distance rom it. Find the minimum distance (in m) between the source and the detector or which the detector detects a maximum o sound. Speed o sound in air = 360 m s –1. 11. A proton o mass m = 1.67 × 10–27 kg moves uniormly in a space where there are uniorm, mutually perpendicular electric and magnetic fields with E = 4.5 × 10 4 V m–1 and B = 40 m at an angle f = 60° with the x -axis in the xy -plane. Find the pitch o the trajectory afer the electric field is switched off. z
x
12. A count rate meter is used to measure the activity o a given sample. At one instant, the meter shows 4750 counts per minute. Five minutes later, it shows 2700 counts per minute. Find the hal-lie (in min) o the sample. (log10 1.760 = 0.2455) Section 3 (Maximum Marks : 18) • • • • • •
This section contains SIX questions of matching type. This section contains TWO tables (each having 3 columns and 4 rows). Based on each table, there are THREE questions. Each question has FOUR options (a), (b), (c), and (d). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories : +3 If only the bubble corresponding to the Full Marks : correct option is darkened. 0 If none of the bubbles is darkened Zero Marks : Negative Marks : –1 In all other cases
Answer Q.13, Q.14 and Q.15 by appropriately matching the information given in the three columns of the following table.
An object is placed at different positions (u) in ront o different lens combinations. Te radius o curvature o curved surace is r and the reractive index o each lens is 1.5. Te object positions u, lens combinations and nature o image ormed by the combined lens are given in column 1, 2 and 3 respectively.
Column 1
(I)
u = 2r
(II) u = r
Column 2
Column 3
(i)
(P) Smaller and real
(ii)
(Q) Smaller and virtual
he switch S is closed at t = 0. At any time t , charge on capacitor R S A is Q and current in branch AB is I . he V C R time t, for which switch R is closed, current I B and charge Q are given in columns 1, 2 and 3 respectively. Column 1
(III) u = r/ 2
(IV) u =
∞
(iii)
(iv)
Column 2
(I)
3RC /2
(i) 0
(II)
∞
(ii)
(R) Larger and real
(S) Larger and virtual
13. Which of the following combinations may be used to correct myopic eye? (a) (III) (iv) (Q) (b) (IV) (iii) (Q) (c) (IV) (i) (P) (d) (II) (iii) (R) 14. Which of the following combinations is correct if the image of an object formed at infinity? (a) (III) (iii) (P) (b) (I) (ii) (P) (c) (II) (iv) (Q) (d) (III) (i) (R) 15. Which of following combinations has magnification greater than 1? (a) (III) (iv) (S) (b) (II) (ii) (S) (c) (III) (ii) (R) (d) (I) (iii) (R) Answer Q. 16, Q.17 and Q. 18 by appropriately matching the information given in the three columns of the following table. In the given circuit diagram, the battery is ideal one with emf V . Te capacitor is initially uncharged.
(III) 3RC
(iii)
(IV) 0
(iv)
Column 3
(P)
CV 2
V 1
1 (Q) 0 − R 2 6e
V
(R)
3R
V
(S)
2R
CV 2
1 1 − e
∞
16. Which of the following combinations is correct regarding minimum charge on the capacitor? (a) (II) (iv) (P) (b) (IV) (i) (Q) (c) (IV) (iii) (Q) (d) (III) (ii) (R) 17. Which of the following combinations is equivalent to charged stored in the capacitor at time which is five times of time constant of given circuit? (a) (III) (iii) (S) (b) (II) (iv) (P) (c) (III) (iv) (P) (d) (II) (ii) (S) 18. Which of the following combinations is possible? (a) (II) (iii) (P) (b) (II) (iv) (S) (c) (III) (ii) (R) (d) (I) (ii) (R)
Paper - II Section 1 (Maximum Marks : 21) • • • •
This section contains SEVEN questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories : +3 If only the bubble corresponding to the Full Marks : correct option is darkened. 0 If none of the bubbles is darkened. Zero Marks : Negative Marks : –1 In all other cases.
1. wo moles of a certain ideal gas at a temperature 300 K were cooled isochorically so that the gas pressure reduced 2.0 times. Ten, as a result of the isobaric process, the gas expanded till its temperature got back to the initial value. Find the total amount of heat absorbed by the gas in both processes. ( g is adiabatic exponent of gas.)
(a)
300R
γ −1
(b)
150R
γ −1
(c) 150 R
PHYSICS FOR YOU
|
(d) 300 R
MARCH ‘18
29
2.
wo spherical planets P and Q have the same uniform density r, masses M P and M Q, and surface areas A and 4 A, respectively. A spherical planet W also has uniform density r and its mass is ( M P + M Q). Te escape velocities from the planets P , Q and W are v P , v Q and v W , respectively. Ten (a) v Q > v W > v P (b) v W > v Q > v P (c) v W > v Q = v P (d) v W < v Q = v P
3.
When an uncharged conducting ball of radius R is placed in an external uniform electric field, a surface charge density s = s0 cos q is induced on the ball’s surface where s0 is a constant and q is a polar angle. Find the resultant electric force acting on an induced charge of the same sign. 2
(a)
πσ0
4.
2
(b)
(d)
4 ε0 2
(c)
R2
πσ0
R2
3ε 0
πσ0
2
An electromagnetic pump designed for transferring molten metals. A pipe section with metal is located in a uniform magnetic field of induction B as shown in the figure. A current I is made to flow across this pipe section in the direction perpendicular both to the vector B and to the axis of the pipe. Find the gauge pressure produced by the pump if B = 0.10 , I = 100 A, l = 1 m and a = 2.0 cm. (a) 500 N (b) 1000 N (c) 1500 N (d) 2000 N PHYSICS FOR YOU
|
MARCH ‘18
(c)
2 gh ln
(b)
l h
2
l h
gh ln
l h
(d) 2 g h ln
• •
This section contains SEVEN questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four options is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories :
Section 2 (Maximum Marks : 28)
• •
Full Marks :
+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.
Partial Marks:
+1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened.
Zero Marks :
•
30
gh ln
A plane loop shown in figure is shaped as two squares with sides a = 20 cm and b = 10 cm and is introduced into a uniform magnetic field at right angles to the loop’s plane. Te magnetic induction varies with time as B = B0 sin wt , where B0 = 10 m and w = 100 rad s–1. Find the amplitude of the current induced in the loop if its resistance per unit length is equal to r = 50 mW m–1. Te inductance of the loop is to be neglected. (a) 5 A (b) 0.5 A (c) 0.2 A (d) 2 A
5ε 0
Shown in the figure is a container whose top and bottom diameters are D and d respectively. At the bottom of the container, there is a capillary tube of outer radius b and inner radius a. Te volume flow rate in the capillary is Q. d l If the capillary is removed the liquid comes out with a velocity of v 0. Te density of the liquid is given as r. Calculate the coefficient of viscosity h.
l h
(a)
7.
R2
πρv 02 d 4 4 πρv 02 d 4 4 (a) a (b) a 1 − 4 1 − 4 4Ql 16Ql D D πρv 02 d 4 4 πρv 02 d 4 4 (c) a (d) a 1 − 4 1 − 4 8Ql 12Ql D D 5.
A chain AB of length l is located in a smooth horizontal tube so that its fraction of length h hangs freely and touches the surface of the table with its end B as shown in the figure. At a certain moment the end A of the chain is set free. With what velocity will t his end of the chain slip out of the tube?
R2
2ε0 πσ0
6.
8.
0 If none of the bubbles is darkened.
Negative Marks : –2 In all other cases. For example, if (a), (c) and (d) are all the correct options for a question, darkening all these three will result in +4 marks ; dark ening only (a) and (d) will result in +2 marks ; and darkening (a) and (b) will result in –2 marks, as a wrong option is also darkened.
A solid body rotates with a constant angular velocity 0.50 rad s–1 about a horizontal axis AB. At time t = 0 the axis AB starts turning about the vertical with a constant angular acceleration 0.10 rad s–2. If
9.
10.
the angular velocity and angular acceleration o the body afer time t = 3.5 s are w and a respectively, then (a) w = 0.6 rad s–1 (b) w = 0.15 rad s–1 (c) a = 0.2 rad s–2 (d) a = 0.1 rad s–2 Light rom a discharge tube containing hydrogen atoms alls on the surace o a piece o sodium. Te kinetic energy o the astest photoelectrons emitted rom sodium is 0.73 eV. Te work unction or sodium is 1.82 eV. Ten choose the correct option(s). (Ionization potential o hydrogen is 13.6 eV; mass o hydrogen atom = 1.6 × 10–27 kg) (a) he energy o the photons causing the photoelectric emission is 2.55 eV. (b) Te quantum numbers ( n) o the two levels involved in the emission o these photons are 2 and 4. (c) Te change in the angular momentum o the electron in the hydrogen atom in the transition is 3h/p. (d) Te recoil speed o the emitting atom assuming it to be at rest beore the transition is 0.814 m s–1. A particle executes simple harmonic motion in a straight line such that in two o its positions the velocities are u and v and the corresponding accelerations are a and b (0 < a < b). Te distance between the positions is x and the period o motion is T . Ten, (a) T = 2π
u2 + v 2 2
β +α
2
(c) x = 11.
2
(b) T = 2π
2
2
u − v α +β
(d) x =
12.
,
is
2
β −α
2
2
u + v α −β
a2 B . 2L
π
(b) Te work perormed during the turn is (c) No current is induced in the ring.
π
2
a 4 B2 . 2L
a 4 B2 . (d) Te work perormed during the turn is 4 L For a certain metal, the K absorption edge is at 0.172 Å. Te wavelength o K a, K b, and K g lines o K π
14.
series are 0.210 Å, 0.192 Å and 0.180 Å respectively. Te energies o K , L and M orbits are E K , E L and E M , respectively. Ten (a) E K = –13.92 keV (b) E L = –8.37 keV (c) E M = –4.06 keV (d) E K = –13.04 keV
u2 − v 2
A body o mass m is hauled rom the Earth’s surace by applying a orce F varying with the height o ascent y as F = 2(ay – 1) mg where a is a positive constant and g is the acceleration due to gravity. When body rises first hal o ascent then (a) work done by orce F is 3mg /4a. (b) work done by orce F is 3mg /a. (c) change in gravitational potential energy is mg /2a. (d) change in gravitational potential energy is mg /a. An observer A is moving directly towards a stationary sound source while another observer B is moving away rom the source with the same velocity. Which o the ollowing statements is/are correct?
13.
(a) Average o requencies recorded by A and B is equal to natural requency o the source. (b) Wavelength o wave received by A is less than that o waves received by B. (c) Wavelength o waves received by two observers will be same. (d) Both the observers will observe the wave travelling with same speed. A superconducting round ring o radius a and inductance L was located in a uniorm magnetic field o induction B. Te plane o ring was parallel to the vector B and the current in the ring was equal to zero. Ten the ring was turned through 90° so that its plane became perpendicular to the field. Choose the correct statement(s). (a) Te current induced in the ring afer the turn
Section 3 (Maximum Marks : 12) • • • • •
This section contains TWO paragraphs. Based on each paragraph, there are TWO questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories : Full Marks :
Zero Marks :
+3 If only the bubble corresponding to the correct option is darkened. 0 In all other cases. PARAGRAPH 1
2 mole o monatomic ideal gas undergoes the shown cyclic process in which path o the p r o c e s s 2 → 3 is a semicircle.
V
2P 0 2
P 0 P 0/2
PHYSICS FOR YOU
3 1
V 0 |
MARCH ‘18
2V 0
V
31
15. Heat given to the system in semicircular process is
(a) (c)
π 7 − 4
P0V 0 2
(b)
π 7 + 4
P0V 0 2
(d)
π 7 + 4
T 4 – T 04 =
π 7 − 4
= 1.404 × 1013 K4. It is clear that the temperature T 0 o the surroundings can be ignored, so we obtain T = 1.94 × 10 3 K.
P0V 0 3
P0V 0 3
A = p × 0.02 × 50 × 10 –6 m2 = 3.14 × 10 –6 m2 1
ρ
=
ε Aσ
0.4 × 3.14 × 10
16. otal heat rejected in one cycle is
(a) (c)
P0V 0 (32 − π) 4
P0V 0 (32 + π) 8
(b)
(d)
P0V 0 (32 + π) 4
P0V 0 (32 − π) 8
PARAGRAPH 2
In a photoelectric setup, a point source o light o power 3.2 × 10 –3 W emits monoenergetic photons o energy 5.0 eV. Te source is located at a distance o 0.8 m rom the center o a stationary metallic sphere o work unction 3.0 eV and o radius 8.0 × 10 –3 m. he eiciency o photoelectron emission is 1 or every 106 incident photons. Assume that the sphere is isolated and initially neutral and that photoelectrons are instantaneously swept away afer emission. 17. Calculate the number o photoelectrons emitted
per second. (a) 103 (b) 104
(c) 5 × 104 (d) 105
18. It is observed that photoelectron emission stops at
a certain time t afer the light source is switched on. It is due to the retarding potential developed in t he metallic sphere due to lef over positive charges. Te stopping potential (V ) can be represented as (e = charge on electron) (a) 2(KEmax/e) (b) (KEmax/e) (c) (KEmax/3e) (d) (KEmax/2e) SOLUTIONS PAPER - 1
1. (a) : Te power radiated by area A o a black body
at temperature T is given by Stean’s law P = sT 4 A where s is the Stean-Boltzmann constant. Tus the power per unit area radiated by a body o emissivity e is esT 4. I the body’s surroundings are at a temperature T 0, it will absorb power per unit area o esT 04 rom them, so the net power emitted is P = es(T 4 – T 04) A Here, e = 0.4, s = 5.67 × 10 –8 W m–2 K–4, P = 1 W 32
PHYSICS FOR YOU
|
MARCH ‘18
2. (a, c) : Given V (r ) =
−6
−8
× 5. 67 × 10
r − K ln
a dV = K/r . dr Gauss’s theorem states that, or a closed region o space, E = −
∫
E ⋅ dS =
1
ε0
∑ Q,
where E is the electric field L vector on the surace o the a region, dS is an element o r the vector area o the surace (pointing outwards), and SQ is the total charge contained within +q –q the region o space. E \ Q = 2prLEe0 = 2pKLe0 So the charge per unit length o d r the wire is q = 2pK e0. Te field at r due to the wire carrying charge per unit length +q is K/r , and the field due to the wire carrying charge per unit length –q = K /(d – r ).
1 + 1 r d − r
E = K
Te potential difference between the wires is ound by integrating the field with respect to r , rom r = a to r = d – a :
+ 1 dr = K [ln r – ln(d – r )]d – a a r d − r
d −a 1
∫
V = K
a
d −a a 3. (a, c) : Since there is no horizontal orce on the system (gun + shell + explosives) the horizontal momentum is conserved. Horizontal relative velocity o the shell = v cos a. I v abs be the absolute horizontal velocity o the shell, then v abs = v cos a – V Here, V is the recoil speed o the gun. By conservation o linear momentum mv cos α m(v cos a – V ) – MV = 0 or V = m + M Te components o the absolute velocity o the shell are (v cos a – V ) and v sin a. = 2K ln
Tus, absolute velocity o the shell 2
=
(v cos α − V )
= v 1 −
+(
m(m + 2 M ) 2
(m + M )
v sin α
\ Velocity o separation = Velocity o approach
v sin α)2
5
cos
2
α
m tan α v cos α − V M 4. (a, d) : Te charge on the capacitor plates at time t is given by Q = CV 0(1 – e–t/ t), CV 0 (1 – e–t/ t ), \ Te charge density on the plates, r = A 2 where A = pa is the area o the plates. Since the displacement current density jd is equal to the rate o change o charge density we must have CV 0 –t/ t jd = (e ) Aτ Since t = CR and A = pa2, we may write this as V 0 −t /τ e \ jd = 2 πa R I we consider a circular loop o radius r concentric with the capacitor, as shown in figure, it links a total displacement current,
tan b =
= 1 +
2
r V 0
I d = pr 2 jd =
a2 R
d a
Amperean loop o radius r Direction o B field
Using Ampere’s law,
∫
B ⋅ dl = µ0 Id
2prB = m0I d B=
µ 0rV 0 2πa
2
R
e −t /τ
5. (b, c, d) : he ball has
4
Conserving linear momentum (o rod + particle) in the direction perpendicular to the rod, 3
mv = mu – mv ′ 5
Conserving angular momentum about point CM as shown in the figure. 3 mv l = ml2 ω − mv′ l
5
4
\ mu ⇒ u =
12
4
2
l
ml
=
4
24v 55
ω
12
,ω=
⇒ u =
l
ω
3
72v 55l
π
ime taken to rotate by p angle, t =
ω
In the same time, distance travelled = ut =
l
π
3
Using impulse-momentum equation on the rod 24mv Ndt = mu =
∫
55
6. (a, c) : Let us consider irst the gravitational
field inside a sphere o density r. At a radius x , the gravitational field is equal to the field due to the mass contained within radius x ,
e −t /τ
Displacement current
ω l = + u + v ′
3v
G \ g =
4 3
3
πx ρ =
4 πGρx , 3
x 2 and since the field is directed radially inwards, we can write this in vector orm as g
4
=−
3
πGρx
Now we can consider the sphere with a spherical cavity as shown in figure. Consider a point P within the cavity such that the vector displacement o R P rom the centre o the cavity r a cavity is r . Te vector P displacement o P rom the centre o the original sphere is thus a + r and the gravitational field at P due to the original sphere is
u
v ′ component o its velocity w B C perpendicular to the length CM A v ′ o the rod immediately afer the collision. u is the velocity o CM o the rod and w is angular velocity o the rod just afer collision. Te ball strikes the rod with a speed o v sin 37° in the perpendicular direction and its component along the length o the rod (i.e., v cos 37°) afer the collision is unchanged.
,
−
4 3
πGρ(a + r )
Te gravitational field due to the material removed to make the cavity is 4 −
3
πGρr ,
PHYSICS FOR YOU
|
MARCH ‘18
33
so the total gravitational field at P is
g = −
4 3
4
πGρ(r + a) +
3
πGρr = −
4 3
πGρa.
Te field within the cavity is thus uniorm. Its magnitude is 4pGra/3 (i.e., depends only on the position, and not the size, o the cavity) and its direction is parallel to the line joining the center o the cavity to the centre o the original space. 2
7. (b, c) : Termal orce = YAad q = Y pr ad q
r 1 = r , r 2 = r 2, r 3 = r 3, r 4 = 2r F 1 : F 2 : F 3 : F 4 = 1 : 2 : 3 : 4 Termal stress = Y ad q As Y and a are same or all the rods, hence stress developed in each rod will be the same. As strain = ad q, so strain will also be the same. E = Energy stored =
1 2
Y (strain)2 × A × L
\ E1 : E2 : E3 : E4 = 1 : 2 : 3 : 4 So, option (b) and (c) are correct. 8. (4) : Let R be the radius o the semicircle. Ten
pR + 2R = l or R =
l
.
2+π
Te magnetic field at the centre o the semicircle = field due to circular part + field due to diametrical part µ I µ I B = 0 + 0 = 0 (2 + π) 4R 4l DF = I DlB sin q 2
∆F µ0 I (2 + π) µ0 I (2 + π) = I = 4l 4l ∆l \
∆F ∆l
=
4 π × 10
−7
2
×8
(2 + π )
4 ×1
r 2
(2/5) Mr
=
5 f 2 Mr
and the clockwise angular velocity at time t will be v 5 f 5 f t = 0 + t ω(t ) = ω 0 + 2 Mr 2r 2 Mr Pure rolling starts when v (t ) = r w(t ) v 5 f i.e., v (t ) = 0 + t 2 2 M
...(ii)
From eqn. (i) and (ii), v (t ) =
2 7
v0
×3
6 =
7
v 0
=
nv 0
\
7
n = 6
10. (3) : he situation is shown in
S
figure. Suppose the detector is placed 2m at a distance o x m rom the source. x Te direct wave received rom the source travels a distance o x meter. D Te wave reaching the detector afer reflection rom the wall has travelled a distance o 2[(2) 2 + x 2/4]1/2 m. Te path difference between the two waves is 1/2 2 x 2 D = 2 (2) + − x m 4 Constructive intererence will take place when D = l, 2l, ... Te minimum distance x or a maximum sound corresponds to D = l ...(i)
Te wavelength is l =
v υ
360 m s =
180 s
−1
−1
=
2 m.
1/2
–4
= 1.03 × 10 N
\ x = 4
2 x 2 Tus, rom eqn. (i), 2 (2) + 4
− x = 2
1/2
9. (6) : Velocity o the centre = v 0 and the angular
v 0 . Tus, v 0 > w0r . Te 2r sphere slips orward and thus the riction by the plane on the sphere will act backward. As the riction is kinetic, its value is mN = m Mg and the sphere will be decelerated by aCM = f / M . f ...(i) \ v (t ) = v 0 – t M Tis riction will also produce a torque t = fr about the centre. Tis torque is clockwise and in the direction o w0. Hence the angular acceleration about the centre will be
velocity about the centre =
34
a = f
PHYSICS FOR YOU
|
MARCH ‘18
x2 x ⇒ 4 + = 1 + \ x = 3 m 4 2 Tus, the detector should be placed at a distance o 3 m rom the source. 11. (1) : Here qE = qvB sin f or v =
E B sin φ
v ′ = v cos f = velocity along the field v ″ = v sin f = velocity perpendicular to field By the dynamics o circular motion mv ″ 2 2πm qv ″B = or qB = mw or T = r qB 2πm E cos φ 2 πm v cos f = \ p = T × v cos f = qB B sin φ qB
or p =
2πmE
qB2 ≈ 1 m
cot φ =
2 π ×1.67 ×10 1.6 × 10
−19
−27
× 4 .5 ×10 4
× 402 × 10 −6
⇒ f = 2r cot 60°
12. (6) : We have, N = N 0e–lt
...(i)
Te activity of the sample is given by dN − λ t = − λN 0e = − λ N dt i.e., activity is proportional to the number of undecayed nuclei. dN = − λ N At t = 0, 0 dt t =0 At t = 5 min,
\
dN = − λ N dt t = 5 min
=e
λ t
or
ln
N 0 N
= λ t
λ = ln
l =
2.3026 5
× 0.2455 = 0.113 min –1
Half-life of sample, T =
0.6931
λ
=
0 .6931 0.113
13. (b), 14. (d), 15 (a)
Using lens maker’s formula,
1 1 = (µ − 1) − R1 R2 f 1
Here, m = 1.5 (i) R1 = r , R2 = –r \
1 1 = (1.5 − 1) = (0.5) 2 + r r r f
⇒ f = r
So,
f ′
=
1
f1
+
1
f 2
=
1
r
+
1
r
=
2
r
⇒ f ′ = (r /2). (ii) R1 = ∞, R2 = –r 1 1 1 0.5 = (1.5 − 1) = = + ∞ r r 2r f 1
1
+
1
f
2
=
f
=
2 2r
(iii) R1 = ∞, R2 = r 1 1 1 = (1.5 − 1) = − − ∞ r 2r f 1
⇒ f = –2r
So,
1
f ′ 1
f ′
=
=
1
f 1
f1
+
+
1
f
=
1
f2
2
=
f 1
r
=
2
−2r
+
1
−2r
⇒ f ′ = −r
=
1 2r
= 6.1 min
For myopic eye, person cannot see far-off objects clearly. o correct such defect of vision, a concave lens must be used whose focal length is equal to the far point of myopic eye. \ u = ∞, f = v and image formed by this lens is virtual and smaller. Te correct combination is IV → iii → Q Image of the object is formed at infinity if the object is placed at focus of the lens. Image formed is very large and real in case of convex lens. So option (d) is correct for question no. 14. Image distance Magnification = Object distance For convex lens : If 2 f < u < f then v > 2 f then image formed will be larger and real. If u is between focus and pole then image will be larger and virtual. So option (a) is correct for question no. 15. 16. (c), 17. (b), 18 (d)
1
1
=
⇒ f ′ = 2r
1 N 0 2.3026 N = × log10 0 t N t N Substituting the given values, we have ⇒
1
f ′ f ⇒ f ′ = r
(iv)
dN dt t = 0 N 0 4750 = = = 1.760 N dN 2700 dt t = 5 min
From eqn. (i), we have N −λ t or N 0 =e N 0 N
So,
Let at any time t charge on capacitor C be Q and currents are as shown in the figure. Since, charge Q will increase with time t .
i
N
A i1
S
O
i – i1 V
M
R
R R B
i
Q
C P
Applying Kirchhoff’s second law in loop MNABM V = (i – i1)R + iR or V = 2iR – i1R ...(i) Similarly, applying Kirchhoff ’s second law in loop MNOPM , we have PHYSICS FOR YOU
|
MARCH ‘18
35
V = i1R +
Q C
PAPER-2
iR
+
...(ii)
Eliminating i from eqns. (i) and (ii), we get V = 3i1R +
1. (d) : Here, n = 2,
For isochoric process, P ∝ T
2Q
P 1
\
C 2Q
V − 2Q or 3i1R = V − or i1 = C C 3R dQ dQ dt 1 2Q = or or V − 2Q dt C 3R 3R 1
T 1
=
P 2
T 2
or
∫
dQ
Q
0
dt
t
2Q
V −
−
= ∫
Q1 = DU 1 =
C
3R
0
C
Tis equation gives charge on capacitor at any time t CV
Q = i1 =
(1 – e–2t /3RC )
2
dQ
V =
dt
3R
e
...(iii)
i =
+
i1R
2t /3 RC
=
2R
V +
3
=
\ Current through AB V 2t /3 RC V + e − V −2t /3 RC 3 I = i – i1 = e − 2R 3R V V −2t /3 RC I = e − 2R 6 R At t = 0, cV (1 e 0 ) 0 Q0 = 2 V V V V V 0 I = (e ) 2R 6R 2R 6R 3R 3RC At t = = τ c −
=
nR∆T
I =
2 V 2R
V −
6R
−5 )
(e
5
−
−
36
V 2R
.
PHYSICS FOR YOU
MARCH ‘18
R
r is same for all planet)
2
4 πRP =
A
1
RQ
⇒
2
= 2
RP
Te spherical planet W has mass M W = M P + M Q
or
3
3
ρπRW = 3
RW
=
3
RP
4 3
+
3
ρπRP +
4 3
3
3
3
3
ρπRQ ⇒ RW = RP + RQ
3
(2RP )
So, RW = (9)1/3R P Terefore, RW > RQ > R P From eqn. (i), v W > v Q > v P
2
3. (a) : An element area dS = 2pRsinq(Rd q) is shown CV 2
in figure. Te force acting on the area element dS of a conductor is
dF |
=
RQ
4
2R
3
...(i)
RP
∴
Tis situation corresponds to time t = ∞ i.e., Q = and I =
3
πR
(
∴
V ≈
R
=
4
Surface area of Q, 4pRQ2 = 4 A
=
2
2GM
2 ⋅ Gρ
Surface area of P ,
CV ≈ CV = 99.99
)
( γ − 1)
v e ∝ R
2
(1 − e
T 1 − RT 1 − T = γ − 1 2 1 (γ − 1) 2R
...(iv)
CV
CV
γ −1
=
R γ T 1
v e =
−1 CV 1 − 1 (1 − e ) = 2 2 e V V −1 V 1 1 − (e ) = I = − 2R 6R R 2 6e 15RC When t = 5tc = Q =
P 1
1 = 2
planet is
2
Q =
2
P 2
2. (b) : he escape velocity from the surface of a
=
−
−
T 1
Hence the total amount of heat absorbed Q = Q1 + Q2 = RT 1 = 300 R
2t /3 RC
e−
2R
−
=
1 = R T − T 1 γ = nR∆T ′ + 1 2 1 γ − 1 2 γ − 1
From eqn. (i) V
P 1
T 1
During the second case (i.e., isobaric process), DW 2 = P DV = nRDT Tus from first law of thermodynamics : nR∆T ′ Q2 = ∆U 2 + ∆W 2 = + nR∆T ′ γ − 1
−
V
×
Now from first law of thermodynamics
=
V
P 2
T 2 =
⇒
=
1 2
EdS
σ
...(i)
\ dF = dF cos q
...(ii)
z
\ dF z =
πσ
2
R2
ε0
6. (c) : Let the length of the chain inside the smooth
horizontal tube at an arbitrary instant is x . From the equation of motion dm ma = F + u dt As u = 0, F = ma
sin θ cos θ d θ
As E = σ ε0
πσ20 R2 π /2 ∫ dF z = ε0 ∫ 0 sin θ cos3 θ d θ ∴
F
=
πσ
F z =
2 0
R2
4 ε0
4. (b) : When the capillary is removed,the liquid comes
out with a velocity of v 0. Density of liquid is r. From Bernouilli’s theorem, ∴
P + P0
+
1 2
2
ρv1 + ρgh =
1
2
ρv0 + P0
2
ρ
2 2 or P + ρgh = (v0 − v1 ) 2
...(i)
By equation of continuity, D
π
A1v1 = A2 v0 , where A1 =
2
4
π
, A2 =
2
d
4
.
2
or
v 1 =
d v D
0
...(ii)
From eqn. (i) and (ii), eliminate v 1
d ∴ P + ρgh = v − v 2 D ρv d ρv d − or P + ρgh = 1 P ∆ or = 1 − 2 2 D D ρ
2
2
2
0
2
2
0
2
4
0
4
4
By Poiseuille’s equation, Q = or
η
=
π( ∆P )a4 8 Ql
or
η
=
4
8 Ql
×
I , aL where L is the length of the section. Te difference in pressure produced must be, I × B × (abL) IB aL ∆P = = ab a
ρv 02
1
2
5. (a) : Te current density is
=
100 × 0.10 2 × 10
−2
=
500 N
2
l or v = = gh ln h
2 gh ln
l h
7. (b) : Te loops are connected in such a way that
8 ηl
π
or
v 2
λ=
(l − h )
0
4
0
π( ∆P )a
m (Since F = T ) ...(i) l Similarly for the overhanging part, u = 0 Tus, ma = F or lha = lhg – T ...(ii) From eqns. (i) and (ii) dv a = v dv l(x + h)a = lhg or, (x + h)v = hg ds ds dv or ( x + h) v = gh, (−dx ) [As the length of the chain inside the tube decreases with time, ds = –dx .] dx or vdv = − gh x+h 0 v dx Integrating ∫ vdv = − gh ∫ x +h or lxa = T where
L
−
4 a D 4 d 4
if the current is clockwise in one, it is anticlockwise in the other. Hence the e.m.f. in smaller loop opposes the e.m.f in larger loop d d e.m.f in larger loop = (a2 B) = a2 ( B0 sin ω t ) dt dt = a2wB0 cos wt Similarly, e.m.f. in smaller loop = b2 B0 wcos wt . Hence, net e.m.f in the circuit = (a2 – b2)B0 w cos wt , as both the e.m.fs are in opposite sence, and resistance of the circuit = 4(a + b) r 2 2 (a − b )B0ω Terefore, the amplitude of the current, I = 4(a + b) ρ Since a = 20 cm = 0.2 m; b = 10 cm = 0.1 m, P = 0.05 Wm–1 B0 = 10 m = 10 –2 and w = 100 rad s –1 \ I = 0.5 A PHYSICS FOR YOU
|
MARCH ‘18
37
8. (a, c)
From eqns. (iii) and (iv), we obtain
9. (a, b, d) : (a) According to Einstein’s photoelectric
equation Incident Energy = work function + Maximum kinetic energy of the photoelectron or E = (1.82 + 0.73) eV or E = 2.55 eV (b) In case of hydrogen atom, En = – (13.6/n2). \ E1 = – 13.6 eV, E2 = – 3.4 eV, E3 = – 1.5 eV, E4 = – 0.85 eV. Te discharge tube contains hydrogen atoms. Since E4 – E2 = – 0.85 – (–3.4) = 2.55 eV Hence the quantum numbers of the two levels involved in the emission of these photons are 4 and 2. Te transition occurs from n = 4 to n = 2 (c) Change in angular momentum in above transition, DL = L2 – L4 h − 4 h −h or ∆L = 2 or ∆L =
2π
2π
π
(d) Momentum is conserved in the process. Momentum of hydrogen atom = mv E Momentum of emitted photon = c E ∴ mv c
MARCH ‘18
2π (β − α)
=
= 2π
(u 2 − v 2 ) (α + β)(β − α)
2
x − x π β − α 2
1
u2 − v 2
= 2π
β2 − α 2
11. (a, c) : First, let us find the total height of ascent.
At the beginning and the end of the path the velocity of the body is equal to zero, and therefore the increment in the kinetic energy of the body is also equal to zero. On the other hand, in accordance with work-energy theorem DK.E. is equal to the algebraic sum of the works W performed by all the forces, i.e., by the force F and gravity, over this path. However, since DK.E. = 0, then W = 0. aking into account that the upward direction is assumed to coincide with the positive direction of the y -axis, we can write
∫
h
∫
W = mg (1 − 2ay )dy = mgh(1 − ah) = 0 0
10. (b, c) : Velocity and acceleration for a particle
|
ω
=
As g acts downwards and dr acts upwards.
or v = 0.814 m s –1 \ Recoil speed of emitting atom = 0.814 m s–1.
PHYSICS FOR YOU
2π
(x2 − x 1 )
∫
−19 2.55 × 1.6 × 10 E or v = or v = 27 8 mc (1.67 × 10− )(3 × 10 )
38
ime period T =
β−α ( x2 − x 1 )
w2 =
⇒
W = (F + mg ) ⋅ dr = (2ay − 1)m( g ⋅ dr )
=
executing S.H.M. are given by the expressions v 2 = w2( A2 – x 2) and a = –w2x Let the particle be at positions x 1 and x 2 at the instant. u2 = w2( A2 – x 21) v 2 = w2( A2 – x 22) a = w2x 1 b = w2x 2 Subtracting eqn, (ii) from eqn. (i), u2 – v 2 = w2(x 22 – x 21) Adding eqns (iii) and (iv), a + b = w2(x 1 + x 2) Dividing eq. (v) by eqn. (vi), u2 − v 2 = x2 − x 1 = x α +β which is separation between particles at the instant.
b – a = w2(x 2 – x 1)
Whence h = 1/a. Te work performed by the force F over the g first half of the ascent is h/2
h/2
∫ F ⋅ dr = 2mg ∫ (1 − ay)dy = 3mg / 4a
W F =
0
given ...(i) ...(ii) ...(iii) ...(iv) ...(v)
given
0
Te corresponding increment of the gravitational potential energy is DU = mgh/2 = mg /2a. 12. (a, c) : Let velocity of each observer be u as shown
in the figure. Ten frequency received by A will be u v + u u1 = υ0 A B S v where u0 is natural frequency of the source and v is sound propagation velocity. Te frequency received by B will be v − u u2 = υ0 v Since (u1 + u2)/2 = u0, therefore option (a) is correct.
13. (b) : In a superconductor there is no resistance. dI d φ Hence, L = + , dt dt ∆φ πa2 B So integrating, I = = L L
∫
π =
d φ
1
∫ dt I dt = 2 LI 2
2 4 2
a B
2L 14. (a, b, c) : Energy of K absorption edge 1242 eVnm Ea = = 73.06 × 103 eV = 73.06 keV 0.017 nm Energy of K a line is 1242 eVnm hc EK = = = 59.14 keV α eλ α 0.021 nm 1242 Similarly, EK = = 64.69 keV β 0.0192 1242 = 69 keV EK = γ 0.0180 Energy of K shell = (EK − Ea ) E K = (59.14 – 73.06) keV = –13.92 keV Energy of L shell = EK − Ea α
β
E L = 64.69 keV – 73.06 keV = –8.37 keV Energy of M shell = EK − Ea γ
E M = 69 keV – 73.06 keV = –4.06 keV
15. (c) : Process 1 → 2 (Isochoric process) W 12 = 0 Q12 = DU 12 = nC v(T 2 – T 1)
3 3 = n R (T2 − T1 ) = [P 0V 0 – (P 0/2)V 0] 2 2 3P V = 0 0 (heat absorbed) 4 Process 3 → 1,W 3 → 1 = Area under process 3 → 1
P 0
W 2 → 3 =
P0V 0
1 P0V0 π 4 2
3P V = − + P 0 ⋅ = − 0 0 2 2 4 Work done during 3 → 1 will be negative as volume is decreasing. 3R 3 P V 9P V (T1 − T 3 ) = 0 0 − P2 2V 0 = − 0 0 DU 3 → 1 = n 2 2 2 4 Q3 → 1 = DU 3 → 1 + W 3 → 1 = –3P 0V 0 (heat rejected) Process 2 → 3, Te temperature will be maximum at x = 2P 0, Since the process in semicircular, volume at x = (3/2) V 0 Using ideal gas equation, T x = 3P 0V 0/2R V0
+
P0V0
2
=
P0V0 π
+ 1 2 4
(π + 4) 4 3 3 (3P V P V ) = 3P 0V 0 DU 2 → x = n R(Tx T2 ) 2 2 0 0 0 0 3 3 3 P0V0 (2P0V0 3P0V0 ) DU x → 3 = n R(T3 Tx ) 2 2 2 −
Because Df = f f – fi, f f = pa2B, fi = 0 Also, the work done is, W = εIdt =
W 2 → x = W x → 3 =
=
−
−
=
−
= −
π + 1 2 4 1 π P V π = P0V 0 3 + + = 0 0 7 +
P V \ DQ2 → x = DU 2 → x + W 2 → x = 3P0V 0 + 0 0
2
8
2
4 (heat absorbed)
DQ x → 3 = DU x → 3 + W x → 3 P V π 3 − P V π = − P0V 0 + 0 0 + 1 = 0 0 2 −
2
4
2
2 4 (heat released)
16. (d) : otal heat rejected in one cycle P V π = 3P0V 0 + 0 0 2 − = 4 P0V 0
πP0V 0
4 8 π P0V 0 (32 − π) = P0V 0 4 − = 8 8 17. (d) : If P is the power of point source of light, the intensity at a distance r is 2
I =
−
P
4 πr 2 Te energy intercepted by the metallic sphere per second is E = intensity × projected area of sphere =
P
× πR
2
=
PR
R
r S
2
4 πr 2 4r 2 If E1 is the energy of the single photon and h the efficiency of the photon to liberate an electron, the number of ejected electron per second is =
η
PR
2
4r 2 E1
(10 6 )(3.2 × 10 3 )(8 × 10 3 )2 −
=
−
−
4 × (0.8)2 × (5 ×1 .6 ×10 5
= 10 electrons s
19
−
)
–1
18. (b) : he emission of electrons from a metallic
sphere leaves it positively charged. As the potential of the charged sphere begins to rise, it attracts emitted electrons. Te emission of electrons will stop when the kinetic energy of the electrons is neutralised by the retarding potential of the sphere. So, we have KEmax eV = KEmax ⇒ V =
e
PHYSICS FOR YOU
|
MARCH ‘18
39
1.
2.
3.
A solid cylindrical wheel of mass M and radius R is pulled by a force F applied to the centre of the wheel at 37° to the horizontal. If the wheel is to roll without slipping, what is the maximum value of |F |? Te coefficients of static and kinetic friction are ms = 0.40 and mk = 0.30 and sin 37° = 3/5. (a) 0.79 Mg (b) 0.98 Mg (c) 0.6 Mg (d) 0.49 Mg A surface will be an equipotential surface, if (a) electric field is tangential to surface at all the points (b) electric potential at all the points is non-zero (c) work done (along any path) in moving a point charge over the surface is non-zero (d) electrical potential energy of a unit positive charge at any point on the surface will be same. A smooth piston of mass m, area of cross-section A P 0 m P 0, V 0 is in equilibrium in a gas jar when the pressure of the gas is P 0. When the piston is disturbed slightly, the angular frequency of oscillation of the piston is (Assume adiabatic change of state of the gas and g is ratio of specific heats of the gas.) 2
(a)
γ P0 A
3mV 0
(b)
(c) 4.
40
mV 0
(c)
(d)
Te kinetic energy versus time graph for a particle is shown in the figure. Te force versus time graph for the particle may be PHYSICS FOR YOU
|
MARCH ‘18
e c r o F
(d)
e c r o F
ime
6.
A rod of length 1000 mm and 1000 mm coefficient of linear expansion –4 a = 10 °C–1 is placed symmetrically between fixed 1001 mm walls separated by 1001 mm. Te Young’s modulus of the rod is 1011 N m–2. If the temperature is increased by 20°C, then the stress developed in the rod is (a) 1011 Pa (b) 1010 Pa (c) 109 Pa (d) 108 Pa
7.
A body is projected up with a speed u and the time taken by it is T to reach the maximum height H . Pick out the correct statement. (a) It reaches height H /2 in time T /2. (b) It acquires velocity u/2 in time T /2. (c) Its velocity is u/2 at H /2. (d) It has same velocity during its journey.
8.
An alternating voltage is given by e = e1 sin wt + e2 cos wt . Ten, the root mean square value of voltage is given by
2 γ P0 A
(a) ime
ime
Te capacitors C 1 and C 2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Ten, (a) 5 C 1 = 3 C 2 (b) 3 C 1 = 5 C 2 (c) 5 C 1 + 3 C 2 = 0 (d) 9 C 1 = 4 C 2
2mV 0
Kinetic energy
e c r o F
5.
γ P0 A
mV 0
(b)
ime
2
e c r o F
ime
2
2
γ P0 A
(a)
(c)
2
2
ε1 + ε 2
ε1ε2 2
(b)
ε1ε2 2
(d)
2
ε1 + ε 2
2
For the arrangement M 2 M 4 M as shown in the figure, y d 5 M find the magnitude 7 M x 3 M O M and direction of the d 7 M net gravitational force 5 M d /2 2 M acting on the central 4 M M d d particle at O, with respect to the axes is 2 2 3GM 3GM (a) i (b) j d 2 d 2 2 3GM 2 3GM (i + j) ( −i − j) (c) (d) 2 2 d d 10. Molar heat capacity for an ideal gas (a) is infinity for adiabatic process (b) is zero for isothermal process (c) is independent of the nature of the gas for a process in which either volume or pressure is constant (d) is equal to the product of molecular weight and specific heat capacity for any process. 9.
11.
A boat crosses a river of width 1 km in shortest path in 15 minutes. If the speed of boat in still water is 5 km h–1, then what is the speed of the river? (a) 1 km h–1 (b) 3 km h–1 (c) 2 km h–1 (d) 5 km h–1
12.
A smooth wire is bent into a vertical w circle of radius a. A bead P can slide A smoothly on the wire. Te circle is a rotated about vertical diameter AB a/2 as axis with a speed w as shown in P figure. Te bead P is at rest w.r.t. the B circular ring in the position shown. Ten w2 is equal to g 3 2 g 2a 2 g (a) (b) (c) (d) a a g 3 a 3
13.
For the circuit as shown in figure, if the value of rms current is 2.2 A, the power factor of the box is (a) (c)
1
2
(d)
100 W
1 2
−
m
ρ −
m
−
m
2
(c)
m
ρ=
2 1 m
1
−
m
ρ −
m
−
m
m
2
1
ρ
1 2
(d)
m
1 2
ρ=
m
2
1
1
ρ
2 1
2
15.
A car is moving along a road with a speed of 45 km h–1. In what direction must a body be projected from it with a velocity of 25 m s –1, so that its resultant motion is at right angles to the direction of car? (a) At an angle of 120° with the direction of motion of car. (b) At an angle of 60° with the direction of motion of car. (c) At an angle of 90° with the direction of motion of car. (d) At an angle of 135° with the direction of motion of car.
16.
A wire of length l is moving with velocity v at an angle q in the region of a uniform magnetic field B such that length of the wire always remains perpendicular to field lines. Te induced emf in wire is (a) 0 (b) l(v ⋅ B) (c) l(v × B) (d) l(B × v )
17.
wo metal spheres are falling through a liquid of density 2 × 103 kg m–3 with the same uniform speed. Te density of sphere 1 and sphere 2 are 8 × 103 kg m–3 and 11 × 103 kg m–3 respectively. Te ratio of their radii is (a)
18.
8
(b)
11
(c)
8
3
(d)
2
3 2
1
2b
2π
m
(b)
2 π
b m
b b 1 (d) m π π 2m An express train is moving with a velocity v 1. Its driver finds another train is moving on the same track in the same direction with velocity v 2. o escape collision, driver applies a retardation a on the train. Te minimum time of escaping collision will be
(c) 19.
11
A particle of mass m moves in potential energy field given by U = bx 2 – ax , where a and b are positive constants. Te frequency of oscillation of the particle is (a)
C
V rms = 220 V, w = 100p s–1
An object weighs m1 in a liquid of density r1 and that in liquid of density r2 is m2. Te density r of the object is m ρ − m ρ − m1ρ1 m ρ 2 2 (a) ρ = 2 2 (b) ρ = 1 1 m
Box
(b) 1
2
3
1/p H
14.
1
PHYSICS FOR YOU
|
MARCH ‘18
41
(a) t
=
(c) t = 20.
v1
−
v 2
(b) t
(d) t =
a v1 + v 2 a
v12
=
−
is 0.25 that o the lef string. At what speed does the transmitted wave travel?
v 22
2a
v12 + v 22
A wire o length L and 3 identical cells o negligible internal resistance are connected in series. Due to the current, the temperature o the wire is raised by DT in a time t . N similar cells is now connected in series with a wire o the same material and crosssection but o length 2L. Te temperature o the wire is raised by the same amount DT in the same time t . Te value o N is (a) 4 (b) 6 (c) 8 (d) 9
21.
A thermometer has wrong calibrations. It reads melting point o ice as –10°C and reads 60°C at a temperature o 50°C. emperature o boiling point noted by this thermometer will be (a) 120 °C (b) 110 °C (c) 90 °C (d) 130 °C
22.
A body o mass 60 kg is resting in a lif which accelerates upwards with acceleration 2 m s–2. Te apparent weight is [ake g = 10 m s–2] (a) 72 N (b) 60 N (c) 48 N (d) 720 N
23.
(a)
25.
42
µ 0 I
2
8π
(b)
µ 0 I
2
(c)
16 π
µ 0 I 4π
An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 10 4 cal heat at higher temperature. Amount o heat converted into work is (a) 4.8 × 104 cal (b) 2.4 × 104 cal (c) 1.2 × 104 cal (d) 6 × 104 cal
27.
A nichrome heating element connected across 230 V supply consumes 1.5 kW o power and heats upto a temperature o 750°C. A tungsten bulb across the same supply operates at much higher temperature o 1600°C. In order to emit light the (a) power o bulb is much higher than that o heating element (b) power o heating element is more than that o bulb (c) power o both is nearly same (d) data is insufficient to reach any conclusion
28.
Variation o temperature T T o two identical bodies A and B with time t , when A they loose heat by radiation only is as shown in the figure. B t Ten, correct relation between emissivity (e) and absorptivity (a) o the two bodies is (a) e A > eB and a A < aB (b) e A < eB and a A > aB (c) e A > eB and a A > aB (d) e A < eB and a A < aB
29.
Find the moment o inertia o a system having a spherical ball o mass m and radius r attached at the end o a thin straight rod o mass M and length L, about the AA′ shown in the figure.
2
(d)
µ 0 I 2π
A certain gas is taken to the five states represented by dots in the graph. Te plotted lines are isotherms. Order o the most probable speed v p o the molecules at these five states is (a) v p at 3 > v p at 1 = v p at 2 > v p at 4 = v p at 5 (b) v p at 1 > v p at 2 = v p at 3 > v p at 4 = v p at 5 (c) v p at 3 > v p at 2 = v p at 4 > v p at 1 = v p at 5 (d) insufficient inormation to predict the result. Te pulse in the figure shown has a speed o 0.1 m s–1. Te linear mass density o the right string PHYSICS FOR YOU
|
MARCH ‘18
(b) 20 cm s–1 (d) 17.5 cm s–1
26.
A long straight wire o circular cross-section is made o a non-magnetic material. Te wire has radius R. Te wire carries a current I which is uniormly distributed over its cross-section. Te energy stored per unit length in the magnetic field contained within the wire is 2
24.
(a) 25 cm s–1 (c) 15 cm s–1
a
L2
m + M + 1 mr 2 2 3 5 M 2 (b) L2 m + + mr 2 3 5 2 M 2 2 2L + mr (c) m + 3 3 5 (a)
(d) None o these
A m
M r A′
L
30.
A spaceship is launched into a circular orbit close to the earth’s surace. Te additional velocity now to be imparted to the spaceship in the orbit to overcome the gravitational pull is (a)
2 gR
(b)
e
τ
(b)
m
m
τ
(c)
e
e m
τ
m
(d)
e
τ
A charged particle moves with a velocity v = ai + dj in a magnetic field B = Ai + Dj . Te orce acting on the particle has magnitude o (a) F = 0, i aD = dA (b) F = 0, i aD = –dA (c) F = 0, i aA = –dD (d) F ∝ (a2 + b2)1/2 × ( A2 + D2)1/2 35. In a common emitter configuration o a transistor, the voltage drop across a 500 W resistor in the collector circuit is 0.5 V when the collector supply voltage is 5 V. I the current gain in the common base mode is 0.96, the base current is
34.
(c) 38.
( 2 + 1) gR
(c) 2 gR (d) ( 2 − 1) gR 31. Let there are two bulbs in your house, one is red and other one is blue. Both bulbs are o same power. I nr and nb are the number o photons per second reaching towards you in a certain time, then (a) nr = nb (b) nr > nb (c) nr < nb (d) nr ⋅ nb = c2 32. Te mean ree path o a gas molecule at SP is 2.1 × 10–7 m. Find the diameter o the molecule. (Boltzmann constant = 1.4 × 10–23 J K–1) (a) 5.2 Å (b) 4.6 Å (c) 3.5 Å (d) 2.0 Å 33. Te mobility o ree electrons (charge e, mass m and relaxation time t) in a metal is proportional to (a)
(a)
ˆ
ˆ
1
r
(c)
20 1 20
µA
(b)
(d)
mA
1 5
µA
1 24
mA
A wheel with 30 metallic spokes each o 0.7 m long is rotated with a speed o 120 rpm, in a plane normal to the horizontal component o earth’s magnetic field H E at a place. I H E = 0.8 G at the place, what is the induced em between the axle and the rim o the wheel? (Given 1 G = 10–4 ) (a) 2.46 × 10–4 V (b) 6.28 × 10–4 V (c) 5.76 × 10–5 V (d) 4.92 × 10–4 V 37. Let r be the distance o a point on the axis o a bar magnet o length 2 l rom its centre (r >>> l ). Te magnetic field at such a point is proportional to 36.
1
r 2
(d) None o these
r 3
A stone o relative density K is released rom rest on the surace o a lake. I viscous effects are ignored, the stone sinks in water with an acceleration o (a) g (1 – K ) (b) g (1 + K ) 1 1 g 1 + (c) g (d) 1 − K K
An a-particle moving horizontally makes a head on collision elastically with a proton (at rest). What are the ratio o de-Broglie’s wavelength’s associated with a-particle and proton just afer collision? (a) 2 : 1 (b) 4 : 3 (c) 1 : 2 (d) 2 : 3 40. wo identical positive point charges Q each are fixed at a distance d = 2a apart in air. A point charge –q lies at mid-point on the line joining the charges. I –q is given a very small lateral displacement, the requency o oscillation is 39.
(a)
Qq
1 2π
2πε 0ma
3
(b)
Qq
1 2π
4 πε0ma
3
ˆ
ˆ
1
(b)
1
(a)
(c)
1 π
Qq πε
0
ma3
(d)
2Qq
1 π
πε
0
ma3
Directions : In the following questions (41-60), a statement of assertion is followed by a statement of reason. Mark the correct choice as
(a) I both assertion and reason are true and reason is the correct explanation o assertion. (b) I both assertion and reason are true but reason is not the correct explanation o assertion. (c) I assertion is true but reason is alse. (d) I both assertion and reason are alse. 41. Assertion : Ionisation energy o isolated pentavalent atom o phosphorus is very large so, in a n-type semiconductor, there are no ree charge carries at room temperature. Reason : When current flows through a p-n junction diode, it becomes hot. 42. Assertion : A point particle o mass m moving with speed v collides with stationary point particle o mass M . I the maximum energy loss possible is 1 then k = M . given as k mv 2 2 M + m Reason : Maximum energy loss occurs when the particles get struck together as a result o the collision. PHYSICS FOR YOU
|
MARCH ‘18
43
43. Assertion : When
the displacement of a body is directly proportional to the square of the time. Te the body is moving with uniform acceleration. Reason : Te slope of velocity-time graph with time axis given acceleration.
44. Assertion : When
stream of a-particles (obtained from a decaying radioactive sample) is made to pass through a region of perpendicular magnetic field, they follow circular paths of some fixed values of radius. Reason : Some of the daughter nuclei may be produced in their excited states. A solid body of density half that of water, falls from a height of 10 m and then enters into water. Te depth to which it will go in water is 10 m. Reason : Depth in water is equal to height from which the body falls.
45. Assertion :
When a body is dropped from a height explodes in mid air, but its centre of mass keeps moving in vertically downward direction. Reason : Explosion occur under internal forces only. External force is zero.
46. Assertion :
NO gate is also called invertor circuit. Reason : NO gate inverts the input order.
47. Assertion :
Te apparent weight of a body in an elevator moving downward with some acceleration is less than the actual weight of a body. Reason: Te part of the weight is spent in producing downward acceleration, when body is in elevator.
48. Assertion:
Fragments produced in the fission of
49. Assertion :
235 92U are not radioactive. Reason : Te
fragmented elements have nearly the same number of protons and neutrons. 50. Assertion :
Te sky waves are not used in the transmission of television signals. Reason : Sky waves are mechanical waves.
51. Assertion :
Te stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. Reason : In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant. Te susceptibility of diamagnetic materials does not depend upon temperature.
52. Assertion
44
:
PHYSICS FOR YOU
|
MARCH ‘18
Reason : Every
atom of a diamagnetic material is not a complete magnet in itself. 53. Assertion :
Energy of K photon is approximated as a
E(K a) = 13.6
×
3 4
2
( Z − 1)
An electron of L-shell is partially screened by a electron of K -shell and so effective nuclear charge is (Z – 1) units. Reason :
54. Assertion : In
an elastic collision of two billiard balls, the total kinetic energy is conserved during the short time of oscillation of the balls (i.e., when they are in contact). Reason : Energy spent against friction does not follow the law of conservation of energy.
55. Assertion : It
is not possible to have interference between the waves produced by two violins. Reason : For interference of two waves the phase difference between the waves must remains constant. If the angles of the base of the prism are equal, then in the position of minimum deviation, the refracted ray will pass parallel to the base of prism. Reason : In the case of minimum deviation, the angle of incidence is equal to the angle of emergence.
56. Assertion :
57. Assertion : A
copper sheet is placed in a magnetic field. If we pull it out of the field or push it into the field, we experience an opposing force. Reason : According to Lenz’s law eddy current produced in sheet opposes the motion of the sheet.
58. Assertion : Equal
masses of helium and oxygen gases are given equal quantities of heat. Tere will be a greater rise in the temperature of helium as compared to that of oxygen. Reason : Te molecular weight of oxygen is more than the molecular weight of helium. If objective and eye lenses of a microscope are interchanged then it can work as telescope. Reason : Te objective lens of telescope has small focal length.
59. Assertion
:
For practical purposes, the earth is used as a reference at zero potential in electrical circuits. Reason : Te electrical potential of a sphere of radius R with charge Q uniformly distributed on
60. Assertion:
the surface is given by
Q 4 πε0 R
.
⇒ Q1 = 120C 1 and Q2 = 200C 2
SOLUTIONS
1.
(a) : N = Mg –
3F 5
= Mg – 0.6 F N +
F
3F 5
4 F
37°
5
msN
= 0. 8
F
Mg
For pure rolling, a = Ra 0.8 F − µ s N
M
(µ N )R = R s 1 2 MR 2
α = τ I
0.8F = 3msN = 3(0.4) ( Mg – 0.6F ) F = 0.79 Mg \ Maximum value of F = 0.79 Mg 2. (d) : At an equipotential surface, V = constant Hence, potential energy of a point charge, qV, is also constant at every point of surface. 3. (c) : In adiabatic process, PV g = C dF = (dP ) A ln P + g ln V = ln C aking differentials on both sides, dP γ dV + =0 P V −γ P dV or dP = ...(i) V where, dV = change in volume of the gas = Adx . dF P = P 0, V = V 0 and dV = Adx in Substituting dP = A the eqn. (i), we have dF γ P 0 = Adx A V 0 ,
⇒
ω =
dF = dx keff m
2
2
γ P0 A
\
V 0
keff =
γ P0 A
V 0
2
=
γ P0 A
mV 0
4.
(b)
5.
(b) : Potential is zero on each capacitor. Ten, net
charge value (Q1 + Q2) must be zero, i.e. magnitude of total charge must be zero. Now, Q1 = C 1V 1, Q2 = C 2V 2
As |Q1| = |Q2| \ 120C 1 = 200C 2 ⇒ 3C 1 = 5C 2 6. (d) : Te increase in length of the rod, due to rise in temperature in the absence of walls is DL = LaDq \ DL = (1000) (10 –4) (20) = 2 mm But the rod can expand upto 1001 mm only. \ Compression, l = 1 mm l 1 = 108 Pa So, stress developed = Y = 1011 1000 L 7. (b) : At maximum height velocity v = 0 We know that v = u + at , hence 0 = u – gT ⇒ u = gT u When v = , then 2
u 2
=
u − gt ⇒ gt =
Hence, at t = 8.
T 2
, it
u
=
2
gT 2
⇒
t =
T 2
acquires velocity
u 2
(d) : Given, e = e1 sin wt + e2 cos wt
rms value of alternating voltage, i.e. ε
1 rms
=
similarly
ε1
and
2 ε ε
erms
ε2
2 rms
=
2
2
2 ε1
Using phasor diagram, Net resultant root mean square of voltage is given by 2
ε rms =
9.
2
2
ε1 + ε 2
2
(a) : Net force on M due to M and M , 2 M and 2 M ,
4 M and 4 M , 5 M and 5 M and 7 M and 7 M is zero. Te force due to 3 M is M 3 M 3GM 2 F =G 2 = along x -axis d d 2 1 dQ 10. (d) : Molar specific heat for a gas, C n dT It is possible to obtain any set to values for DQ and DT by proper selection of process. As, DQ = 0 is adiabtic process \ C = 0 and DT = 0 is isothermal process \ C=∞ =
⋅
11. (b) : Te boat follows the shortest path. It is
perpendicular to the bank of river. PHYSICS FOR YOU
|
MARCH ‘18
45
For the resultant motion to be upwards. v cos q + v C = 0 v 25 / 2 1 cos q = − C = − =− ⇒ θ = 120 ° v 25 2 16. (d) : When conductor moves a distance dx (in direction of v ) in time dt , the component of displacement dx moved perpendicular to field is dx sin q. And hence, induced emf dx sin θ = − Blv sin θ = −l(v × B) ε = − Bl dt or ε = l( B × v )
\ Velocity of boat Distance =
1 km =
Time =
1 × 60 km 15 h
15 min =
4 km h
−1
In the right angled triangle ABC , AC 2 = AB2 + CB2 (5)2 = (4)2 + (CB)2 or (CB)2 = 9 \ CB = 3 km h–1 = velocity of river a 12. (b) : As cos θ = 2a \ N sin 60° = mg 2
N cos 60° = m \ tan 60° =
ω
⇒
q = 60°
2 2
q
a
2 g
N
2
⇒ ω
=
mw2 mg
2 g
a 3 a 13. (a) : As, rms value of current, I rms = V rms /Z So, net impedance across LCR circuit,
=
=
(100)2 + 100 × π ×
2
(100)
vT =
2 R 2 (ρ − σ) g 9
η
where h is the coefficient of viscosity of the liquid. L − XC )2
+ ( ω
1 ( X C ) −
17. (d) : Te terminal velocity of the spherical body of radius R, density r falling through a liquid of density s
is given by
ω
Z = R2 + ( XL − XC )2
a 2
∴ vT = 1
2 R12 (ρ1 − σ ) g 9η
and vT 2
According to the given problem,
2
\
2
2
R1 (ρ1 − σ ) = R2 (ρ2 − σ ) or
=
2 R22 (ρ2 − σ) g 9η
vT
=
1
2 R1
vT
2
=
ρ2 − σ
For the resultant motion to be upwards. v cos q + v C = 0 v 25 / 2 1 cos q = − C = − =− ⇒ θ = 120 ° v 25 2 16. (d) : When conductor moves a distance dx (in direction of v ) in time dt , the component of displacement dx moved perpendicular to field is dx sin q. And hence, induced emf dx sin θ = − Blv sin θ = −l(v × B) ε = − Bl dt or ε = l( B × v )
\ Velocity of boat Distance =
1 km =
Time
15 min
1 × 60 km
=
=
15 h
4 km h
−1
In the right angled triangle ABC , AC 2 = AB2 + CB2 (5)2 = (4)2 + (CB)2 or (CB)2 = 9 \ CB = 3 km h–1 = velocity of river
a 12. (b) : As cos θ = 2a \ N sin 60° = mg 2
N cos 60° = m \ tan 60° =
ω
q = 60°
⇒
N
q
a
2
2 g 2
2
⇒ ω
=
mw2
a 2
mg
2 g
is given by
a 3 a 13. (a) : As, rms value of current, I rms = V rms /Z So, net impedance across LCR circuit, ω
Z= R
=
2
+(
2
X L − XC )
2
(100)
=
(100)2 + 100 × π ×
2
η
∴ vT = 1
2
2
R1
2
(8 × 103
−
2 × 103 ) 2 × 103 )
2
1
F=
dU a dx
−
=
2
−
vT
2
2 R1 2
R2
=
ρ2 − σ ρ1 − σ
...(i) ...(ii)
=
9 6
=
dU dx
3 2
=
or
2bx
R1
3 =
2
R2
−
a
Now, k =
dF dx x =
\ w=
k m
a 2b a 2b
⇒
=
2b
u =
1 2π
k m
=
1
2b
2π
m
19. (a) : As the trains are moving in the same direction.
So the initial relative speed (v 1 – v 2) and by applying retardation final relative speed becomes zero. v v From v = u – at = 0 ⇒ 0 = (v 1 – v 2) – at ⇒ t 1 2 a
m1ρ2 − m2ρ1
−
m1 − m2
=
–1
15. (a) : v C = 45 km h =
25
ms
20. (b)
−1
2
21. (d) : For a faulty thermometer, if reading is y
whereas correct reading is x then, x (LFP)correct scale y (LFP)incorrect sccale
v
−
q
v C 48
=
1
2bx
When F = 0, x =
m2
or m2r – m2r1 = m1r – m1r2 or r(m1 – m2) = m1r2 – m2r1 ρ=
−
18. (a) : U = bx – ax ⇒
m1
9η
vT
2
(11 × 103
=
Divide (i) by (ii), we get
or
=
R1 (ρ1 − σ ) = R2 (ρ2 − σ ) or
R2
14. (d) : Let V be the volume of the object. \ V (r – r1) g = m1 g and V (r – r2) g = m2 g
ρ − ρ2
and vT 2
2 R22 (ρ2 − σ) g
Substituting the given values, we get
⇒ X C = 100 W
X C 100 = =1 R 100 ⇒ q = tan–1(1) or q = 45°
=
9η
According to the given problem, \
As, tan q =
ρ − ρ1
2 R12 (ρ1 − σ ) g
2
2
Power factor, cos q =
9
where h is the coefficient of viscosity of the liquid. L − XC )
220 = (100) 2 + (100) − ( X ) 2 ( C ) 2.2
2 R 2 (ρ − σ) g
vT =
+ ( ω
1 ( X C ) − π
17. (d) : Te terminal velocity of the spherical body of radius R, density r falling through a liquid of density s
PHYSICS FOR YOU
|
MARCH ‘18
(UFP
−
−
LFP)correct scale
=
(UFP-LFP)incorrect scale
Tat means we can view them as two different scales. So,
60 ( 10) −
50
−
=
T ( 10) −
−
0
⇒ T = 130°C
100 0
−
−
22. (d) : When
a lif accelerates upwards with acceleration a, the apparent weight will be W app = m( g + a) = 60 (10 + 2) = 720 N 23.
dU B2 (b) : Energy density dV = 2µ0 2 B
R
∫
i.e., U =
2
2
16 π
24. (a) : States
1 and 2 are at the same temperature. Also states 4 and 5 are at same temperature. As v p is more at higher temperature and same at all states at equal temperature. v p at 3 > vp at 1 = vp at 2 > vp at 4 = vp at 5
µ1 4
(given), T 1 = T 2
\ m2v 22 = m1v 12 ⇒ v2
µ1
=
µ2
v1 =
4 v1 = 2v1
or v 2 = 20 cm s–1 26. (c) : Here, temperature o source, T 1 = 227°C = 500 K emperature o sink, T 2 = 127°C = 400 K Heat absorbed rom the source, Q1 = 6 × 104 cal Heat rejected to the sink Q2 = ? As
Q1 Q2
=
T 1 T 2
4
cal
Q2
or
Q2
4 =
5
×
\ W = Q1 – Q2 27. (b) : Steady
4
cal
= 6 × 10
= 4
4 .8 × 10
4
M
2
+ mr 2 3 5
circular orbit close
dU Energy emitted = dt Time
Hence, E = P Dt Also, E = number o photons × energy o 1 photon So, number o photons N =
E P ∆t = h υ hυ
\ Number o photons emitted by source per unit time N P n= = ∆t hυ υ n As ub > ur and r = b > 1 nb υr ⇒ nr > nb 1
32. (d) : Mean ree path, λ =
πd
2
n where n is the number density and d is the diameter o 2
the molecule. As PV = NkBT
∴
n=
1
P πd kBT
N P = V kBT kBT
=
2
2
πd P 2
or d 2 =
kBT 2πP λ
Here, kB = 1.4 × 10–23 J K–1, l = 2.1 × 10–7 m, At SP, T = 0°C = 273 K, P = 1 atm = 1.01 × 105 N m–2
cal 4
5
mr 2 + mL2 = L2 m +
to the earth’s suace is given by vo = gR We know that, ve = 2 gR Additional velocity required to escape = v e – v o = ( 2 − 1) gR
2
400 K 6 × 10
3
2
+
30. (d) : Te speed o spaceship in a
500 K =
ML2
Tus, λ =
Substituting the given values, we get 6 × 10
(I AA′)system =
31. (b) : As power, P =
µ 0 I
25. (b) : m2 =
12
2
L ML2 + M = 2 3
5
dV
µ2 I 2r 2 2πrdr R µ0 I 2 1 3 U = ∫ 0 2 4 = ∫ 0 × 4 × r dr 0 4π 4 π R 2µ 0 R R µ 0 I 2 r 4 µ 0 I 2 R 4 = × = × × −0 4πR 4 4 0 4π R 4 4
∴
29. (b) : (I AA′)rod =
ML2
2
R
U =
but B loses heat much rapidly than A. So, all o temperature o B is more than that o A in same time. Hence, e A > eB. As good emitters are also good absorbers so, a A > aB.
(I AA′)sphere = mr 2 + mL2
µ0 µ Ir Magnetic field in a straight wire, i.e., B = 0 2 2πR dV = 2 p r dr where, r is a radius o circular cross-section o wire. 0
28. (c) : Initially, both A and B are at same temperature
4
cal − 4.8 × 10 cal = 1.2 × 10 cal
state temperature o resistor depends not only on the power but also on surace area, emissivity, etc. Bulb is o less power but acquires more temperature (∼ 1600°C).
\
2
d
1.4 × 10
=
−23
× 273
2 × 3.14 × 1.01 × 10
5
× 2 .1 × 10
−7
= 4.06 × 10–20 m2 or d = 2.01 × 10–10 m ≈ 2.01 Å 33. (a)
34. (a) : F ∝ (v × B) = k(aD − dA)
PHYSICS FOR YOU
|
MARCH ‘18
49
35. (d) : Given : a = 0.96 0.96 α ∴ β= = = 24 1− α 1 − 0.96
Eliminating v i from there, we get
Collector current,
Now, de-Broglie wavelength of proton and a-particle h h are λ p = and λ α = m p v p mα v α
I C
=
Voltage drop across RC Resistance RC
Also
β=
I C
∴
I B
I B
=
4v p – 4v a = 4v a + v p or vα 0.5 =
I C
β
=
500
= 1 × 10
1 mA 24
=
1 24
−3
A
=
1 mA
mA
\
36. (a) : Here, radius of the wheel, R = 0.7 m 120 =
60
rps = 2 rps
=
Magnetic field, B = H E = 0.8 G = 0.8 × 10–4 Induced emf across the ends of a spoke, 2 –4 2 –4 e = BpR u = 0.8 × 10 × 3.14 ×(0.7) × 2 = 2.46 × 10 V As all the 30 spokes are connected in parallel, therefore emf induced across each spoke is same. Tus, the induced emf between the axle and the rim of the wheel is same as across any spoke i.e. 2.46 × 10–4 V. Te number of spokes is immaterial. 38. (c) : Relative density of stone, Density of stone Density of water
=
ρstone
...(i)
ρ water
Let V be volume of the stone. Weight of the stone, W = V rstone g Buoyant force (upthrust) on the stone due to water, B = V rwater g Net downward force on the stone, F = W – B = V rstone g – V rwater g
ρwater = V ρstone g 1 − ρ stone = V ρstone g 1 − = mg 1 −
K 1
K 1
(Using (i))
2KQq (a
2
+
2
x )
⋅
x
F
a2 + x 2
Q +
2KQq
and frequency =
2π
q
+
Q
q
2F sinq
Qq
1
2πε0ma
3
in a n-type semiconductor phosphorus atoms are not isolated and there are enough free charge carriers in a semiconductor at room temperature. 42. (b) : Loss of energy is maximum when collision is
inelastic. Maximum energy loss =
mM 2 u 2 ( M + m) 1
43. (b) 44. (b) : Radius of path of a-particle in region of a
uniform perpendicular magnetic field is 2 Km 2eB
As a-particles have only discreate and quantised energy values (K ), so, they follow circular path of fixed radii.
(Q Mass of stone, m = V rstone)
1 mg 1 − K = g 1 − 1 F a= = K
a-Source
B
m
39. (d) : Let v i be initial speed of a-particle and v a and v p are final speeds of a-particle and proton. Ten, 4m p . v i = 4m pv a + m pv p [Q ma ≈ 4m p]
Collision being elastic, v i – 0 = v p – v a 50
m v 1 8 2 = p ⋅ p = × = mα v α 4 3 3
⋅ x a3 (Q x is small as compare to a) F 2 \ Acceleration, a = –w x = m =
r =
Acceleration of the stone,
m
m p v p h × mα v α h
v p
41. (d) : Assertion is incorrect as in a p-n junction or
37. (c)
K =
λ p
=
8
40. (a) : Restoring force = 2 F sin q
Frequency of rotation of the wheel, u = 120 rpm
λ α
3 =
PHYSICS FOR YOU
|
MARCH ‘18
45. (b) : As density of water is twice the density of ball, \ upthrust = weight of water displaced
= 2 × weight of ball
PHYSICS FOR YOU
|
MARCH ‘18
51
– 2mg 2mg = = –mg –mg \ Net upward force on the ball = mg – therefore, inside water a = – g . Hence ball will stop at the same depth as height from which it started. 46. (a) : Explosion is due to internal forces. As no
external force is involved, the vertical downward motion of centre of mass is not affected.
54. (d) : In an elastic collision, linear momentum and
kinetic energy, both, are always conserved. Energy transformation occurs. Due to friction, mechanical energy may be converted into heat energy. 55. (a) : Since the initial phase difference between
opposite order, means for high input it give low output and for low input it gives high output. For this reason NO gate is known as invertor circuit.
the two waves coming from different violins changes, therefore, the waves produced by two different violins does not interfere because two waves interfere only when the phase difference between them remain constant throughout.
48. (b) : Te apparent weight of a body in an elevator
56. (a) : If m is the refractive index of glass with respect
47. (a) : A NO gate puts the input condition in the
moving with downward acceleration a = m( g – – a). 49. (d) 50. (c) : .V.
signals are of high frequencies (100 MHz to 200 MHz). Tey cannot be reflected to earth by ionosphere, whereas skywaves are reflected from ionosphere. Hence sky waves are not used for the transmission of .V. signals. Te waves which are reflected from ionosphere (frequency in range 2 MHz to 20 MHz) are called sky waves or ionospheric propagation. 51. (a) : As the stream falls down, its speed will increase
and cross-section area will decrease, it will become narrow. Similarly as the stream will go up, speed will decrease and cross-section area will increase, it will become broader. 52. (c) : Diamagnetic substance are composed of atom
which have no net magnetic moment (i.e. (i.e.,, all the orbital shells are filled and there are no unpaired electrons). When exposed to a field, a negative magnetization is produced and thus the susceptibility is negative. Behaviour of diamagnetic material is that the susceptibility is temperature temperature independent. M
c = constant
Z eff = (Z (Z – – 1) and so for n = 2 → n = 1 transition, t ransition,
52
ke2 2a0
×
3 4
2
( Z − 1)
magnetic field or push it into the magnetic field, magnetic flux linked with the plate changes. As a result of this eddy currents are produced in the plate which oppose its motion (according to Lenz’s law). 58. (b) : Helium is a monatomic gas, while oxygen is
diatomic. Terefore, the heat given to helium will be totally used up in increasing the translational kinetic energy of its molecules; whereas the heat given to oxygen will be used up in increasing the translational kinetic energy of the molecule and also in increasing the kinetic energy of rotation and vibration. Hence there will be a greater rise in the temperature temperature of helium. 59. (d)
60. (b)
=
PHYSICS FOR YOU
T
•
Suravi Sahitya Kutir- Dibrugarh
• • •
Prime Books And Periodicals- Guwahati Ph: 0361-2733065; Mob: 9177439700
• • • •
Manika Books- Guwahati Mob: 8876881519, 9864344372
Ph: 0373-2322085; Mob: 9435002641, 9435031727 Book Emporium- Guwahati Ph: 0361-2635094; Mob: 9864057226 Book Land Publisher & Distributors- Guwahati Ph: 2520617, 263772, 2511617; Mob: 9864267457, 9864091521
− ke2 (Z − 1)2 −ke2 ( Z − 1)2 = − 2 2 2a0 2a0 1 Z =
57. (a) : When we pull a copper plate out of the
ASSAM at
53. (a) : Due to partial screening,
α
Because, for minimum deviation i = e, hence r = = r ′.
c
M = cH H Slope = c
EK
to air, then according to Snell’s law for the refraction of light, sin i µ= (At the point of incidence) sin r sin e and, µ = (At the point of emergence) sin r ′
|
13.6 ×
3 4
MARCH ‘18
2
( Z − 1)
Sarkar Book Stall- Nagaon Mob: 9954104236, Book World- Silchar Mob: 9401252440, Haque Book Stall- Tezpur Ph: 03712-231129; Mob: 9435081257
FULL LENGTH PRACTICE PAPER Exam date : 16th to 31st May 2018
SECTION-I (PHYSICS) 1.
In the circuit as shown in the figure, the value of each resistance is R, then equivalent resistance between points A and B is A
R
2.
3.
In Young’s double slit experiment, the fringe width is b. If the entire arrangement is placed in a liquid of refractive index m, the fringe width b ecomes β β β (a) mb (b) (c) (d) µ−1 µ+1 µ
7.
Voltage of modulating wave of 5 V with 10 MHz frequency was superimposed on carrier wave of frequency 20 MHz and voltage 20 V then the modulation index is (a) 0.25 (b) 1.25 (c) 2.43 (d) 64.0
(b)
2 3
8.
A capacitor of capacitance C is is charged to a potential difference V 0 and is then discharged through a resistance R. Te discharge current gradually decreases, with a straight line 1 corresponding to this process, as shown in figure where time is along x -axis -axis and the logarithm of the current on y -axis. -axis. Later on, one of the three parameters V 0, R or C , is changed (keeping (keeping the other two unchanged) unchanged) in such a manner that the ln I versus versus t dependence dependence is represented represented by the straight line 2. Which option correctly ln I represents the change? (a) V 0 is decreased (b) R is decreased 1 (c) R is increased 2 t (d) C is is decreased. O
9.
At 0 K, the quantity which is zero for a gas is (a) Potential energy (b) Kinetic energy (c) Internal energy (d) Vibrational energy
10.
An isolated and charged spherical soap bubble has a radius r and the pressure inside is equal to
R
R
R
(c)
1 2
R
(d) R
Te maximum heights of a projectile for projection angles q and (90 – q) are H 1 and H 2. R is the range in each case. Te relation between R, H 1 and H 2 can be expressed as 2
2
(a) R = H1H 2
(b) R = H 1 + H 2
(c) R = H 1 + H 2
(d) R = 4 H1H 2
2
wo spheres of radii R1 and R2 have densities r1 and r2 and specific heats C 1 and C 2 respectively. If they are heated to the same temperature, the ratio of their rates of cooling will be (a) (c)
4.
7
R
6.
R
R
5
If the radius of earth shrinks by 1.5% (mass remaining same), then the value of acceleration due to gravity changes by (a) 1% (b) 2% (c) 3% (d) 4%
B R
(a)
5.
R2 ρ1 C 1 R1 ρ2 C 2 R1 ρ2 C 2 R2 ρ1 C 1
(b)
(d)
R1 ρ1 C 1 R2 ρ2 C 2 R2 ρ2 C 2 R1 ρ1 C 1
An electron orbiting around a nucleus has angular momentum L. Te magnetic field produced by the electron at the centre of the orbit can be expressed as (a) B =
µ 0e L 3 8 πmr
(b) B =
µ0e L 3 4 πmr
(c) B =
µ0e L πmr 3
(d) B =
e L 4 πε mr 3 0
PHYSICS FOR YOU
|
MARCH ‘18
53
atmospheric pressure. If T is the surface tension of soap solution, then charge on drop is
11.
(a)
2
(c)
8 πr
13.
ε0
(b)
rT ε0
(d)
8 πr
2rT ε0
8 πr
2rT ε
P V ( γ − 1)
(d)
V P ( γ − 1)
A transverse wave is described by the equation x y = y 0 sin 2 π υt − . Te maximum particle
wo tuning forks A and B vibrating together produce 5 beats per second. Frequency of B is 512 Hz. It is seen that if one arm of A is filed, then the number of beats increase. Frequency of A will be (a) 502 Hz (b) 507 Hz (c) 522 Hz (d) 517 Hz Te kinetic energy of a particle moving along a circle of radius R is K.E. = bx 2 where x is the distance covered. Te force acting on the particle is PHYSICS FOR YOU
|
MARCH ‘18
(2)
(3)
(4) (b) 2
(c) 3
(d) 4
19.
A photosensitive metallic surface has work function, huo. If photons of energy 2hu0 fall on this surface, the electron come out with a maximum velocity of 4 × 106 m s–1. When the photon energy is increased to 5hu0, then maximum velocity of photo electron will be (a) 2 × 107 m s–1 (b) 2 × 107 m s–1 (c) 8 × 10 5 m s–1 (d) 8 × 106 m s–1
20.
Te potential energy of a mass m is given by the following relation U = E0 for 0 ≤ x ≤ 1 = 0 for x > 1 If l1 and l2 are the de-broglie wavelengths of the mass in the region 0 ≤ x ≤ 1 and x > 1 respectively and the total energy be 2E0 then find the value of
λ
B are two identical spherical charged bodies which repel each other with force F , kept at a finite distance. A third uncharged sphere C of the same size is brought in contact with sphere B and removed. It is then kept at mid-point of A and B. Find the magnitude of force on C . (a) F /2 (b) F /8 (c) F (d) zero
(1)
(a) 1
14. A and
54
R2
Which of the following figure is a correct representation of deviation and dispersion of light by a prism ?
2
16.
(d) 2bxR2
x 2
18.
velocity is four times the wave velocity if πy (a) λ = 0 (b) l = 2 p y 0 4 πy 0 (c) l = p y 0 (d) λ =
15.
(c) 2bx 2R
2βx 1 +
A uniform chain of length l is placed on a rough table with length l /n (where n > 1), hanging over the edge. If the chain just begins to slide off the table by itself from this position, the coefficient of friction between the chain and the table is (a) 1/n (b) 1/(n – 1) (c) 1/(n + 1) (d) (n – 1) (n + 1)
According to zeroth law of thermodynamics, if the systems in contact are in thermal equilibrium, (a) the temperature of two systems is zero (b) one system provides heat to the other system (c) one system absorbs heat from the other system (d) no heat flows between them.
(b)
17.
0
If the ratio of specific heat of a gas at constant pressure to that at constant volume is g , the change in internal energy of a gas, when the volume changes from V to 2V at constant pressure P is (V − 1) PV (a) (b) ( γ − 1) PV (c)
12.
2rT
(a) 2bx (R2 – x 2)
λ 1 λ 2
(a) 21.
? 2
(b)
1 2
(c) 2
A particle is projected from the inclined plane at angle 37° with the inclined plane in upward direction with speed 10 m s–1. Te
(d)
1 2
angle o inclined plane with horizontal horizontal is 53°. Ten the maximum height attained by the particle rom the incline plane will be (a) 3 m (b) 4 m (c) 5 m (d) zero 22.
Te ollowing circuit is in the state o resonance. Which o the ollowing statement is correct? i
(a) (b) (c) (d) 23.
C
R 27.
A large number o water droplets, each o radius r , combine to have a drop o radius R. I the surace tension is T and and mechanical equivalent o heat is J is J , the rise in heat energy per unit volume will be 3T (a) 2T (b) rJ rJ
1 − 1 (c) r R J
25.
L
1 − 1 (d) J r R 2T
A cylinder with a fixed volume contains hydrogen gas H2 at 25 K, then we add heat to the gas at constant rate until its temperature reaches 500 K. Does the temperature o the gas increases at a constant rate ? (a) Yes. (b) No, it is increasing rapidly at the end o the process. (c) No, it is increasing rapidly at at the beginning o the process. (d) No, but its variable rate can’t be described. A vertical ring o radius r and and resistance R alls vertically. It is in contact with two vertical rails which are joined at the top. Te ring moves without riction and resistance. Tere is a horizontal uniorm magnetic field o magnitude B perpendicular to the plane o the ring and the rails. When the speed o the ring is v , the current in the section PQ is PQ is 8 Brv 2 Brv 4 Brv (a) zero (b) (c) (d) R R R
A long straight straight wire carries a charge charge with with linear density l. A particle o mass m and a charge q is released at a distance r rom rom the wire. Te speed o the particle as it crosses a point at a distance 2r 2r is is qλ ln 2 qλ ln r (a) (b) πm ε0 πm ε0
28.
qλ ln 2 2 πm ε0
qλ ln r 2 πm ε0 Te bob o a simple pendulum o mass m and total energy E will have maximum linear momentum equal to (a) 2mE 2mE (b) mE2 2E (c) 2mE (d) m A tuning ork o length l , thickness t is made o material having Young’s modulus Y and and density r. Frequency u o a tuning ork is given by (Here k is a constant) (c)
Value o i depends upon the value o L, C and and R. Maximum current flows in circuit. Minimum current flows in circuit. Power actor is 0.
3T
24.
26.
(a)
(c)
(d)
kt Y l ρ kt Y
(b)
kt 2
Y
l 2
ρ
kt 2 (d) l
Y
ρ l 2 ρ 29. A particle o mass m = 5 units is moving with a uniorm speed v = 3 2 units in the X the XO OY plane plane along the line y =x = x + + 4. Te magnitude o the angular momentum o the particle about the origin is units ts (a) 60 units (b) 40 2 uni (c) zero (d) 7.5 units 30.
An electron is travelling along the x -direction. -direction. It encounters a magnetic field in the y -direction. -direction. Its subsequent motion will be (a) straight line along the x -direction -direction (b) a circle in the xz plane plane (c) a circle in the yz the yz plane plane (d) a circle in the xy plane. plane.
31.
A beam o light is converging towards a point I on on a screen. A plane parallel plate o glass o thickness t (in the direction o the beam) and reractive index o m is inserted in the path o the beam. Te convergence point is shifed by
away µ
(b) t 1 +
nearer µ
(d) t 1 +
(a) t 1 − (c) t 1 −
1
1
away µ
nearer µ
PHYSICS FOR YOU
|
1
1
MARCH ‘18
55
32.
In a common emitter transistor the base current I B = 2 mA, a = 0.9, then I C = (a) 18 mA (b) 20 mA (c) 22 mA (d) 24 mA
33.
Te time dependence o a physical quantity p is
39.
2
given by p = p0 e–at , where a and p0 are constants and t is is the time. Te constant a (a) is dimensionless (b) has dimensions [ –2] (c) has dimensions [ 2] (d) has dimensions o p 34.
Te moment o inertia o a solid cylinder o mass M and and radius R about a line parallel to the axis o the cylinder out lying on the surace o the cylinder is (a) M (L2 + R2) (b) MR2
L2 R2 (c) M + 12 4 35.
40.
3
(d)
MR
(a) electron density (c) drif velocity
2
1 3
41.
rd
o the velocity o sound in air, the number o waves per second received by the observer will be (a) 300 (b) 192 (c) 384 (d) 200
37.
38.
R22
=
R13
Electric dipole moment o combination shown in the figure is q q a 2 q a a q q a + + (a) 2
(c) (d)
2qa
a
–3q
) qa .
2 +1
PHYSICS FOR YOU
|
MARCH ‘18
III
V
(a) II < IV < III < V < I (b) III < IV < I < V < II (c) V < IV < III < I < II (d) IV < V < I < III < II 42.
Phenol and cyclohexanol can be chemically distinguished by all the given reagents except (a) Br2/H2O (b) anhydrous ZnCl2/conc. HCl (c) neutral FeCl3 (d) metallic sodium.
43.
I l0 is the threshold wavelength or photoelectric emission, l is the wavelength o light alling on the surace o a metal and m is the mass o the electron, then the velocity o ejected electron is given by 1/2
2h (a) (λ0 − λ ) m
1/2
2hc (λ0 − λ ) (b) m
a 1/2
2qa
(
II
IV
R23
540 g o ice at 0°C is mixed with 540 g o water at 80°C. Te final temperature o the mixture is (latent heat o usion o ice is 80 cal g –1 and specific heat capacity o water 1 cal g –1 oC –1) (a) 0°C (b) 40°C (c) 80°C (d) 25°C
(b)
56
=
g 1 (d) g 2
Choose the correct comparison o heat o hydrogenation or the ollowing alkenes : I
wo planets have the same average aver age density and their radii are R1 and R2. I acceleration due to gravity on these planets be g 1 and g 2 respectively, then g 1 R2 g 1 R1 = = (b) (a) g 2 R1 g 2 R2 R12
(b) current density (d) electric field
SECTION-II (CHEMISTRY)
A whistle sends out 256 waves in a second. I the
g 1 (c) g 2
A constant current I is is flowing along the length o a conductor o variable cross-section as shown in the figure. Te quantity which does not depend upon the area o cross-section is I
2
whistle approaches approaches the observer with velocity
36.
A hydrogen atom and a Li++ ion are both in the second excited state. I l H and l Li are their respective electronic angular momenta, and EH and ELi their respective energies, then (a) l H > l Li and |EH| > |ELi| (b) l H = l Li and |EH| < |ELi| (c) l H = l Li and |EH| > |ELi| (d) l H < l Li and |EH|<|ELi|.
a
q
λ − λ (c) 2hc 0 m λ0 λ
1/2
2h 1 1 (d) − m λ0 λ
44.
Te weight of 1 g-equivalent of V 2O5 used in the reaction : Zn + V2O5 ZnO + V is (At. wt. of V = A) A A + 80 (a) (b) 5
(c) 45.
50.
At what temperature will a 5% solution (weight/volume) of glucose develop an osmotic pressure of 7 atm? (a) 33.94 °C (b) 54.76 °C (c) 24.55 °C (d) 47.32 °C
51.
Which of the following ions does not involve pp–d p bonding? (a) SO2– (b) PO43– 3 (c) NO3– (d) XeOF4
52.
An ideal gas with pressure P , volume V and temperature T is expanded isothermally to a volume 2V and and a final pressure is P I. If the same gas is expanded adiabatically to a volume 2V , and the final pressure is P II and the ratio of specific heats for the gas is 1.67, then the ratio of P II/P I is
5
2 A + 80 5
(d)
2 A + 80 10
Which of the following is likely to be the e.m.f. of the cell (chloride based) shown below? Ag(s)|AgCl( K sp)| Cl–(x M)||AgCl(K sp)|Ag(s) x (a) 0.059 log K sp (b) 0.059 log [( x )( )(K sp)] (c) 0.059 log
K sp
x (d) Data is insufficient for calculation. 46.
47.
48.
Which of the following values of stability constant K corresponds to the most unstable complex compound? (a) 1.6 × 10 7 (b) 4.5 × 1014 (c) 2.0 × 10 27 (d) 5.0 × 1033
53.
l l e
c ° E
–1.0
(a) OD (c) OC
(d) F2C
C D
0.553 g of boron-hydrogen compound created a pressure of 0.658 atm in a bulb of 407 mL at 373 K. If the compound has 85.7% boron, what will be the molecular formula of the compound? (Given atomic mass of B = 10.8) (a) B2H6 (b) B3H8 (c) B5H9 (d) B6H14
2
−0.67
Te monomer(s) of dacron is(are)
HOOC
In K
(b) OB (d) none of these
1
(d)
COOH COOH
(c) HOCH2 CH2OH and
B
–0.5
(2)
(b)
54.
O
(b) (2)0.67
0.67
(a) HOCH2 CH2OH and
Given : DG° = –nFEcell ° and DG° = – RT ln lnK Te value of n = 2 will be given by the slope of which line in the given figure? 0.5
1
−0.67 1 (c) − 2
Which of the following mixtures will give a buffer solution when dissolved in 500.00 mL of water? (a) 0.200 mol of aniline and 0.200 mol of HCl HCl (b) 0.200 mol of aniline and 0.400 mol of NaOH (c) 0.200 mol of NaCl NaCl and 0.100 mol of HCl HCl (d) 0.200 mol of aniline and 0.100 mol of HCl
1.0
49.
(a)
55.
COOH
CF2
o avoid the precipitation of hydroxides of Ni 2+, Co2+, Zn2+ and Mn2+ along with those of Fe 3+, Al3+ and Cr3+, the third group solution should be (a) heated with a few drops of conc. HNO3 (b) treated with excess of NH 4Cl (c) concentrated (d) none of these. A fixed mass of a gas is subjected to transformations tr ansformations of state from K to to L to to M to N and and back to K as as shown in the figure. Te pair of isochoric processes among the transformations of state is K L (a) K to to L and L to M (b) L to M and and N to to K P N M (c) L to M and and M to to N V (d) M to to N and and N to to K PHYSICS FOR YOU
|
MARCH ‘18
57
56.
57.
A nuclide o an alkaline earth metal undergoes radioactive decay by emission o a-particle in succession to give the stable nucleus. Te group o the periodic table to which the resulting daughter element would belong is (a) group 4 (b) group 6 (c) group 16 (d) group 14.
63.
Te order o decreasing stability o the given carbanions is (I) (CH3)3C– (II) (CH3)2CH– (III) CH3CH2– (IV) C6H5CH2– (a) I > II > III > IV (b) IV > III > II > I (c) IV > I > II > III (d) I > II > IV > III
64.
In the reaction, H O
Te rate o change o concentration o ( A) or d [ A] k [ A]1 / 3 B is given by reaction: A dt Te hal-lie period o the reaction will be
2 HCHO + CH3MgI → A → B + Mg(OH)I A and B are respectively (a) CH3OMgI and CH3 – OH (b) CH3CH2OMgI and C2H5 – O – C2H5 (c) CH3CH2OMgI and CH3 – CH2 – OH (d) CH3 – CH2 – I and CH3 – CH2 – OH
−
=
2/3
(a)
3[ A0 ]
5/3
3
(b) 2
k
2/3
2/ 3
[ A0 ]
(d) 58.
59.
60.
61.
62.
58
[(2)
2 /3
[(2) 5/3
(2) 3
65.
Propan-1-ol can be prepared rom propene by (a) H2O/H2SO4 (b) Hg(OAc) 2/H2O ollowed by NaBH4 (c) B2H6 ollowed by H2O2 (d) Br2/H2O
66.
Te coagulation value in millimoles per litre o electrolytes used or the coagulation o As2O3 are as given : I. NaCl = 52 II. KCl = 5 III. BaCl2 = 0.69 IV. MgSO 4 = 0.22 Te correct order o their flocculating power is (a) I > II > III > IV (b) I > II > III = IV (c) IV > III > II > I (d) IV = III > II > I
67.
Under two different conditions an element could be made to exist in bcc and fcc arrangements with exactly same interatomic distance. Te ratio o the densities o bcc to fcc arrangements is (a) 1 : 1 (b) 0.919 : 1 (c) 1 : 0.919 (d) 0.2 : 1
68.
Te hormone that controls the contraction o the uterus afer child birth and releases milk rom the mammary glands is (a) oxytocin (b) vasopressin (c) thyroxine (d) adrenaline.
69.
In the balanced equation : H2SO4 + x HI → H2S + y I2 + z H2O the values o x , y and z are (a) 3, 5, 2 (b) 4, 8, 5 (c) 8, 4, 4 (d) 5, 3, 4
70.
From the ollowing equations, what is the heat o a hypothetical reaction, P 2Q ? (i) P R; DH 1 = x (ii) R S; DH 2 = y
− 1]
k
3[ A0 ] 2
− 1]2
(2)
2/3
(c)
2 /3
[(2)
3/2
[ A0 ]
− 1]
k 2/ 3
[(2)
− 1]
k A compound contains atoms X , Y and Z . Te oxidation number o X is +3, Y is +5 and Z is –2. Te possible ormula o the compound is (a) XYZ 2 (b) Y 2( XZ 3)2 (c) X 3(YZ 4)3 (d) X 3(Y 4Z )2 Te standard ree energy change, DG° is related to equilibrium constant, K p as e ∆G° (a) K p = – RT ln DG° (b) K p = RT ∆G° (c) K p = (d) K p = e − ∆G°/ RT RT Which o the ollowing gases does not contribute to greenhouse effect? (a) O3 (b) H2O vapour (c) O2 (d) N2O Te correct order o bond angles o NO2+, NO2 and NO2– is (a) NO2+ < NO2 < NO2– (b) NO2+ = NO2– < NO2 (c) NO2+ > NO2 > NO2– (d) NO2+ > NO2 < NO2– Which o the ollowing organic compounds does not give iodoorm test? (b) CH3CH2CH2CH2OH (a) CH3CH2OH (d) CH3COCH3 (c) (CH3)2CHOH PHYSICS FOR YOU
|
MARCH ‘18
(iii)
1 2
S
Q; DH 3 = z
(a) x + y – 2z (c) x + y + 2z
(b) x + 2 y – 2z (d) x – y + 2z
71. During the ormation o the N2O4 dimer rom two molecules o NO2, the odd electrons, one in each o the nitrogen atoms o the NO 2 molecules, get paired to orm a (a) weak N–N bond, two N–O bonds become equivalent and the other two N–O bonds become non-equivalent. (b) weak N–N bond and all the our N–O bonds become equivalent. (c) weak N–N bond and all the our N–O bonds become non-equivalent. (d) strong N–N bond and all the our N–O bonds become non-equivalent. 72. Chemical ’A’ is used or water sofening to remove temporary hardness. ’A’ reacts with sodium carbonate to generate caustic soda. When CO2 is bubbled through ’ A’ , it turns cloudy. What is ’A’ ? (a) CaCO3 (b) CaO (c) Ca(OH)2 (d) Ca(HCO3)2 73. Te numbers o monochloro derivatives obtained with Cl2 /hv rom the ollowing :
77. I K 1 and K 2 are the ionization constants o +
+
H3N CHRCOOH and H3N CHRCOO– respectively, the pH o the solution at the isoelectric point is (a) pH = pK 1 + pK 2 (b) pH = (pK 1 pK 2)1/2 (c) pH = (pK 1 + pK 2)1/2 (d) pH = (pK 1 + pK 2)/2 78. Which o the ollowing will give maximum number o isomers? (a) [Co(NH3)4Cl2] (b) [Ni(en)(NH3)4]2+ (c) [Ni(C2O4)(en)2] (d) [Cr(SCN)2(NH3)4]2+ 79. An optically active amine ( A) o molecular ormula C4H11N is subjected to Hoffmann’s exhaustive methylation process and ollowing hydrolysis an alkene (B) is produced which upon ozonolysis and subsequent hydrolysis yields ormaldehyde and propanal. Te amine ( A) is (a) CH3 CH CH2CH3 NH2
(b)
CH3
NH
CH
CH3
C2H 5
(c) (d) CH3CH2CH2CH2 NH2 +
1
2
3
are
(a) 1, 1, 1 (c) 2, 2, 1
(b) 1, 2, 1 (d) 2, 2, 2
2NH3( g ); 74. For the reaction : N2( g ) + 3H2( g ) –1 DH = –93.6 kJ mol , the concentration o H2 at equilibrium can be increased by (a) lowering the temperature (b) increasing the volume o the system (c) adding N2 at constant volume (d) all o these. 75. Which o the ollowing sets is named as errous metals? (a) Fe, Ru, Os (b) Fe, Co, Ni (c) Fe, Mn, Cr (d) Fe, Rh, Pt 76. Which o the ollowing structures or a nucleotide is not correct? (a) Cytosine-Ribose-Phosphate (b) Uracil-2-Deoxyribose-Phosphate (c) Uracil-Ribose-Phosphate (d) Tymine-2-Deoxyribose-Phosphate
80. Te maximum concentration o M ions that can be attained in a saturated solution o M 2SO4 at 298 K is (K sp = 1.2 × 10 –5)
(a) 7.0 × 10 –3 M (c) 2.88 × 10 –2 M
(b) 3.46 × 10 –3 M (d) 14.4 × 10–3 M
SECTION-III (ENGLISH AND LOGICAL REASONING) 81. Complete the sentence : Te higher you go, the more difficult it ........ to breathe. (a) is becoming (b) became (c) has become (d) becomes Direction : In the following sentence, choose the word opposite in meaning to the bold word to fill in the blanks. 82. Absolute control o the firm is what he wanted, but he ended up with ......... powers. (a) complex (b) limited (c) little (d) ew Pick up the correct synonym. 83. imid (a) Veteran (c) Cowardly
(b) Fearul (d) Plucky PHYSICS FOR YOU
|
MARCH ‘18
59
Direction: In the given question, out of the four alternatives, choose the one which can be substituted for the given words/sentence. 84. A light sailing boat built especially or racing (a) Yacht (b) Frigate (c) Dinghy (d) Canoe Direction : In the given question, a word has been written in four different ways out of which only one is correctly spelt. Find the correctly spelt word. 85. (a) emperature (c) empereture
(b) amperature (d) emparature Direction : In the following question, find out which part of the sentence has an error. 86. I you are great at ideas but not very good at getting into (a) the nitty gritty / (b) o things and implementing them, then you work on a team / (c) that has someone who can implement. (d) No error Directions (Question 87 – 89) : Read the passage and answer the following questions. he low unit o gas is a real temptation to anyone choosing between gas and electrical processes. But gas-fired processes are ofen less efficient, require more floor space, take longer and produce more variable product quality. he drawbacks negate the savings many businesses believe they make. By contrast, electricity harnesses a unique range o technologies unavailable with gas. And many electric processes are well over 90 percent efficient, so ar less energy is wasted with benefits in terms o products quality and overall cleanliness, it can so ofen be the better and cheaper choice. Isn’t that tempting? 87. Te passage can be described as an. (a) advertisement or electricity and its efficiency (b) extract rom a science journal (c) account o the growth o technology (d) appeal not to use gas. 88. What does the writer mean by ‘variable quality’? (a) Te quality o the products cannot be assessed. (b) Products rom gas-fired processes are inefficient. (c) Te kind o products vary rom time to time. (d) Te quality o the products is not uniorm. 89. “Electricity harnesses a unique range o technologies” - What does the writer mean? Electricity ____________ . (a) has developed new technologies (b) ensures power or electricity and its efficiency 60
PHYSICS FOR YOU
|
MARCH ‘18
(c) depends on new kinds o technology (d) makes use o several technologies Direction : In the following question, choose the alternative which can replace the word printed in bold without changing the meaning of the sentence. 90. A bone got stuck in his gullet. (a) Chest (b) Troat (c) Stomach (d) Molars Directions (Question 91 – 95) : In each of the following questions, a sentence is given with a blank to be filled with an appropriate word. Four alternatives are suggested for each question. Choose the correct alternative. 91. He did not register his ........ to the proposal. (a) disavour (b) dissent (c) deviation (d) divergence 92. Will you, like the ........ gentleman and solider you are, leave at once beore he finds you here? (a) chivalrous (b) luminous (c) barbarous (d) ostentatious 93. In these days o inflation, the cost o consumer goods is ........ (a) climbing (b) raising (c) ascending (d) soaring 94. Te Committee’s appeal to the people or money ........ little response. (a) gained (b) provided (c) evoked (d) provoked 95. Te manager tried hard to ........ his men to return to work beore declaring a lockout. (a) encourage (b) permit (c) motivate (d) persuade 96. A letter series is given with two terms missing as shown by(?). Choose the missing term out o the given alternatives Z,L,X,J,V,H,,F,?,? (a) R,D (b) R,E (c) S,E (d) Q,D 97. A matrix carrying certain numbers is given. Tese numbers ollow a certain trend, row wise or column wise. Find out the missing number accordingly.
(a) 3
(b) 4
(c) 5
7
4
5
8
7
6
3
3
?
29 19
31
(d) 6
98. Which o the ollowing options will continue the same figure as established by problem figures. Problem Figures – + – + – + + + – –
A
B
C
D
E
(a)
–
(c)
+
– +
105.Find
out which of the figures (a), (b), (c) and (d) can be formed from the pieces given in fig. ( X ).
– +
(b)
– +
(d)
Fig.( X ) 99.
Choose the odd numeral pair in the following question. (a) 15 : 46 (b) 12 : 37 (c) 9 : 28 (d) 8 : 33
100. Tere
are five persons P , Q, R, S and T . One is football player, one is chess player and one is hockey player. P and S are unmarried ladies and do not participate in any game. None of the ladies plays chess or football. Tere is a married couple in which T is the husband. Q is the brother of R and is neither a chess player nor a hockey player. Who is the football player? (a) P (b) Q (c) R (d) S
101.In
a certain coding system, ‘816321’ means ‘the brown dog frightened the cat ’; ‘64851’ means ‘the frightened cat ran away ’; ‘7621’ means ‘the cat was brown’; ‘341’ means ‘the dog ran’. What is the code for ‘the dog was frightened ’? (a) 5438 (b) 8263 (c) 8731 (d) none of these
102.Victory : Encouragement : : Failure :
(a) Sadness (c) Anger
(b)
103.If
‘+’ means ‘divided by’, ‘_’ means ‘added to’, ‘×’ means ‘subtracted from’ and ‘ ÷’ means ‘multiplied by’, then what is the value of 24 ÷ 12 – 18 + 9 ? (a) – 25 (b) 0.72 (c) 15.30 (d) 290 the following question, find out which of the answer figures (a), (b), (c) and (d) completes the figure matrix ?
(c)
(d)
SECTION-IV (MATHEMATICS)
chord of the hyperbola 4x 2 – 9 y 2 = 36 is bisected at the point (3, 5). Te distance of the origin from the chord is
106.A
63
(a)
50
(b)
241
13
(c)
19
241
13
(d)
190
107.Triangles are formed with
vertices of a regular polygon of 20 sides. Te probability that no side of the polygon is a side of the triangle, is 25
(a)
57
(b)
108.Let
30 57
(c)
35 57
40
(d)
57
a b c d be such that (a × b) × (c × d ) = 0. Let p1 and p2 be the planes determined by the pairs of vectors, a , b and c , d respectively. Te angle between the planes p1 and p2 is
,
,
(a) 0
(b)
109.Let f (x ) = [x ] sin
(a) Z
,
?
(b) Defeat (d) Frustration
104.In
(a)
π
4
π
[x + 1]
(c)
π
3
(d)
π
2
. Te points of discontinuity are
(b) Z – {0} (c) Z – {1} (d) Z– {–1}
110.Te
normal at any point P (x , y ) on a curve meets the x -axis at N . If OP = PN , where O is the origin, then the curve is a/an (a) circle (b) parabola (c) ellipse (d) none of these
111.wo mappings f : R → R and g : R → R are defined in
0, when x is rational 1, when x is irrational
the following ways : f (x ) =
− 1, when x is rational 0, when x is irrational Ten the value of ( gof )(e) + ( fog )(p) is (a) –1 (b) 1 (c) 0 (d) 2 x 2 −1 b 4 − x f − 1 sin 2b 2 112.If dx = dx then which is 2 2a 4 x − −1 a sin incorrect ? g (x ) =
?
(a)
(b)
(c)
(d)
∫
∫
PHYSICS FOR YOU
|
MARCH ‘18
61
1 (a) f = 2 1 (c) f = 4
2 3
1 3 (b) f = 3 2 2
4
(d) none of these
121.In a triangle ABC , angle A is
greater than angle B. If the measures of angles A and B satisfy the equation 3 sinx – 4sin3x – k = 0, 0 < k < 1, then the measure of angle C is
15
(a)
113.Te
number log2 7 is (a) an integer (c) rational
(b) prime (d) irrational
π
3
−
115.Find
the general solution for |sinq – cos2q| ≥ |sin2q – 3sinq + 3| + 4|1 – sin q|. (a) np; n∈I (b) 2np; n∈I (d) none of these
2
116.Te
number of integers greater than 6000 that can be formed with 3, 5, 6, 7 and 8, where no digit is repeated, is (a) 120 (b) 192 (c) 216 (d) 72 2
(tan–1 2) + cosec2 (cot–1 3) is equal to (a) 1 (b) 5 (c) 10 (d) 15
117.sec
π 118.
∫
2 0
(sin x + cos x )
2
1 + sin 2 x
(a) 0
dx =
(b) 1
119.If tan 25° =
(c) 2
a, then the value of
in terms of a is 2 1+ a (a) 2 a −1 2 1−a (c) 2 1+ a
(b) (d)
1+
(d) 3 tan 205° − tan 115 ° tan 245° + tan 335°
a2
1−a
2
a2
−1
1+
a2
b are roots of ax 2 + 2bx + c = 0 and g , d are the roots of px 2 + 2qx + r = 0. If a, b, g , d are in A.P., b2 − ac
120.a,
then (a) (c) 62
q a2
p
2
2
−
pr
equals
(b)
c2
2
(d) none of these
r 2 PHYSICS FOR YOU
|
123.If f (x ) =
MARCH ‘18
(c)
3
(d)
5π 6
x , x ≠ 1, then ( fofo.....of )(x ) is equal to x − 1 19 times
19
x (a) x − 1
(c) 124.If
19 x
x − 1
x (b) x − 1
(d) x
dy x − y , y (1) = 1, then ( y(0))2 = = dx x + y
(a) 1
(b) 2
(c) 3
(d) 4
a and b be the roots of the equation x 2 + x + 1 = 0. Te equation whose roots are a19, b7 is (a) x 2 – x – 1 = 0 (b) x 2 – x + 1 = 0 (c) x 2 + x – 1 = 0 (d) x 2 + x + 1 = 0
125. Let
126. Te
value of n ∈I for which the function sin nx has 4p as its period, is f (x ) = sin(x / n) (a) 2 (b) 3 (c) 4 (d) 5
127. Tere
are 3 copies each of 4 different books. Te number of ways they can be arranged in a shelf is (a) 369600 (b) 400400 (c) 420600 (d) 440720
128. Line
L has intercepts a and b on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q, then
(a) a2 + b2 = p2 + q2 (b) (c) a2 + p2 = b2 + q2 (d)
b2 q
2
number of integral values of k for which the equation 7 cos x + 5 sin x = 2k + 1 has a solution is (a) 4 (b) 8 (c) 12 (d) 10
=
(c) (4n + 1) ; n ∈I
(b)
122.Te
2 1 y dy determines 114.Te differential equation dx y a family of circles with (a) variable radii and a fixed centre (0, 1) (b) variable radii and a fixed centre (0, –1) (c) fixed radius 1 and variable centres along the x -axis (d) fixed radius 1 and variable centres along the y -axis
π
2π
π
m
129. The
sum
1
a
2
1
a2
+
+
10 20
1 2
b
1
p2
1
=
p
=
2
1
b2
+
+
1
q2 1
q2
p
∑= i m − i (where q = 0 if p < q) i 0
is maximum where m is (a) 5 (b) 10 (c) 15
(d) 20
130. If
−3
cosq =
5
and
cosecθ + cot θ sec θ − tan θ
(a) 1/6
p
<
3π
<
q
2
, then the value of
137. If
and are the direction cosines of two rays OP and OQ enclosing an angle q, then the d.c. of the bisector of ∠POQ are l +l m + m2 n1 + n2 , (a) < 1 2 , 1 >
is
(b) 1/7
(c) 1/5
(d) 1/2
2 cos (θ / 2) 2 cos (θ / 2) 2 cos (θ / 2)
2
131. If xy = m
– 9 be a rectangular hyperbola whose branches lie only in the second and fourth quadrant, then (a) |m| ≥ 3 (b) |m| < 3 (c) m ∈ R – {|m|} (d) none of these P be the product and R be the sum of the reciprocals of n terms of a G.P. Ten RnP 2 : Sn is equal to (a) (first term) 2 : (common ratio) n (b) 1 : 1 (c) (common ratio) n : 1 (d) None of these
(b)
<
(c)
<
(d)
<
l1 + l2
133. Let
l1 − l2
π
3
and c is
π
6
, then the value of
β1
β2
β3 equals
γ1 n
| a | | b | (a) 2×3
140.
γ 3
(c) 3 | b | | c | 2
(c)
∫
,
(a) (b) (c) (d)
dx x − 3 x
(c) 3 | a |2 (d) 4 | a |2
=
2x 1/2 + 3x 1/3 + x 1/6 + ln x + C 2x 1/2 + 3x 1/3 + 6x 1/6 + 6ln x + C 2x 1/2 + 3x 1/3 + 6 x 1/6 + 6ln |x 1/6 – 1| + C none of these
136. Te
first 12 letters of English alphabet are written in a row at random. Te probability that there are exactly four letters in between A and B is (a)
5 66
(b)
1 22
(c)
7 66
(d)
+
y2 )dy ( y ≠ 0) is x y
(b) e = x + C (d) e y = x + C ≤
x 1
≤
3,
x
⋅
3
(ln 2)
+
C
x
(b) 22 ⋅ (ln 2)2 + C
x
2
3
(ln 2)
+
C
(d) none of these
any discrete series when all values are not same, the relation between M.D. about mean and S.D. is (a) M.D. = S.D. (b) M.D. ≥ S.D. (c) M.D. < S.D. (d) M.D. ≤ S.D.
135.
x y
141.In
>
⋅ 22 ⋅ 2 x dx =
22
(d) none of these
(a) 2 | a |2 (b) | a |2
x
(a) 22
any vector a prove that 2 | a × i | + | a × j |2 + | a × k |2 is equal to
∫
2
2 2
n
n1 − n2
,
of the differential equation
2 x
n/2
| c | | b | 3 × 2
134. For
γ2
(b)
n
α3
>
Z = 10x 1 + 25x 2, subject to 0 0 ≤ x 2 ≤3, x 1 + x 2 ≤ 5. (a) 80 at (3, 2) (b) 75 at (0, 3) (c) 30 at (3, 0) (d) 95 at (2, 3)
α2
n1 − n2
,
139. Maximize
with the plane of b & c and angle between b α1
>
2 sin (θ / 2) 2 sin (θ / 2) 2 sin (θ / 2)
(a) e = y + C (c) ex + y = C
m1 − m2
,
x y
c = γ 1 i + γ 2 j + γ 3 k and | a |= 2 2 makes the angle
m1 − m2
,
l1 − l2
ye dx = ( xe
a = α1 i + α2 j + α3 k, b = β1 i + β2 j + β3 k,
n1 + n2
,
2 cos (θ / 2) 2 cos (θ / 2) 2 cos (θ / 2)
138. Solution x y
m1 + m2
,
2 sin (θ / 2) 2 sin (θ / 2) 2 sin (θ / 2)
132. Let S be the sum,
1 11
142.Te relation S defined on the set N × N by (a, b) S(c, d )
a + d = b + c is an (a) equivalence relation (b) reflexive but not symmetric (c) only transitive (d) only symmetric ⇒
all n ∈ N , 41n – 14n is a multiple of (a) 26 (b) 27 (c) 25 (d) none of these
143. For
144. If
x
y
∫ a f (t )dt − ∫ g (t )dt = b, then the value of 0
dy at (x0 , y 0 ), is dx PHYSICS FOR YOU
|
MARCH ‘18
63
f (x 0 )
(a)
g ( y 0 )
(b)
(c) g (x 0) – f ( y 0) 145.Evaluate
g (x 0 ) f ( y 0 )
SOLUTIONS
(b) : Te simplified circuit is given here.
1.
(d) f (x 0) – g ( y 0)
tan x
∫ sin x ⋅ cos x dx
(a)
3
tan x + c
(b)
tan
(c)
2
tan x + c
(d)
tan x
x +c +
2
c 2
146.Te curve described parametrically by x = t + t + 1,
y = t 2 – t + 1 represents (a) a pair of straight lines (b) an ellipse (c) a parabola (d) a hyperbola.
3
Since R′ and R′′ are connected in series, the equivalent resistance between A and B is given by R R 2R Req = R′ + R′′
147. Which of the following is/are true?
(i) Te principal value of cos–1
3 2
is
π
. π
.
4
(iii) Te principal value of tan–1 ( − 3 ) is
−π
H1H 2 =
equal to
2
+
2
=1
(a) (b) (c) (d)
2
is always touching the circle
\
2
4.
c2 (a4 y 2 + b4 x 2) = a4 b4 c2 (a4 x 2 + b4 y 2) = a4 b4 c2 (a4 y 2 + b4 x 2 – a3 b3) = a4 b4 none of these
x →
π
4
(a) e 64
(b) e PHYSICS FOR YOU
(c) e |
MARCH ‘18
=
u 4 (4 sin2 θ cos2 θ) 2
=
R2 16
16 g
⇒
R = 4 H1H 2
(d) : By Stefan’s law, rate of heat emission,
∆T = ρVC ∆T = ρ 4 πR3C ∆T = mC dt dt 3 dt 4 dT πR13ρ1C 1 2 3 (dT / dt )1 R2 ρ2 C 2 dt 1 R1 = 2 ⇒ = 4 dT R2 (dT / dt )2 R1 ρ1 C 1 3 πR2ρ2C 2 dt 2 3 (b) : Magnetic field at the centre of current carrying circular
given by –1
3
µ0 I
2r Magnetic moment m of electron revolving in circular orbit is
equals 2
2
coil, B =
tan 2x
150. lim (tan x )
=
Rate of heat loss is given by
a b x + y = c , then locus of pole is 2
3
dQ 4 4 = Aσ (T − T ) 0 dt Since T , T 0 and s are same for the two bodies , hence ∆Q 2 dt 1 R1 = ∆Q R22 dt 2 dQ dt
(b) {–2, –3} (d) {1, 2} y 2
+
u2 sin2 θ u2 cos2 θ × 2 g 2 g
u 4 sin2 θ cos2 θ
3.
0 2 5 x − 1 x 2 − 2 3 4 1 is 0, then x is A = x − 3 1 −1 −2 2 2 x − 6 0 4
x 2
3
4 g
148. If the trace of the matrix
149. If polar of
=
.
3
(b) (ii), (iii) (d) (i), (ii), (iii)
(a) {2, 3} (c) {–3, 2}
=
u2 sin2 (90 − θ) u2 cos2 θ u2 sin2 θ = 2. (d) : H 1 = , H 2 = 2 g 2 g 2 g 2 u sin 2θ R= g
6
(ii) Te principal value of cosec–1 (2) is
(a) (i), (ii) (c) (i), (iii)
In section I, three resistors each of resistance R are connected in parallel. Te equivalent resistance R′ is given by R 1 1 1 1 = + + or R ′ = R′ R R R 3 Similarly in section II, three resistors each of resistance R are connected in parallel. Te equivalent resistance R′′ is given by R R′′ = 3 3
(d) e
–2
I =
µ
L
e L 2m π r 2
=
e 2m \
∴
I × πr 2 L
B=
µ0
=
eL
4 πr
3
m
e 2m
GM
(c) : As g =
5.
\
∆ g
g
R
∆R
= −2
2
= –2 × 1.5 = –3%
R
(d) : When the entire arrangement of Young’s double slit experiment is placed in a liquid of refractive index m, then the fringe width will become λ′D λ D β λ D β′ = = = β = d d µd µ 7. (a) : Given, Am = 5 V, Ac = 20 V A 5 = 0.25 \ Modulation index, µ = m = Ac 20 6.
(d) : Capacitor gets discharged as per the relation,
8.
V I 0 e R =
−
V 0 t ln I ln R RC V In both cases log of initial current is same that means 0 R V is constant. o keep 0 constant, both V 0 and R have to be R changed whereas it is stated that only one parameter out of V 0, R and C is changed. Terefore only C has been changed and to match the straight line in the graph, it is decreased. 9.
−
(b) : At zero degree kelvin, molecular motion stops. Velocity
is zero. Hence kinetic energy is zero. 10. (b) : Inside pressure must be
4T
greater than outside
r pressure in bubble. Tis excess pressure is provided by charge on bubble. 4T
r
σ
2
⇒
=
2ε0
Q = 8πr
4T
r
Q2
=
2 4
16 π
=
2
m aR + aT
⋅
2
2βx 2βx x = m⋅ + = 2βx 1 + R mR m 2
2
=
2
2 2
17. (b)
18. (c)
19. (d)
r
× 2ε0
Q σ = 2 4 πr
For x > 1, K.E. = 2E0 λ 1 h / p1 p K.E.2 = = 2= = λ 2 h / p2 p1 K.E.1
2 E0
E0
=
2
21. (a) : Maximum height from inclined plane is 2 u⊥ H = 2a⊥
2
=
(10 sin 37°)
2 g cos 53°
= 3m
22. (b) 23. (c) : Equate volumes,
4
R3 = n ×
4
π
3
⇒ ∆U =
P ∆V ( γ − 1)
=
R nDT γ −1
P (2V − V ) ( γ − 1)
=
PV γ − 1
12. (d) 13. (d) : For the given wave, wave-velocity, v = lu
Maximum particle velocity, v p= aw = 2p u y 0 where a = maximum amplitude v p 2 π υ y 0 2πy 0 = = v υλ λ y πy 0 2π 0 2πy 0 λ
⇒ λ =
=
4
2
14. (c) 15. (d) : Either u A = uB + 5 or u A = uB – 5
By filing, u A increases. Since beats per second increase, the choice is only u A = uB + 5 = 512 + 5 = 517 Hz 16. (b) : K.E. =
1 2
mv 2 = bx 2 or v 2 =
2
2βx
m
r 3
π
3
\ R3 = nr 3 Change in surface energy = T [n.4pr 2 – 4 pR2] = 4 pT (nr 2 – R2)
Change in energy 4 π T (nr 2 − R 2 ) = 4 volume 3 πR 3 2 3T 1 1 = 3 (nr 2 − R2 ) = 3T nr 3 − 1 = 3T − R R r R R 3T
Rise in heat energy per unit volume =
J
2rT ε 0
11. (b) : DU = nC V DT =
∴ 4=
=
20. (a) : For 0 ≤ x ≤ 1, K.E. = 2E0 – E0 = E0
t / RC
=
dv 4βx Differentiate w.r.t. to x , 2v = dx m dv 2 βx or v = dx m v 2 2βx 2 Radial acceleration = aR = = R mR dv dv angential acceleration = aT v dt dx 2βx ∴ aT = m otal force = mass × acceleration
24. (c) : C V for H2 gas is varying in given temperature range, so the temperature of the gas is not increasing at constant rate. Here, in the shown diagram, temperature is on the logarithmic scale. dQ dT = nC V
1 − 1 r R
C
V
Since for the initial portion C V is small, so dT is greatest. 25. (d) : When a ring moves in a magnetic field perpendicular
to its plane, replace the ring by a diameter perpendicular to the direction of motion. Te emf is induced across this diameter. Induced emf, e = B(2r )v In the question, current flow in the ring will be through the two semicircular portions, in parallel. Resistance of each half of the ring = R/2 As these are in parallel, the equivalent resistance = R/4 B(2r )v 8 Brv Current in the circuit = = (R/4) R PHYSICS FOR YOU
|
MARCH ‘18
65
2r
26. (b) : Work done by electric field = q
2k
38. (b) : From vector diagram shown in figure,
λ
∫ r
dr = q2kl ln2
r
λ q ln2 1 2 ∴ 2k λ q ln 2 = mv ⇒ v = mπε0 2 27. (c) : Maximum linear momentum = m × maximum velocity pmax = m × v max or px = mv x
E = total energy = Maximum K.E.=
\
2
2
1 2
m ( v x )
2
2
px 1 mpx px = = m 2 m2 2m
1
E= m. 2
px 2 =
H
2qa
1 2
qa
=2
2qa
=
l Li =
3
h 2π
40. (a) : When a constant current is flowing through a
Here v v x i and B B y j =
F = evx B y (i × j) = ev x B y k
Tereore the subsequent motion o the charged particle will be a circle in the xz plane. 31. (a) 32. (a) : Given : a = 0.9 0.9 α = =9 1− α 1 − 0.9
conductor o non-uniorm cross-section, electron density does not depend upon the area o cross section, while c urrent density, drif velocity and electric field all vary inversely with area o cross-section. 41. (c) : Greater the stability o alkene, lower is the heat o hydrogenation. Out o cis and trans isomers, trans isomer is more stable than cis isomer in which two alkyl groups lie on the same side o the double bond and hence cause steric hindrance, thereore, heat o hydrogenation o trans isomer is less than that o cis isomer.
I C \ I C = bI B = 9(2 mA) = 18 mA I B 2 − αt 33. (b) : Given : p = p0e at 2 is a dimensionless β=
As
[α] =
1
1
=
2 2 [t ] [T ]
= [T
−2
35. (c) : From Doppler’s effect, υ′ =
v −
v
=
256 v × 3 2v
OH
(v − v s )
42. (d) :
= 384 Hz
GM
g 1 g 2
R2 =
3
= G⋅
4 πR ⋅ ρ 3
R2
=
2 4 πρG 3
or
37. (a) : Te possible quantity o heat that will be released by
540 g o water at 80°C cooling down to 0°C, is 540 × 1 × 80 = 540 × 80 cal. o melt 540 g o ice at 0°C heat required = 540 × 80 cal. Hence the mixture will remain at 0°C. |
MARCH ‘18
c c − λ λ 0
mv 2 = h(υ − υ0 ) = h
1 1 λ − λ = hc − = hc 0 λ λ 0 λ0λ
⋅R
R1 R2
PHYSICS FOR YOU
OH and
1
43. (c) :
3
36. (a) : g =
hydrogens
Both will liberate H2 gas with Na metal.
3
256 (v )
hydrogens
υ (v − v o )
Here v o and v s are in direction o v . v u = 256 Hz, v s = , v o = 0 υ′ =
hydrogens
]
34. (d)
66
qa +
Z H = 1, Z Li = 3, E ∝ Z 2 \ |ELi| = 9|EH| or |EH < |ELi|
∴
p = p2x + p2y = 2
∴ l
orce on a charged particle e due to a magnetic field is given by F = e(v × B) .
∴ β=
=
39. (b) : In the second excited state, n = 3
30. (b) : Te
∴
2qa
p y = p1 + p2 cos 45º = 2 qa
29. (a)
=
=
px = p3 + p2 cos 45º
2mE
28. (c)
p1 = qa = p3, p2
∴
2mE or px =
–3q
2
v
=
2hc
m
λ 0 − λ λ λ 0
44. (d) : 10e– + (V+5)2
\
E=
M 10
=
2 A + 80 10
or
2hc λ0 − λ 1/2 v = m λ0λ 2V0
hydrogens
1
45. (a) : At anode : Cl –
At cathode :
1 2
2
Cl2 + e–
Cl–
[Cl–] at anode = x M At cathode, [Ag+][Cl–] = K sp [Cl–] =
⇒
0.059
=0–
( [Ag +] = [Cl–])
K sp
Ecell = E°cell –
[Cl ]cathode
log
n
55. (b) : For transormations L → M and N → K , volume is
−
[Cl ]anode
K sp
log
γ …(ii) = 2 P II Dividing equation (i) by (ii) P II 2 P 1 1 or II = = = γ 1 0.67 γ − P I 2 P I 2 (2) 53. (c) : Dacron (terylene) is a polymer o ethylene glycol and terephthalic acid.
54. (b)
−
0.059
P
PV g = P II(2V ) g or
Cl2 + e–
constant.
x
= + 0.059 log
56. (d) : Alkaline earth metals are group 2 elements. Emission
in terms o the stability constant. Higher the value o stability constant, more stable will be the compound. Tus, the complex with stability constant value 1.6 × 10 7 is the least stable compound.
o a-particle (24He) will reduce its atomic number by 2 units and thus, displaces the daughter nuclei two positions lef in the periodic table, thus to group 18, 16, 14, 12, 10, 8, 6 or 4, etc. As the last stable daughter nuclei ormed could be either Pb (group 14) or Bi (group 15) thereore, the daughter nuclei would belong to group 14.
47. (d) : In a buffer, there are (i) weak acid and its conjugate
57. (c)
1
x
K sp
46. (a) : Te stability o a complex compound is measured
base or (ii) weak base and its conjugate acid In (d), 0.100 mol o HCl will convert 0.100 mol o aniline into anilinium ion, mixture will contain 0.100 mol o aniline and 0.100 mol o anilinium salt. Hence it is a buffer. 48. (b) : – nFE°cell = – RT ln K or E °cell
Plot o E°cell vs ln K will have slope
1
=
RT nF
407
60. (c) : Oxygen gas does not contribute to greenhouse effect. 61. (c) : Te bond angles o NO2, NO2+ and NO2– are in the
RT
.
order NO+2 > NO2 > NO2–. Tis is because NO+2 has no unshared electron and hence it is linear. NO2 has one unshared electron while NO2– has one unshared electron pair.
2
L
=
1000
M
=
In all other compounds, the sum o oxidation numbers is a finite number. 59. (d)
ln K
F 49. (c) : P = 0.658 atm, T = 373 K, w = 0.553 g V
58. (c) : Sum o oxidation numbers o all the atoms in X 3(YZ 4)3 is zero i.e., 3 × (+3) + 3 [1 × (+5) + 4 × (–2)] = 9 – 9 = 0
+
O
wRT PV
=
0.5 53 × 0.0 82 1× 3 73 0.658 × 407 / 1000
=
N
63.23
N +
100 g compound has B = 85.7 g \
63.23 g compound has B
=
85.7 × 6 3.2 3 100
=
O
54.2
7×
100 1000
5 =
180
=
M
⋅
132°
O
–
115°
O
O
since it does not contain CH3CH(OH)— or CH 3CO— group. 63. (b) : Order o stability o carbanions is 1° > 2° > 3°, but
50. (a) : Given; w = 5 g, V = 100 mL, p = 7 atm, M = 180 g mol –1
Applying pV
–
N
62. (b) : CH3CH2CH2CH2OH does not give iodoorm test
54.2 g atom of B = 5 g atom o B 10.8 Formula is B5Hx x = 9.23 9 \ 5 × 10.8 + x = 63.23 or \ Mol. ormula o the compound is B5H9. w
O
Bond angle = 180°
RT
C6H5CH2 – is most stable due to resonance stabilisation. Tus, order o stability is : C6H5CH2– > CH3CH2– > (CH3)2CH– > (CH3)3C– (IV)
(III)
(II)
(I)
64. (c) : 7 ×1 00 ×180
× 0.0821 ×
T
⇒ T =
= 306.94 K or 33 .94 °C 1000 × 5 × 0.0821
51. (c) 52. (a) : For an isothermal process PV = constant P 1V 1 = P 2V 2
PV = P I × 2V or
P
=
P I For an adiabatic process PV g = constant
2
65. (c) : Hydroboration-oxidation gives
…(i)
anti - Markownikoff ’s product.
( g = 1.67) PHYSICS FOR YOU
|
MARCH ‘18
67
66. (c) : Smaller the coagulation value of an electrolyte, greater
is its coagulating or precipitating power. Terefore, the correct order is IV > III > II > I. 67. (b) : For bcc, 4r =
3 a 1 and
Z 1 = 2
For fcc, 4r = 2 a2 and Z 2 = 4 For equal interatomic distance, 3 a 1 =
2+
78. (d) : [Cr(SCN) 2(NH3)4] shows geometrical and linkage
2 a 2
isomerism.
MZ 1 a2 = a1
=
3 2
a3Z d 1 a3N = 1 0 = 2 1 MZ 2 d 2 a13Z 2 a23N 0
;
3 × 2
At the isoelectric point, [H2NCHRCOO–] = [H3N+CHRCOOH] K 1K 2 = [H+]2 ; 2log [H+] = logK 1 + logK 2 –2log [H+] = –logK 1 – logK 2 2pH = pK 1 + pK 2 or pH = (pK 1 + pK 2)/2
2
3
4
=
(
3
1.5
)
=
2
0.919 1
68. (a) 69. (c) :
[Cr(SCN) 2(NH 3) 4] 2+ and [Cr(NCS) 2(NH 3) 4] 2+ are linkage isomers. cis- and trans-forms are possible for both linkage isomers. 79. (a) 80. (c) : M 2SO4
Balanced equation is H2SO4 + 8HI → H 2S + 4I2 + 4H 2O x = 8, y = 4, z = 4
2–
2–
70. (c) : DH for P
2Q is obtained using Hess’s law, by adding eqn. (i), eqn. (ii) and 2 × eqn. (iii ); DH = x + y + 2z . 71. (b) : In N2O4 dimer, N–N bond is formed by pairing of odd
electrons on each nitrogen atom in NO 2 and all four N–O bonds become equivalent.
2CaCO3↓ + 2H2O
72. (c) : Ca(OH)2 + Ca(HCO3)2 (A)
(Clark’s method)
CaCO3↓ + 2NaOH
Ca(OH)2 + Na 2CO3
+
2 M + SO4
( A)
Caustic soda
K sp = [ M + ]2[SO 4 ] = (2s)2(s) = 4s3 or
\
s = 1.2 × 10 4
−5 1/3
= 1.44 × 10−2 +
Concentration of M ions = 2s = 2.88 × 10 –2 M ANSWER KEYS
81.
(d)
82.
(b)
83.
(c)
84.
(a)
85.
(a)
86. 91.
(b) (b)
87. 92.
(b) (a)
88. 93.
(d) (d)
89. 94.
(a) (a)
90. 95.
(b) (d)
96.
(a)
97.
(c)
98.
(d)
99.
(d)
100. (b)
101. (c) 106. (a)
102. (d) 107. (d)
103. (d) 108. (a)
104. (b) 109. (b)
105. (b) 110. (a)
73. (c) : (I) contains 1° and 3° carbon atoms, (II) contains 2°
111. (a)
112. (d)
113. (d)
114. (c)
115. (c)
and 3° carbon atoms and (III) contains only 2° carbon atoms. Hence, numbers of monochlorination products obtained from them are 2, 2 and 1.
116. (b) 121. (c)
117. (d) 122. (b)
118. (c) 123. (a)
119. (b) 124. (b)
120. (a) 125. (d)
126. (a) 131. (b)
127. (a) 132. (b)
128. (b) 133. (c)
129. (c) 134. (a)
130. (a) 135. (c)
136. (c)
137. (a)
138. (a)
139. (d)
140. (c)
141. (d) 146. (c)
142. (a) 147. (c)
143. (b) 148. (c)
144. (a) 149. (a)
145. (c) 150. (c)
Ca(OH)2 + CO2 ( A)
CaCO3 + H2O Calcium carbonate (cloudy)
74. (b)
75. (b)
76. (b)
77. (d) :
MPP CLASS XI
+
K 1
=
−
+
[H3 NCHRCOO ][H ] +
[H3NCHRCOOH]
; K 2 −
Tus, K1K 2
68
=
=
+
+
+
[H3NCHRCOOH]
|
−
[H3NCHRCOO ]
[H2 NCHRCOO ][H ]
PHYSICS FOR YOU
+
−
[H2 NCHRCOO ][H ]
MARCH ‘18
1. (c)
2.
6. (a)
7.
11. (b)
12.
16. (d)
17.
21. (c,d) 22. 26. (4)
27.
(a) (d) (b) (b) (a,d) (a)
3. 8. 13. 18. 23. 28.
ANSWER
(c) 4. (a) 9. (c) 14. (c) 19. (b,c,d) 24. (b) 29.
(d) (a) (c) (c) (3) (d)
KEY
5. 10. 15. 20. 25. 30.
(a) (d) (c) (a,c) (8) (b)
Class XI
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Total Marks : 120
Time Taken : 60 min NEET / AIIMS
5.
Only One Option Correct Type 1.
wo identical containers A and B have frictionless pistons. Tey contain the same volume of an ideal gas at the same temperature. Te mass of the gas in A is m A and that in B is mB. Te gas in each cylinder is now allowed to expand isothermally to double the initial volume. Te change in the pressure in A and B, respectively, is DP and 1.5 DP . Ten (a) 4m A = 9mB (c) 3m A = 2mB
2.
(a) (c) 3.
4.
5 1 3
g / 2
(b)
g / 2
(d)
3 1 5
(c)
−
(b)
1 1 , 2 2
(d)
−
1
1 2 ,
,
−
2
1
−
2
1
2
7.
A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the mass of the block, the tension in the middle of the rope will be (a) F (b) 2F /3 (c) 3F /5 (d) 5F /6 B Figure shows a uniform solid b a block of mass M and edge lengths a, b and c. Its moment c C of inertia about an axis through A one edge and perpendicular D (as shown) to the large face of the block is M 2 2 (a) M (a2 + b2) (b) (a + b )
g / 2
An observer standing on a railway crossing receives frequencies 2.2 kHz and 1.8 kHz when the train approaches and recedes from the observer. Find the velocity of the train (speed of sound in air is 300 m s–1). (a) 10 m s–1 (b) 20 m s–1 (c) 25 m s–1 (d) 30 m s–1
1 1 , 2 2
A cubical block of wood 10 cm along each side floats at the interface between an oil and water with its lowest surface 2 cm below the interface. If the heights of oil and water columns are 10 cm each and ρoil = 0.8 g cm–3, find the mass of the block. (a) 840 g (b) 940 g (c) 1040 g (d) 1500 g
g / 2
A bucket full of hot water cools from 75 °C to 70 °C in time T 1, from 70 °C to 65 °C in time T 2 and from 65 °C to 60 °C in time T 3, then (a) T 1 = T 2 = T 3 (b) T 1 > T 2 > T 3 (c) T 1 < T 2 < T 3 (d) T 1 > T 2 < T 3
(a)
6.
(b) 2m A = 3mB (d) 9m A = 4mB
Te maximum height reached by projectile is 4 m. Te horizontal range is 12 m. Te velocity of projection in m s–1 is ( g is acceleration due to gravity)
Te velocity of a body falling freely under gravity varies as g ahb, where g is the acceleration due to gravity and h is the height. Te value of a and b respectively are
8.
4
3
(c) 7 M (a2 + b2) 12
(d)
M 12
PHYSICS FOR YOU
(a
|
2
+
b2 )
MARCH ‘18
69
9.
10.
What should be the lengths o steel and copper rods respectively so that the length o steel rod is 5 cm longer than copper rod at all temperatures? [a or copper = 1.7 × 10–5 °C–1 and a or steel = 1.1 × 10 –5 °C–1]. (a) 14.17 cm, 9.17 cm (b) 19 cm, 14 cm (c) 17 cm, 12 cm (d) 9.17 cm, 4.17 cm
15. Assertion : Te
Four moles o hydrogen, 2 moles o helium and 1 mole o water vapour orm an ideal gas mixture. What is the molar specific heat at constant pressure o mixture? 16 7R 23 (a) (c) R (d) R (b) R
16.
7
11.
12.
16
time o flight o a body becomes n time the original value i its speed is made n time. Reason : Te range o the projectile becomes n times when speed becomes n times. JEE MAIN / JEE ADVANCED
Only One Option Correct Type
between the block and the ground is
A mass M is lowered with the help o a string by a distance h at a constant acceleration g /2. Te work done by the string will be
(c)
Mgh 2
3 Mgh 2
17.
Mgh
(b)
−
(d)
−
2
3 Mgh
(a)
12 m s
(c)
10 m s
the ollowing questions, a statement o assertion is ollowed by a statement o reason. Mark the correct choice as : (a) I both assertion and reason are true and reason is the correct explanation o assertion. (b) I both assertion and reason are true but reason is not the correct explanation o assertion. (c) I assertion is true but reason is alse. (d) I both assertion and reason are alse. 13. Assertion :
I the earth suddenly stops rotating about its axis, then the acceleration due to gravity will become the same at all the places. Reason : Te
value o acceleration due to gravity is independent o rotation o the earth. 14. Assertion : Te
orce o riction is dependent on normal reaction and the ratio o orce o riction and normal reaction cannot exceed unity. Te coefficient o riction can be greater than unity. Reason :
|
MARCH ‘18
−1
(b)
4.2 m s
−1
(d)
6.4 m s
−1 −1
A U-shaped tube contains a liquid o density ρ and it is rotated about the lef dotted line as shown in the figure. Find the difference in the levels o the liquid column. 2 2
(a)
2
ω
L
2 2 g 2 2
Directions : In
PHYSICS FOR YOU
.
5
Find maximum value o v 0, so that afer pressing the spring, the block does not return back but stops there permanently.
Assertion & Reason Type
70
1
µ=
7
A stone is dropped into a well and its splash is heard afer an interval o 1.45 s. Find the depth o top surace o water in the well. Given that the velocity o sound in air at room temperature is 332 m s –1. (a) 5 m (b) 11 m (c) 20 m (d) 30 m
(a)
A block o mass m = 2 kg is moving with velocity v 0 towards a massless unstretched spring o orce constant k = 10 N m–1. Coefficient o riction
(b)
L 2 g
ω
2 2
(c)
L g
2ω
2 2
(d) 18.
2 2 ω
L
g
Te molar heat capacity C or an ideal gas going a through a process is given by C = where a is a T C constant. I γ = P the work done by one mole o C V gas during heating rom T 0 to h T 0 will be (a) a ln (h) ,
,
(b)
1
a ln(η)
η − 1 RT γ − 1 0
(c) a ln(η) −
(d) a ln (h) – ( γ – 1) RT 0
19.
A particle perorms SHM in a straight line. In the first second, starting rom rest, it travels a distance a and in the next second it travels a distance b in the same direction. Te amplitude o the SHM is 2a − b (a) a – b (b)
1
(c) at equilibrium, mixture contains 13 kg o 3
water 2
(d) at equilibrium, mixture contains 1 kg o
3
(c)
2a
2
(d) None o these
3a − b
24.
More than One Options Correct Type 20.
21.
A driver in a stationary car blows a horn which produces sound waves o requency 1000 Hz normally towards a reflecting wall. Te sound reflected rom the wall approaches the car with a speed o 3.3 m s–1. (a) Te requency o sound reflected rom wall and heard by the driver is 1020 Hz. (b) Te requency o sound reflected rom wall and heard by the driver is 980 Hz. (c) Te percentage increase in requency o sound afer reflection rom wall is 2%. (d) Te percentage decrease in requency o sound afer reflection rom wall is 2%. wo cities A and B are connected by a regular bus service with buses plying in either direction every T seconds. Te speed o each bus is uniorm and equal to v b. A cyclist cycles rom A to B with a uniorm speed o v c. A bus goes past the cyclist in T 1 second in the direction A to B and every T 2 second in the direction B to A. Ten (a) (c)
22.
T 1 =
T 1
=
vbT vb
+ v c
vbT vb
−
v c
(b) (d)
=
T 2
=
vbT vb
−
v c
vbT vb
+ v c
A double star is a system o two stars o masses m and 2m, rotating about their centre o mass only under their mutual gravitational attraction. I r is the separation between these two stars then their time period o rotation about their centre o mass will be proportional to 3
(a) 23.
T 2
r 2
1
1
(b) r
(c)
m2
(d)
m
−
2
5 kg o steam at 100 °C is mixed with 10 kg o ice at 0 °C. Ten (Given swater = 1 cal g–1 °C, LF = 80 cal g–1, –1 LV = 540 cal g ) (a) equilibrium temperature o mixture is 160 °C (b) equilibrium temperature o mixture is 100 °C
3
steam
25.
26.
Integer Answer Type Te value o γ = C P /C V is 4/3 or an adiabatic process o an ideal gas or which internal energy U = K + n PV . Find the value o n (K is a constant). A block o mass m = 2 kg is resting on a rough inclined plane o inclination 37° as shown in figure. Te coefficient o riction between the block and the plane is m = 0.5. What minimum orce F (in N) should be applied perpendicular to the plane on the 37° block so, that the block does not slip 3 on the plane? sin37° = 5 A horizontal orce F = 14 N acts at the centre o mass o a sphere C F o mass m = 1 kg. I the sphere rolls without sliding, find the rictional orce (in N). Form IV
1. Place of Publication 2. Periodicity of its Publication 3. Printer’s Name 3a. Publisher’s Name Nationality Address
: : : : : :
New Delhi Monthly HT Media Ltd. MTG Learning Media Pvt. Ltd. Indian 406, Taj Apartment, New Delhi - 110029 4. Editor’s Name : Anil Ahlawat Nationality : Indian Address : 19, National Media Centre, Gurgaon, Haryana - 122002 5. Name and address of individuals who : Mahabir Singh Ahlawat own the newspapers and partners or 64, National Media Centre, shareholders holding more than one Nathupur, Gurgaon percent of the total capital : Krishna Devi 64, National Media Centre, Nathupur, Gurgaon : Anil Ahlawat & Sons 19, National Media Centre, Nathupur, Gurgaon : Anil Ahlawat 19, National Media Centre, Nathupur, Gurgaon I, Mahabir Singh, authorised signatory for MTG Learning Media Pvt. Ltd. hereby declare that particulars given above are true to the best of my knowledge and belief. For MTG Learning Media Pvt. Ltd. Mahabir Singh Director PHYSICS FOR YOU
|
MARCH ‘18
71
Comprehension Type
30.
A monatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in figure with V B/V A = 2. emperature T A at A is 27 °C
Column I gives some systems whose moment of inertia are listed in column II about the shown axis. Match column I with column II. Column I
Column II
(A) (P)
R 30°
27.
28.
Te work done during the process A → B is (a) 1200 R (b) 1500 R (c) 1600 R (d) 1000 R
Matrix Match Type 29.
A block of mass m is stationary with respect to a rough wedge as shown in figure. Starting from rest in ti me t , work done by various force is given in the columns. Match the column I with column II. ( g = 10 m s –2, m = 1 kg, a = 2 m s–2, t = 4 s). Column I
(B) By normal reaction
(Q)
32 J
(C) By friction
(R)
160 J
(D) By all the forces
(S)
48 J
D
P S P R
Q Q Q P
S P R S
R R S Q
R Uniform triangular plate of mass M
(R)
(S)
144 J
C
R 60° R
R
2R
(P)
B
(Q)
Uniform semicircular ring of mass M . Axis is perpendicular to its plane.
11
MR2 12
13 MR
2
8
MR2 8
Uniform disk of initial mass M from which a circular portion of radius R is then removed. Axis is perpendicular to its plane.
Column II
A
M
(D)
(A) By gravity
(a) (b) (c) (d)
R
(C)
2
Uniform rod of mass M and length R
(B)
Te temperature at B, T B is (a) 600 K (b) 450 K (c) 400 K (d) 900 K
8 MR
(a) (b) (c) (d)
A
B
C
D
Q Q S R
R P P S
P S R Q
S R Q P
Keys are published in this issue. Search now!
Check your score! If your score is > 90%
EXCELLENT WORK !
You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
GOOD WORK !
You can score good in the final exam.
No. of questions correct
……
74-60%
SATISFACTORY !
You need to score more next time.
Marks scored in percentage
……
< 60%
NOT SATISFACTORY!
Revise thoroughly and strengthen your concepts.
72
PHYSICS FOR YOU
|
MARCH ‘18
J
7
Time Allowed : 3 hours Maximum Marks : 70 GENERAL INSTRUCTIONS
(i) (ii) (iii) (iv) (v) (vi) (vii)
All questions are compulsory. Q. no. 1 to 5 are very short answer questions and carry 1 mark each. Q. no. 6 to 10 are short answer questions and carry 2 marks each. Q. no. 11 to 22 are also short answer questions and carry 3 marks each. Q. no. 23 is a value based question and carry 4 marks. Q. no. 24 to 26 are long answer questions and carry 5 marks each. Use log tables if necessary, use of calculators is not allowed.
SECTION - A
1. What is the work done in moving a test charge q through a distance of 1 cm along the equatorial axis of an electric dipole? 2. A car battery is of 12 V. 8 simple cells connected in series can give 12 V; but such cells are not used in starting a car; why? 3. A thin prism of angle 60° gives a minimum deviation of 30°. What is the refractive index of the material of the prism? 4. Here three lenses have been given. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope? Lenses
Power (P )
Aperture ( A)
L1
3D
8 cm
L2
6D
1 cm
L3
10 D
1 cm
5. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. If the
frequency is halved and intensity is doubled, what happens to photoelectric current? SECTION - B
6. A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? 7. For an electromagnetic wave propagating along x -axis, an oscillating electric field along y -axis in free space has a frequency of 6.0 × 10 8 Hz and also an amplitude of E0 = 27 V m–1. Write the equations →
→
for E and B of the given electromagnetic wave. 8. A convex lens of focal length 30 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. Te two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Determine the nature and position of the image formed. 9. Draw the output waveform at X , using the given inputs A and B for the logic circuit shown here. Also, identify the logic operation performed by this circuit. PHYSICS FOR YOU
|
MARCH ‘18
73
(c) Te bar magnet is replaced by a solenoid of
10. In the given block diagram of a receiver, identify the
boxes labelled as X and Y and write their functions. ece v ng Antenna
Amplifier
Detector X
Received signal
Y
Output
OR
A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth is 6.4 × 106 m. SECTION - C
11. n tiny drops, all of same size, are given equal charges.
If the drops coalesce to form a single bigger drop, then what will be the new potential of the drop? What is the surface charge density of the bigger drop? 12. Using Gauss' law establish that the magnitude of electric field intensity, at a point, due to an infinite plane sheet, with uniform charge density s is independent of the distance of the field point. 13. Determine the current in each branch of the network
shown in figure. B
0 1
5 5
D
10
C
0 1
10 V
14. A short bar magnet placed with its axis at 30° with
an external field of 600 G experiences a torque of 0.015 N m. (a) What is the magnetic moment of the magnet? (b) What is the work done in moving it from its most stable to most unstable position? 74
PHYSICS FOR YOU
|
MARCH ‘18
17. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double slit experiment. What is the least distance from the central maximum where the bright fringes due to the both the wavelengths coincide? Te distance between the slits is 2 mm and the distance between the plane of the slits and screen is 120 cm. 18. In a single slit diffraction experiment, when a tiny
5
A
cross-sectional area 10–4 m2 and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid. 15. wo long parallel horizontal → M B rails, distance d apart and ⊗ each having a resistance l R F d per unit length, are joined at one end by a resistance R. N A perfectly conducting rod MN of mass m is free to slide along the rails without friction as shown in figure. Tere is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that as the rod moves, constant current flows through R. Find the velocity of the rod and the applied force F as a function of the distance x of the rod from R. 16. Figure shows how the reactance of an inductor varies with frequency. (a) Calculate the value of ) the inductance of the ( 8 inductor using the ecn 6 a information given in tca 4 e the graph. R 2 (b) If this inductor is 100 200 300 400 connected in series Frequency (Hz) to a resistor of 8 Ω, find what would be the impedance at 300 Hz?
circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why? State two points of difference between the interference pattern in Young’s double slit experiment and diffraction pattern due to a single slit. 15 19. Radiations of frequency 10 Hz are incident on two photosensitive surfaces A and B. Following observations are recorded : Surface A : No photoemission takes place. Surface B : Photoemission takes place but photoelectrons have zero energy.
PHYSICS FOR YOU
|
MARCH ‘18
75
Explain the above observations on the basis of Einstein's photoelectric equation. How will the observation with surface B change when wavelength of incident light is decreased ?
SECTION - E
24. Draw a neat and labelled diagram of a cyclotron. State
the underlying principle and explain how a positively charged particle gets accelerated in this machine. Show mathematically that the cyclotron frequency does not depend upon the speed of the particle.
OR
Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.(mass of neutrons m = 1⋅675 × 10–27 kg, Boltzmann constant k = 1⋅38 × 10–23 J K–1) 20. Obtain an expression for the frequency of radiation
emitted when a hydrogen atom de-excites from level n to level (n – 1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit. 21. A 1000 MW fission reactor consumes half of its fuel
in 5.00 years. How much 235 92 U
did it contain initially? Assume that all the energy generated arises from the fission of 235 92 U and that this nuclide is consumed only by the fission process. Te energy released per fission is 200 MeV. 22. Define the term modulation index for an AM wave.
What would be the modulation index for an AM wave for which the maximum amplitude is a while the minimum amplitude is b?
OR
State Biot-Savart law. Use it to obtain the magnetic field, at an axial point, distance z from the centre of a circular coil of radius a, carrying a current I . Hence, compare the magnitudes of the magnetic field of this coil at its centre and at an axial point for 3 a. which z =
25. Derive the lens-maker’s formula in case of a double
convex lens. State the assumptions made and conventions of signs used. OR (a) Draw a labelled ray diagram of an astronomical
telescope used in the normal adjustment position. Write the expression for its magnifying power. (b) Write expression for magnifying power of astronomical telescope when final image is formed at least distance of distinct vision. 26. What is a p-n junction? Explain with the help of a diagram, how depletion layer is formed near the junction. Explain also what happens to this layer when the junction is (i) forward biased and (ii) reverse biased. OR
Give the symbols of npn and pnp transistor. Show the biasing of an npn transistor and explain its action.
SECTION - D
23. Shivansh was doing an experiment of comparison
of emf using potentiometer in Physics lab. He observed that the galvanometer was showing one side deflection by touching at any point of potentiometer wire. His classmate ‘Risha Mushir’ who was also doing her experiment checked the circuit and suggested to increase the voltage of the battery eliminator from 2 V to 6 V connected in the standard calibrating circuit of the potentiometer. Shivansh did the same and was able to observe two side deflection along the length of potentiometer wire and could get null point with the help of galvanometer. He was thankful to ‘Risha Mushir’ and asked for the cause of the same. (a) What are the values displayed by both Shivansh and Risha Mushir? (b) State the reason why galvanometer showed same side deflection. (c) Distinguish between emf and potential difference. 76
PHYSICS FOR YOU
|
MARCH ‘18
SOLUTIONS
1. Work done is zero because potential at any point on
the equatorial line of an electric dipole is zero. Hence, work done, W = q(V 2 – V 1) = 0 2. Te internal resistance of simple cells is high;
therefore the series combination of simple cells does not give enough current required to start a car. A m sin 60 30 sin
3.
2
A
sin
2
1/ =
2
1/ 2
=
2
=
sin 45 sin 30 60 sin 2 2
1.41
4. Since the aperture of lens L1 is largest, it is used as
objective for a telescope. Te lens L3 is used as eyepiece since its focal length is smaller.
5.
If the frequency is halved, the frequency of incident
light will become
1.5
= 0.75 times (i.e., less than 1) the
2
threshold frequency. Hence, photoelectric current will be zero. 6. For the free oscillations, the angular frequency should be resonant frequency. Resonant angular frequency of oscillation of the circuit, 1 ω r =
10
1 =
LC
27 × 10
−3
× 30 × 10
−6
=
9
3
= 1.1 × 10
rad s
−1
As per question, direction of propagation of electromagnetic wave is along x -axis. E0 = 27 V m–1 and υ = 6.0 × 10 8 Hz 8 9 –1 \ w = 2pυ = 2p × 6.0 × 10 = 1.2p × 10 rad s
\
k=
=
=
4π rad m
Terefore, the given logic circuit acts as OR gate. Hence, output is high when both or one of them is high. Accordingly the waveform of output is shown in figure. 10. X = Intermediate frequency (IF) stage
Y = Amplifier/Power amplifier IF Stage : IF stage changes the carrier frequency to a lower frequency. Amplifier: Increases the strength of signals. OR
For LOS mode, dm =
O is at 2 f of lens, so it will form image at 2 f on other side, i.e., 60 cm from lens. Hence, position of virtual object from mirror is at (60–15) cm = 45 cm behind the mirror. Te ray diagram briefing the image formation is shown here. 8.
I1
I2
O
45 cm
For mirror, R = 20 cm, f = +10 cm, u = +45 cm, v = ?
\
v
+
u
v = +
=
90 7
1
1 ⇒
f cm
v
1 +
45
1 =
=
An virtual image is formed at a distance behind the mirror. \
9.
Boolean expression of this combination is,
cm
5
2
2 × 64 × 10 × 50 3
64 × 10 ×
10 + 8 × 10
2
10 = 144 × 10 ×
10 m
Let each drop be having a radius r and charge q. Ten, potential at the surface of each drop, q V 4 0 r q and surface charge density, 2 4 r When n drops coalesce to form a single bigger drop of radius R, total volume remains unchanged. 11.
4
R3 = n ×
4
π
3
r 3 ⇒ R = (n)1/3r
π
3
and total charge on the bigger drop, Q = nq \ Potential of bigger drop,
10
90
5
d m = 45 .5 km
V
(Behind the mirror)
2hR R
2 × 64 × 10 × 32 +
Hence,
60 cm 15 cm
2hT R +
=
1
X =Y = A+ B = A+ B
−1
8 c 3 × 10 → Equation for electric field, E = E0 sin (wt – kx ) j^ → E = 27 sin [1.2 p × 109 t – 4p x ] j^ V m–1
1
and
9
1.2π × 10
ω
E 27 8 Further, B0 0 9 10 T 8 c 3 10 Moreover, magnetic field should be along the z -axis. → ^ Hence, B = B0 sin (wt – kx ) k → ^ B = 9 × 10–8 sin [1.2 p × 109 t – 4p x ] k
As,
A+B
=
4
7.
\
Y
Q
nq
4 0 R
2/3
1/3
4 0 (n)
r
(n)
V
and new surface charge density,
Q 4 R
7
12.
2
nq
1/3
2 /3 2
4 n
r
(n)
According to Gauss' theorem total electric flux
through a closed surface is enclosed by the surface, i.e.,
1 ε
times the net charge
0
PHYSICS FOR YOU
|
MARCH ‘18
77
q
→ da
0
P
φE = ∫ E . da = ε A
→ E
Consider a thin infinite r → E s plane sheet of charge + + + + + with uniform surface + + + + + charge density s. o + + + + calculate electric field at a point (P ) at distant r from r the sheet, we imaging a symmetrical Gaussian surface in such a way that → → E da the point lies on it. Here we assume a cylinder of cross-sectional area A and length 2r with its axis perpendicular to the sheet. Flux through the curved surface of the cylinder ( f1) : As electric lines are parallel to the curved surface, flux through it is zero. (∵ θ = 90°) φ1 = E . da = 0 Flux through the plane faces of the cylinder (f2): (∵ θ = 0°) φ2 = 2 E . ds = 2 EA otal flux through the cylindrical Gaussian surface is f = f1 + f2= 2EA. otal charge enclosed by the surface q = s A q According to Gauss' theorem, φ =
∫
∫
i.e., 2EA =
A
σ
ε
0
ε0
or E =
σ
2ε0
= constant
Let us first distribute the current in different branches. Now, equations for different loops using Kirchhoff’s second law, 13.
Loop 1 10I 1 + 5I g – 5I 2 = 0 or 2I 1 + I g – I 2 = 0 ...(i) Loop 2 5I g + 10(I 2 + I g ) – 5(I 1 – I g ) = 0 or 2I 2 + 4I g – I 1 = 0 ...(ii) Loop 3 5I 2 + 10(I 2 + I g ) + 10I = 10 or 15I 2 + 10I g + 10I = 10 or 3I 2 + 2I g + 2I = 2 …(iii) 78
PHYSICS FOR YOU
|
MARCH ‘18
Solving equations (i) and (ii), I 1 = –2I g …(iv) Equations (i) and (iv) I 2 = –3I g …(v) Now using the equation (v) in equation (iii), – 3[3I g ] + 2I g + 2I = 2 or 2I – 7I g = 2 …(vi) Using Kirchhoff’s law, I = I 1 + I 2 I = –5I g (using (iv) and (v)) So, equation (vi), 2[–5I g ] – 7I g = 2 or – 17I g = 2 So, I g = Also
−2
17
I 1 =
I 1 – I g
6
=
17
A 4 17
and
I =
+10
17
A
6
A,
A , I 2 + I g
I 2 = 4 =
17
A
A
17
(a) orque experienced by the bar magnet t = MB sin q 0.015 = M (600 × 10 –4) sin 30° M = 0.5 A m2 (b) Most stable position when qi = 0° and most unstable when q f = 180° Work done, W = MB [cos qi – cos q f ] W = 0.5 × 600 × 10 –4 [cos 0° – cos 180°] W = 0.06 J (c) A solenoid provides dipole moment M = NIA 0.5 = 1000 × 10–4 I I = 5 A 14.
Let the distance from R to MN be x . Ten the area of the loop between MN and resistance R is xd and the magnetic flux linked with the loop is B x d . As the rod moves, the emf induced in the loop is given by d dx | ε | = (B x d) = B d = B v d dt dt where v is the velocity of MN . Te total resistance of the loop is R + 2lx . Te current in the loop is given by | | Bvd I R 2 x R 2 x (i) Force acting on the rod, B2d 2 F = IBd = v R + 2λ x 15.
dv B2d 2 dx B2d 2 dx ∴ m = ⋅ or dv = ⋅ dt R + 2λ x dt m R + 2λ x On integrating both sides, we get v
B2d 2 R 2 x ln 2 m R
19. Einstein's photoelectric equation is ...(i) hυ = W + Ek ⇒ Ek = hυ – W or Ek = hυ – hυ0 where W is work function of metal, υ is frequency of
B2 d 2 B2 d 2 F = ⋅ R + 2 λx 2λ m
R + 2λ x ln R 4 4 B d R + 2λ x = ln (2 λm)(R + 2 λ x ) R
X L 1 X L 2 2 8 4 4 From the graph, Slope 400 200 200 16. X L = 2p υL, L
−
=
=
=
−
(a) L =
1 2π
× (slope ) =
1 = 2π 50
1
1
1 50
H
100
(b) At 300 Hz, X L = 6 Ω Impedance, Z = R2 + X L2 = 82 + 62 = 10 Ω 17. For least distance of coincidence of fringes, there must be a difference of 1 in order of l1 and l2. As l1 > l2, n1 < n2 If n1 = n, n2 = n + 1 \
( y n)ll = ( y n + 1)l2 ⇒
⇒
nl1 = (n + 1)l2
⇒
n
(n + 1)D λ 2
=
2
(650 520) nm
or n
130
4 –3
Here D = 120 cm = 1.20 m, d = 2 mm = 2 × 10 m \ Least distance, y min =
nDλ 1 d
=
4 × 1.2 × 650 × 10 2 × 10
hc
− W =
λ
−3
18. When a tiny circular obstacle is placed in the path
of light from a distant source a bright spot is seen at the centre of the shadow of the obstacle because of the constructive interference of diffracted rays of light by the circular obstacle. Interference pattern Diffraction pattern (Due (By Young’s double slit to a single slit.) experiment.)
(i) All the bright and (i) Central bright fringe dark fringes are of same is twice the width of any width. other secondary bright or dark fringe. (ii) All the bright fringes (ii) Intensity of central are of same intensity. bright fringe is maximum and it decreases with increase in the order of secondary bright fringes.
(υ0 = 1015 Hz)
− hυο
h kT
Here mass of neutron, m = 1⋅675 × 10–27 kg T = 27°C = (27 + 273) K = 300 K \
6 63 10
(3 1 675 10
27
6 63 10
34
1 38 10
23
34
48
300)
6 63 10
(3 1 675 1 38 3) 10 –10
–3
= 1.56 × 10 m = 1.56 mm
λ
de Broglie wavelength associated with thermal neutrons,
−9
m
hc
OR
3m
520
Ek =
λ=
d
520 nm
2 1
nDλ1 d
incident light and υ0 is threshold frequency. Surface A : As no photoemission takes place; energy of incident photon is less than the work function. Surface B : As photoemission takes place with zero kinetic energy of photolectrons (i.e., Ek = 0), then equation (i) gives W = hυ or υ0 = υ. i.e., energy of incident photon is equal to work function. When wavelength of incident light is decreased, the energy of incident photon becomes more than the work function, so photoelectrons emitted will have finite kinetic energy given by
10
4 56
= 1⋅45 × 10 m = 1⋅45 Å Tis is comparable with interatomic spacing in a crystal. Tus thermal neutrons may be diffracted by crystals and hence are a suitable probe for diffraction experiments. High energy neutron beam has smaller wavelength and has larger probability of passing through the crystal spacing without suffering any diffraction. Tat is why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments. MPP CLASS XII
1. (b) 6. (a) 11. (b) 16. (b) 21. (b,d) 26. (3)
2. 7. 12. 17. 22. 27.
(b) (b) (b) (b) (b, d) (c)
3. 8. 13. 18. 23. 28.
ANSWER
(a) 4. 9. (c) (d) 14. 19. (b) (a,c,d) 24. (c) 29.
PHYSICS FOR YOU
(b) (c) (d) (c) (9) (d) |
KEY
5. 10. 15. 20. 25. 30.
MARCH ‘18
(b) (c) (b) (a,b,c) (2) (a) 79
20. Let us first find the frequency of revolution of electron in the orbit classically. In Bohr’s model velocity of electron in nth orbit is
n2h2ε0 nh v = , where radius r = 2 2πmr πme Tus orbital frequency of electron in nth orbit is v nh / 2πmr nh nh me2 υ= = 2 2 2 2 2 2πr 2πr 4 mr 4 m n h 0 or
me 4
υ=
3 3
2
me 4 1 1 me 4 − 2 3 2 3 2 2 n n 8ε0h f i 8 0 h
or
υ=
1 1 2 2 (n 1) n
me 4 2 n − 1 2 3 2 2 8ε0h n (n − 1)
For large n, 2n – 1 ≈ 2n and n – 1 ≈ n, frequency, me 4 2n υ= 2 3 4 8ε0h n
me 4 3 3
4n
h
2
...(ii)
0
Equations (i) and (ii) are equal, hence for large value of n, the classical frequency of revolution of electron in nth orbit is same as frequency of radiation when electron de-excite from level n to (n – 1). 21. Power of reactor = 1000 MW = 10 9 W= 109 J s–1 Energy generated by reactor in 5 years = 5 × 365 × 24 × 60 × 60 × 10 9 J Energy generated per fission = 200 MeV = 200 × 1.6 × 10 –13 J Number of fission taking place or number of U235 nuclei
required
5 365 24 60 60 10 200 1.6 10
9
= 49.275 × 10 26
13
Mass of 6.023 × 10 23 nuclei of U = 235 g = 235 × 10 –3 kg Mass of 49.275 × 1026 nuclei of U
235 10
3
26
23
49.275 10
2
Am Ac
=
24. Refer to point 5, Page no. 173 (MTG Excel in Physics). OR Refer to point 3.1 (1, 3), Page no. 169 (MTG Excel in Physics). 25. Refer to 6.6 (1, 2), Page no. 374 (MTG Excel in Physics). OR Refer to point 2, Page no. 382 (MTG Excel in Physics). 26. Refer to point 9.3 (1, 2, 4, 5), Page no. 587 (MTG Excel in Physics). OR Refer to point 9.4 (1, 3, 4), Page no. 592 (MTG Excel in Physics).
Solution Senders of Physics Musing 1. Suhas Sheikh, Mumbai
22. (a) In AM, modulation index is the ratio of amplitude of modulating signal to the amplitude of |
MARCH ‘18
a −b a +b
23. (a) Sharing of knowledge, caring for each other and open for discussion and learning from each other. (b) Tis happens when the emf of the driving cell is not greater than the emf of experimental cell or the positive ends of all cells are the connected to the same end of the wire. (c) Te maximum potential difference between terminals of a cell in an open circuit is called as emf and the potential difference between terminals when some current is drawn is called terminal potential or potential difference.
SET-55
otal fuel = 3846 kg
PHYSICS FOR YOU
µ=
1923 kg
of total fuel = 1923 kg
80
Ac (b) Since AM wave is given by C m(t ) = ( Ac + Am sin wmt ) sin wmt So, maximum amplitude is Ac + Am = a and minimum amplitude is Ac – Am = b a +b Adding them, we get Ac =
~
6.0203 10 1
Am
So, modulation index ...(i)
υ=
µ=
2
h ε0 2 Te frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1), 4n
carrier wave
2. Anita Dabbas, Uttar Pradesh 3. Vipul Gorai, West Bengal
Contd. from Page No. 26 7.
As cos ic
Free body diagram of cylinder and wedge are shown in figure.
=
Since, sinic
1 =
−
1 µ
sin2 ic ∴
cosic
1−
=
1 µ
2
1 1 ∴ f = 1 − 1 − 2 2 µ 9.
Equation of motion of cylinder, m1 g sina – N 3cosa = m1a1 ...(i) Equation of motion of wedge; N 3cosa – m2 g sina = m2a2 ...(ii) Since a1 = a2 ...(iii) On solving, equations (i), (ii) and (iii), we get 2m1m2 N 3 = g tan α m1 + m2 8.
…(ii)
(using (ii))
Consider an infinitesimal length dx of tube at a distance x from one end of capillary tube of crosssectional area A. \ Volume of this section = Adx Here, T
T − T = T 0 + L 0 x
L Volume of the capillary, V = AL, is constant. Applying ideal gas equation to this differential volume, we get T − T 0 P ( Adx ) = dnRT = dnR T 0 + L L x
Te light energy which escapes is combined within cone of apex angle 2ic, where ic is critical angle for total internal reflection. Consider source of light is at the centre of sphere and surface area A of the sphere is inside the cone.
L
n
dx R or dn = PA − T T 0 x 0 T + L 0 0 L L L TL − T 0 nR ln T + x = (TL − T 0 ) 0 L 0 PA
∫
∫
Solving, we get L
T − T 0 nR ln T 0 + L L x 0 = PA (TL − T 0 ) L
10.
Fraction of energy passing out f =
A
...(i) 4 πR2 Area of the differential ring of angular thickness d q on the sphere is dA = 2pRsinq(Rd q), On integrating,, we get, \
2
A = 2 πR
∫ sin θdθ = 2πR 0
A 2πR
ic
2
=
(1 − cos ic ) or
or
2 f = (1 – cos ic)
\
f
(1 cos ic ) 2 −
=
2
(1 − cos ic )
2 A 4 πR2
=
(1 − cos ic ) (using (i))
Using conservation of energy DK.E. = DP.E. ⇒ 0 = mgy – qEx y qE ...(i) ⇒ x mg y From figure, sin θ = l ⇒ y = l sinq l − x x and cos θ = or x = l (1 − cos θ) ⇒ cos θ = 1 − l l y l sin θ 2 sin(θ/2) cos(θ/2) 1 = = = \ x l (1 − cos θ) tan(θ/2) 2 sin2 (θ/2) x mg θ tan = = (Using eqn. (i)) 2 y qE −1 mg = 2 tan−1 2mg ε0 q = 2 tan qE qσ =
PHYSICS FOR YOU
|
MARCH ‘18
81
Class XII
T
his specially designed column enables students to self analyse their extent of understanding of complete syllabus. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Total Marks : 120
Time Taken : 60 min 4.
NEET / AIIMS
Only One Option Correct Type 1.
2.
Wavelengths of photons n = 3 2 in transitions having serial n = 2 1 3 numbers 1, 2, and 3 of a n = 1 hydrogen like atom are l1, l2 and l3 respectively. Ten, mark the correct option. (a)
l1 = l2 + l3
(c)
λ 3 =
λ1 + λ 2 λ1 λ 2
(b)
λ 1 =
(d)
l2 = l1 – l3
λ 3 + λ 2
π . Te power consumption in = I0 sin ω t − 2
the circuit is given by ε I 0 0 (a) 2
(c)
82
λ3 λ 2
For an AC circuit, the voltage applied is e = e0 sin wt . Te resulting current in the circuit is I
3.
5.
ε0
I 0
2
(b) zero (d)
2ε0
I 0
In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude a and of wavelength l. In other experiment with the same set up, the two slits are sources of equal amplitude a and wavelength l, but are incoherent. Te ratio of intensity of light at the mid point of the screen in the first case to that in the second case is (a) 2 : 1 (b) 1 : 2 (c) 3 : 4 (d) 4 : 3 PHYSICS FOR YOU
|
6.
MARCH ‘18
Te given circuit It is equivalent to (a) OR gate A (b) AND gate (c) NOR gate B (d) NAND gate A conducting rod moves r with constant velocity v perpendicular to the long, I straight wire carrying a l current I as shown in the figure. Te emf generated between the ends of the rod is µ 0vIl µ vIl (a) 0 (b) 2 πr πr µ0 vIl 2µ 0 vIl (c) (d) 4 πr πr If two dielectrics are placed + K 1 in a parallel plate capacitor + as shown. Ten, find + K 2 the equivalent dielectric + constant of the system K + K 2 2 (a) K eq = 1 (b) K eq = K1 + K 2 2 (c) K eq = K 1 + K 2
7.
Y
(d) K eq
=
v
– – – –
1
K1 + K 2 If refractive index of material of a prism with prism A angle A is µ = cot , then angle of minimum 2
deviation will be (a) 180° – A (c) 180° – 3 A
(b) 180° – 2 A (d) 180° – 4 A
8.
Four charges each with charge +q are placed at the q our corners o a square o side l , a charge is −
4
at centre o the square. Force on the charge at the centre due to other charges is (where K = 1/4 πε0 ) (a)
Kq2 l 2
(b) K
(d) K
(c) zero 9.
2q
(c) I assertion is true but reason is alse. (d) I both assertion and reason are alse. I a rod has a resistance 4 W is turned into semicircle, then its resistance along its diameter is 1.0 W.
13. Assertion :
2
On bending a rod, its length decreases and hence resistance decreases. Reason :
l 2 2
2q
2
l 2 A charged particle moves in a magnetic field B 10 i with initial velocity u = 5 i + 4 j . Te path o the particle will be (a) straight line (b) circle (c) helical path (d) None o these
14. Assertion : Charge never flows rom a condenser o
higher capacity to the condenser o lower capacity. Reason : Direction o flow o charge is determined
=
10.
11.
by the difference in charge in the two condenser. 15. Assertion : When radius o a circular wire carrying
current is doubled, its magnetic moment becomes our times. Reason : Magnetic moment depends on area o the loop.
–1
A light beam, E = 100 [sin( w1t ) + sin(w2t )] V m with w1 = 5 × 10 15 rad s–1 and w2 = 8 × 10 15 rad s–1, alls on a metal surace o work unction 2.0 eV. Maximum kinetic energy o emitted photoelectrons is (a) 3.20 eV (b) 1.5 eV (c) 3.27 eV (d) 2.1 eV Te electric field part o an electromagnetic wave in a medium is represented by Ex = 0 ; E y = 2 ⋅ 5N C −1 cos 2 π × 106 rad s −1 t − π × 10−2 rad m −1 x Ez = 0. Te wave is (a) moving along the x -direction with requency 106 Hz and wavelength 100 m (b) moving along x -direction with requency 106 Hz and wavelength 200 m (c) moving along – x direction with requency 106 Hz and wavelength 200 m (d) moving along y -direction with requency 2p × 106 Hz and wavelength 200 m
(
JEE MAIN / JEE ADVANCED
Only One Option Correct Type 16.
Tere is a stream o neutrons with a kinetic energy o 0.0327 eV. I the hal-lie o neutrons is 700 s, what raction o neutrons will decay beore they travel a distance o 10 m? Given, mass o neutron = 1.676 × 10 –27 kg. (a) 4.8 × 10–5 (b) 3.9 × 10–6 (c) 8.4 × 10–5 (d) 2.3 × 10–6
17.
A thin circular ring o area A is held perpendicular to a uniorm field o induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance o the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is
)
(
)
90% o the active nuclei present in a radioactive sample are ound to remain undecayed afer 1 day. Te percentage o undecayed nuclei lef afer two days will be (a) 85% (b) 81% (c) 80% (d) 79% Assertion & Reason Type Directions : In the ollowing questions, a statement o assertion is ollowed by a statement o reason. Mark the correct choice as : (a) I both assertion and reason are true and reason is the correct explanation o assertion. (b) I both assertion and reason are true but reason is not the correct explanation o assertion. 12.
(a) 18.
BR A
(b)
AB R
(c) ABR
(d)
B2 A
R2 A ray o light enters a rectangular glass slab o reractive index 3 at an angle o incidence 60°. It travels a distance o 5 cm inside the slab and emerges out o the slab. Te perpendicular distance between the incident and the emergent rays is (a)
5 3 cm
(c)
5
3
cm
(b)
5
cm
2
(d) 5 cm
2 PHYSICS FOR YOU
|
MARCH ‘18
83
19.
An electromagnetic wave of frequency u = 3.0 MHz passes from vacuum into a dielectric medium with permittivity e = 4.0. Ten (a) wavelength is doubled and the frequency remains unchanged (b) wavelength is doubled and frequency becomes half (c) wavelength is halved and frequency remains unchanged (d) wavelength and frequency both remain unchanged.
22.
(a) dmin (c) x = 23.
More than One Options Correct Type 20.
21.
84
A charged particle enters Q into a region which offers a resistance against its P motion and a uniform magnetic field exists in the region. Te particle traces a spiral path as shown in figure. Which of the following statements are correct? (a) Component of magnetic field in the plane of spiral is zero. (b) Te particle enters the region at Q. (c) If magnetic field is outward, then the particle is positively charged. (d) If magnetic field is outward, then the particle is negatively charged. wo circular coils P and I 1 I 2 Q are fixed coaxially and carry currents I 1 and I 2 respectively as shown in P Q figure. Mark the correct options. (a) If I 2 = 0 and P moves towards Q, a current in the same direction as I 1 is induced in Q (b) If I 1 = 0 and Q moves towards P , a current in the opposite direction to that of I 2 is induced in P . (c) When I 1 ≠ 0 and I 2 ≠ 0 are in the same direction then the two coils tend to move apart. (d) When I 1 ≠ 0 and I 2 ≠ 0 are in opposite directions then the coils tends to move apart. PHYSICS FOR YOU
|
MARCH ‘18
Te minimum value of d so that there is a dark fringe at O is d min. For the value of d min, the distance at which the next bright fringe is formed is x . Ten =
λ D
d min 2
P B d
x O′ O
A D
(b) d min
=
D λ D 2
(d) x = d min
A cubical region of side a has its centre at the origin. It encloses three fixed point charges,
0, −a , 0 , + 3q 4
–q at
at (0, 0, 0) and – q at
0, + a , 0 . Choose the correct option(s). 4 (a) Te net electric flux crossing the plane x = +
a 2
is equal to the net electric flux crossing the plane x
=
−
a 2
.
(b) Te net electric flux crossing the plane y = +
a 2
is more than the net electric flux crossing the a plane y =
−
2
(c) Te net electric flux crossing the entire region q is ε
0
(d) Te net electric flux crossing the plane z = +
a 2
is equal to the net electric flux crossing the a plane x = + 2
Integer Answer Type 24.
A square loop of side a = 6 cm carries a current I = 1 A. Calculate magnetic induction B (in m) at point P , lying on the axis of loop and at a distance x =
25.
7 cm
from the center of loop.
A solid sphere of radius R has a charge Q distributed in its volume with a charge density r = kr a, where k and a are constants and r is the distance from its centre. If the electric field at r = R/2 is 1/8 times that at r = R, find the value of a.
26.
A silver ball o radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light o wavelength 200 nm is incident on the ball or some time during which a total light energy o 1.0 × 10–7 J alls on the surace. Assuming that on the average, one photon out o ten thousand photons is able to eject a photoelectron, find the electric potential (in × 10–1 V) at the surace o the ball assuming zero potential at infinity. Comprehension Type
In a mixture o H – He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transer their total excitation energy to He+ ions (by collisions). Assume that the Bohr model o atom is exactly valid. 27. Te quantum number n o the state finally populated in He+ ions is (a) 2 (b) 3 (c) 4 (d) 5 28.
29.
Column I
(A) At P, beore pasting transparent paper (B) At O, beore pasting transparent paper (C) At P , afer pasting the transparent paper (D) At O, afer pasting the transparent paper (a) �b� �c� �d� 30.
Te wavelength o light emitted in the visible region by He+ ions afer collisions with H atoms is (a) 6.5 × 10–7 m (b) 5.6 × 10–7 m (c) 4.6 × 10–7 m (d) 4.0 × 10–7 m Matrix Match Type
A
B
C
D
P Q P S
R R Q P
S S S Q
Q P R R
Column I
2 mm
P
10 mm
20 mm O S2
1m
Column II
(A) Charge at rest experience a orce. (B) A charge in motion goes undeviated with same velocity. (C) A charge in motion goes undeviated with varying speed. (D) A charged particle undergoes helical motion. A
S
(P) Bright ringe o order 80 (Q) Bright ringe o order 262 (R) Bright ringe o order 62 (S) Bright ringe o order 280
Column I shows the state o motion o a charged particle. Column II shows the possible combination o electric field and magnetic field under which the path in column I is possible. Match column I with column II.
In Young’s double-slit experiment, the point source S is placed above the central axis as shown in figure and the intererence pattern was obtained. Now he pasted a transparent paper o thickness 0.02 mm and reractive index 1.45 in ront o slit S1 and again obtained the pattern. Column II contains the nature and order o ringe and Column I contains positions on the screen. I l = 500 nm, then match column I with column II. S1
Column II
(a) �b� �c� �d�
B
C
Q,S P,Q,R Q,S P,R R,S Q,R P,R Q,S Q,R Q P,S R,Q
(P) E = 0, B = 0 (Q) E ≠ 0, B ≠ 0
(R) E = 0, B ≠ 0
(S)
E ≠ 0, B = 0
D
Q,R R,S Q,S Q,S
2m
Keys are published in this issue. Search now !
J
Check your score! If your score is > 90%
EXCELLENT WORK !
You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
GOOD WORK !
You can score good in the final exam.
No. of questions correct
……
74-60%
SATISFACTORY !
You need to score more next time.
Marks scored in percentage
……
< 60%
NOT SATISFACTORY!
Revise thoroughly and strengthen your concepts.
PHYSICS FOR YOU
|
MARCH ‘18
85
86
PHYSICS FOR YOU
|
MARCH ‘18
PHYSICS FOR YOU
MARCH ’18
87
88
PHYSICS FOR YOU
MARCH ’18
PHYSICS FOR YOU
MARCH ’18
89
90
PHYSICS FOR YOU
MARCH ’18
