IIT-JAM 2015 Solutions

fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES JAM JOINT ADMISSION TEST FOR MSc 2015 SECTION – A: MCQ Q1 – Q10 carry one mark each.

Q1.

A system consists of N number of particles, N >> 1 . Each particle can have only one of the two energies E 1 or E 1 + ε (ε > 0 ) . If the system is in equilibrium at a temperature T , the average number of particles with energy E 1 is (a)

Ans:

N

2

(b)

e

ε / kT

+1

(c)

N e

− ε / kT

+1

(d) Ne −ε / kT

(c)

Solution: N = Ne Q2.

N

−( E2 − E 1 ) kT

− ⎡⎣( E1 + ε ) − E 1 ⎦⎤

=e

kT

− ε

⇒ N = Ne kT

A mass m , lying on a horizontal, frictionless surface, is connected to one end of a spring. The other end of the spring is connected to a wall, as shown in the figure. At t = 0 , the mass is given an impulse.

m

Impulse

The time dependence of the displacement and the velocity of the mass (in terms of nonzero constants A and B ) are given by

Ans:

(a) x(t ) = A sin t , v(t ) = B cos t

(b) x(t ) = A sin t , v(t ) = B sin t

(c) x(t ) = A cos t , v (t ) = B sin t

(d) x(t ) = A cos t , v(t ) = B cos t

(a)

Solution: At time t = = 0 , the mass ‘ m ’ is at rest. Thus, displacement will be zero at time t = = 0 .

∴ Velocity is v =

x = A sin (ω t )

dx

= Aω cos ωt = B cos ω t dt Thus, x = A sin ω t and V ( t ) = B cos ω t

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Q3.

A particle with energy E is incident on a potential given by

⎧ 0,

x < 0

⎩ V0 ,

x≥0

V ( x ) = ⎨

.

< V 0 in the region x > 0 (in terms of positive The wave function of the particle for E < constants A, B and k ) is (a) Ae kx + Be − kx Ans:

(b) Ae − kx

(c) Ae ikx + Be − ikx

(d) Zero

(b)

Solution: For x > 0 ; −



2

2m

d 2ψ ΙΙ d

+ V0ψ ΙΙ = Eψ ΙΙ ;

ψ ΙΙ = Be kx + Ae − kx where k = ψ ΙΙ → 0 as x → ∞

E < V0

2m (V0 − E ) 

2

⇒ A = 0 ⇒ ψ ΙΙ = Ae− kx

⎡ π ⎞⎤ ⎛ The electric field of a light wave is given by E = E 0 ⎢iˆ sin (ω t − kz ) + jˆ sin ⎜ ω t − kz − ⎟⎥ . 4 ⎠⎦ ⎝ ⎣ 

Q4.

The polarization state of the wave is

Ans:

(a) Left handed circular

(b) Right handed circular

(c) Left handed elliptical

(d) Right handed elliptical

(c)

⎛ ⎝

Solution: E x = E0 sin (ωt − kz ) , E y = E0 sin ⎜ ω t − kz −

π ⎞

⎟.

4⎠

Thus resultant is elliptically polarized wave.

⎛ ⎝

At z = 0, E x = E0 sin (ωt ) , E y = E0 sin ⎜ ω t − When ω t = 0, E x = 0, E y = −

E 0

2

π ⎞

⎟

4⎠

and when ω t =

π

4

, E x =


E 0

2

, E y = 0



Q3.

A particle with energy E is incident on a potential given by

⎧ 0,

x < 0

⎩ V0 ,

x≥0

V ( x ) = ⎨

.

< V 0 in the region x > 0 (in terms of positive The wave function of the particle for E < constants A, B and k ) is (a) Ae kx + Be − kx Ans:

(b) Ae − kx

(c) Ae ikx + Be − ikx

(d) Zero

(b)

Solution: For x > 0 ; −



2

2m

d 2ψ ΙΙ d

+ V0ψ ΙΙ = Eψ ΙΙ ;

ψ ΙΙ = Be kx + Ae − kx where k = ψ ΙΙ → 0 as x → ∞

E < V0

2m (V0 − E ) 

2

⇒ A = 0 ⇒ ψ ΙΙ = Ae− kx

⎡ π ⎞⎤ ⎛ The electric field of a light wave is given by E = E 0 ⎢iˆ sin (ω t − kz ) + jˆ sin ⎜ ω t − kz − ⎟⎥ . 4 ⎠⎦ ⎝ ⎣ 

Q4.

The polarization state of the wave is

Ans:

(a) Left handed circular

(b) Right handed circular

(c) Left handed elliptical

(d) Right handed elliptical

(c)

⎛ ⎝

Solution: E x = E0 sin (ωt − kz ) , E y = E0 sin ⎜ ω t − kz −

π ⎞

⎟.

4⎠

Thus resultant is elliptically polarized wave.

⎛ ⎝

At z = 0, E x = E0 sin (ωt ) , E y = E0 sin ⎜ ω t − When ω t = 0, E x = 0, E y = −

E 0

2

π ⎞

⎟

4⎠

and when ω t =

π

4

, E x =


E 0

2

, E y = 0



Q5.

Consider the coordinate transformation x ′ =

x + y

2

, y ′ =

x − y

2

. The relation between the

area elements d x ′d y ′ and dxdy is given by d x ′d y ′ = jdxdy . The value of j is (a) 2 Ans:

(b) 1

(c) − 1

(d) − 2

(c)

Solution: x′ =

x + y

2

, y′ =

x− y

2

∵ dx′dy′ = J dxdy

⎛ ∂ x′ ⎜ ∂ x ⇒ J = ⎜ ⎜ ∂ y′ ⎜ ∂ x ⎝

Q6.

∂x′ ⎞ ⎛ 1 1 ⎞ ⎟ ⎜ ⎟ ∂y 1 1 2 ⎟ ⎟=⎜ 2 = − − = −1 ∂y ′ ⎟ ⎜ 1 1 ⎟ 2 2 − ⎜ ⎟ ⎟ ∂y ⎠ ⎝ 2 2⎠ The trace of a 2 × 2 matrix is 4 and its determinant is 8 . If one of the eigenvalues is 2(1 + i ) , the other eigenvalue is

(a) 2(1 − i ) Ans:

(b) 2(1 + i )

(c) (1 + 2i )

(d) (1 − 2i )

(a)

Solution: λ1 = 2 + 2i , λ2 = 2 (1 − i ) ⇒ λ1 + λ2 = 4 and λ1 ⋅ λ 2 = 8 Q7.

Temperature dependence of resistivity of a metal can be described by (a)

(b) R

R

T

T

(c)

R

(d)

R

T

Ans:

T

(a)




Solution: Electrical resistivity of metal varies as ρ ∝ T 5

(For T << θ D )

ρ ∝ T

(For T >> θ D )

where θ D is the Debye temperature. Thus, correct answer is option (a) Q8.

A proton from outer space is moving towards earth with velocity 0.99 c as measured in earth’s frame. A spaceship, traveling parallel to the proton, measures proton’s velocity to be 0.97 c . The approximate velocity of the spaceship in the earth’s frame, is (a) 0.2 c

Ans:

(b) 0.3 c

(c) 0.4 c

(d) 0.5 c

(d)

Solution: Velocity of proton w.r.t. spaceship = 0.97 c ∵ u x′

= 0.99 c, v = −v, u x = 0.97 c

s′

E −v u x′ + v 0.99 c − v ⇒ u x = ⇒ 0.97 c = ⇒ v = 0.5 c 0.97v u x′ v

1+

Q9.

c

1−

2

c

s

p = 0.99 c

p = 0.99 c

A charge q is at the center of two concentric spheres. The outward electric flux through the inner sphere is φ while that through the outer sphere is 2 φ . The amount of charge contained in the region between the two spheres is (a) 2q

Ans:

(c) − q

(d) − 2q

(b)

Solution: φ = Q10.

(b) q

q

ε 0

, φ ′ = 2φ =

q + q′

ε 0

⇒ q′ = q

At room temperature, the speed of sound in air is 340 m/sec. An organ pipe with both ends open has a length L = 29 cm . An extra hole is created at the position L / 2 . The lowest frequency of sound produced is (a) 293 Hz

Ans:

(b) 586 Hz

(c) 1172 Hz

(d) 2344 Hz

(c)


Branch office Anand Institute of Mathematics, 28‐B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi‐16


Solution: The fundamental frequency in organ pipe with both end open is f = L/2

v

2 L

L

with additional rate at f ′ =

v

2 L′

=

L

2

, the fundamental frequency becomes

v v 340 m / sec = = = 1172 Hz 2 L L 29 × 10 −2 m 2

Q11 – Q30 carry two marks each.

Q11.

A system comprises of three electrons. There are three single particle energy levels accessible to each of these electrons. The number of possible configurations for this system is (a) 1

Ans:

(b) 3

(c) 6

(d) 7

(c)

Solution: For electron spin is

1 2

. So in one single state two electrons can be adjusted the number

of ways are Ground

First

Second

1

2

1

0

2

2

0

1

3

1

2

0

4

1

0

2

5

0

1

2

6

0

2

1

So, number of ways are 6 .




Q12.

A rigid and thermally isolated tank is divided into two compartments of equal volume V , separated by a thin membrane. One compartment contains one mole of an ideal gas A and the other compartment contains one mole of a different ideal gas B . The two gases are in thermal equilibrium at a temperature T . If the membrane ruptures, the two gases mix. Assume that the gases are chemically inert. The change in the total entropy of the gases on mixing is (a) 0

Ans:

(b) R ln 2

(c)

3 2

R ln 2

(d) 2 R ln 2

(d)

Solution: For A , number of microstate after mixing is 2 For A , number of microstate before mixing is 1

A

B

⇒ ΔS A = R ln 2 − R ln1 = R ln 2 Similarly, for B ⇒ ΔS B = R ln 2

⇒ ΔS = ΔS A + ΔS B = 2R ln 2 Q13.

A Zener regulator has an input voltage in the range 15V − 20V and a load current in the range of 5 mA − 20 mA . If the Zener voltage is 6.8V , the value of the series resistor R S

should be

V 0

+ 15 − 20 V

(a) 390 Ω Ans:

6.8V

−

(b) 420 Ω

(c) 440 Ω

(d) 460 Ω

Some data is missing. (No answer is possible)




Q14.

The variation of binding energy per nucleon with respect to the mass number of nuclei is shown in the figure.

9

r e 8 p y 7 g r e ) 6 n V e e 5 g M n ( i d n o 4 n l e i b c 3 e u g n 2 a r e v 1 A 0

20 40 60 80 100 120 140 160 180 200 220 240

Number of nucleons in nucleus, A Consider the following reactions: (i)

238 92

206 U →82 Pb + 10 P + 22n

(ii)

238 92

U→

Pb + 8 2 He + 6e

206 82

4

−

Which one of the following statements is true for the given decay modes of

238 92

U ?

(a) Both (i) and (ii) are allowed

(b) Both (i) and (ii) are forbidden

(c) (i) is forbidden and (ii) is allowed

(d) (i) is allowed and (ii) is forbidden

Ans:

(c)

Q15.

A rigid triangular molecule consists of three non-collinear atoms joined by rigid rods. The constant pressure molar specific heat C p ) of an ideal gas consisting of such molecules is (a) 6 R

Ans:

(c) 4 R

(d) 3 R

(c)

Solution: D.O.F = 6 ⇒ U = Q16.

(b) 5 R

⎛ ∂U ⎞ ⇒ CV = ⎜ ⎟ = 3R ⇒ CP = CV + R = 4 R 2 ⎝ ∂T ⎠V

6 RT

A satellite moves around the earth in a circular orbit of radius R centered at the earth. A second satellite moves in an elliptic orbit of major axis 8 R , with the earth at one of the foci. If the former takes 1 day to complete a revolution, the latter would take (a) 21.6 days

Ans:

(b) 8 days

(c) 3 hours

(d) 1.1 hour

(a)



fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2

3

⎛T ⎞ ⎛ R ⎞ 3/ 2 ⇒ T2 = ( 8 ) T1 ≈ 22 days Solution: ⎜ 1 ⎟ = ⎜ ⎟ ⎝ T2 ⎠ ⎝ 8 R ⎠ Q17.

A positively charged particle, with a charge q , enters a region in which there is a uniform 



electric field E and a uniform magnetic field B , both directed parallel to the positive y -axis. At t = 0 , the particle is at the origin and has a speed v 0 directed along the

positive x - axis. The orbit of the particle, projected on the x - z plane, is a circle. Let T be the time taken to complete one revolution of this circle. The y -coordinate of the particle at t = T is given by (a)

Ans:

π 2 mE

2qB 2

2π 2 mE

(b)

qB

2

(c)

π 2 mE qB

2

+

v0π m qB

(d)

2π mv 0 qB

z

(b) 2

1 1 qE ⎛ 2π m ⎞ 2π 2 mE 2 Solution: y = u y t + a y t ⇒ y = ⎜ ⎟ = 2 2 m ⎝ qB ⎠ qB 2

 

E , B

x

Q18.

y

v0

Vibrations of diatomic molecules can be represented as those of harmonic oscillators. Two halogen molecules X 2 and Y 2 have fundamental vibrational frequencies v X = 16.7 ×10

12

Hz and vY = 26.8 × 10

12

Hz , respectively. The respective force constants

are K X = 325 N / m and K Y = 446 N / m . The atomic masses of F , Cl and Br are 19.0, 35.5 and 79.9 atomic mass unit respectively. The halogen molecules X 2 and Y 2 are

Ans:

(a) X 2 = F 2 and Y 2 = Cl 2

(b) X 2 = Cl 2 and Y 2 = F 2

(c) X 2 = Br 2 and Y 2 = F 2

(d) X 2 = F 2 and Y 2 = Br 2

(b)

Solution: The oscillation frequency of diatomic molecule with reduce mass ‘ μ ’ is f =

1 2π

k

μ

⇒ μ =

1 k where k is force constant. 4π 2 f 2




For X 2 molecule: μ =

⇒ m x =

m x mx m x + mx

=

mx

2

k 1 1 325 N / m × x2 = × 2 2 2 2π f x 2 × ( 3.14 ) (16.7 ×1012 Hz )

⇒ m x = 59.07 ×10−27 kg = 35.5 ×1.67 ×10 −27 kg = 35.5 a.m.u. This is the atomic mass of chlorine ( Cl ) . For Y 2 molecule: μ =

⇒ m y =

m y m y m y + m y

=

my

2

k y 1 1 446 N / m × = × 2π 2 ( f )2 2 × ( 3.14 )2 ( 26.8 ×1012 Hz )2 y

⇒ m y = 31.73 ×10−27 kg = 19 × 1.67 × 10−27 kg = 19 a.m.u . This is the atomic mass of F . Thus, correct answer is option (b) Q19.

A hollow, conducting spherical shell of inner radius R1 and outer radius R2 encloses a charge q inside, which is located at a distance d (< R1 ) from the centre of the spheres. The potential at the centre of the shell is (a) Zero

(c) Ans:

(b)

⎛q q ⎞ ⎜ − ⎟ 4π ∈0 ⎝ d R1 ⎠ 1

(d)

1

R1 q d

R2

q

4π ∈0 d

⎛q q q ⎞ ⎜ − + ⎟ 4π ∈0 ⎝ d R2 R2 ⎠ 1

(d)

Solution: V =

1 ⎛q

q q ⎞ ⎜ − + ⎟ 4πε 0 ⎝ d R1 R2 ⎠




Q20.

Doppler effect can be used to measure the speed of blood through vessels. Sound of frequency 1.0522 MHz is sent through the vessels along the direction of blood flow. The reflected sound generates a beat signal of frequency 100 Hz. The speed of sound in blood is 1545 m/sec. The speed of blood through the vessel, in m/sec, is (a) 14.68

Ans:

(b) 1.468

(c) 0.1468

(d) 0.01468

(d)

Solution: Consider Vb , V sound are velocities of blood cell and sound in blood. The sound of frequency

( f 0 ) is traveling towards blood cell where blood cell is moving away with

velocity V b f 0 V sound

V b

Frequency of sound observed on blood cell is

⎛ Vsound − V b ⎞ V sound ⎟⎠

f ′ = f 0 ⎜ ⎝

(i)

Sound from blood cell of frequency f ′ reflect back. f ′

observer

V b

⎛ V sound ⎞ The frequency observed by observer is f = f ′ ⎜ ⎝ Vsound + V b ⎟⎠

(ii)

⎛ Vsound − Vb ⎞ ⎛ V sound ⎞ Vsound ⎟⎠ ⎜⎝ Vsound + V b ⎟⎠

From equation (i) and (ii), we get f = f 0 ⎜ ⎝

⎛V − V ⎞ ⇒ f = f 0 = ⎜ sound b ⎟ ⎝ Vsound + V b ⎠ ⎛ Vsound − V b ⎞ ⎟ ⎝ Vsound + V b ⎠

Now, Δ f = f 0 − f = f 0 − f 0 ⎜

(iii)

⎛ 2V b ⎞ = f 0 ⎜ ⎟ ⎝ Vsound + V b ⎠




⇒

2V b Vsound + Vb

=

Δf f 0

⇒

Vsound + Vb Vb

=

2 f 0

Δf

⇒ V b =

V sound

⎛ 2 f 0 ⎞ ⎜⎝ Δ f − 1⎟⎠

Given Vsound = 1545 m / sec, f 0 = 1.0522 × 10 6 Hz, Δf = 100 Hz

∴V b =

1545

⎛ 2 × 1.0522 × 106 ⎞ − 1⎟ ⎜⎝ 100 ⎠

=

1545 21043

= 0.073 ⇒ Vb = 0.073 m / sec

Thus the best suitable answer is option (d). Q21.

Which of the following circuits represent the Boolean expression S = P + QR + Q P

(a) P

(b) P S

Q

S

Q

(c) P

(d) P

Q

Q

S

R

S

R

Ans:

(b)

Q22.

A conducting wire is in the shape of a regular hexagon, which is I

inscribed inside an imaginary circle of radius R , as shown. A current I flows through the wire The magnitude of the magnetic field at the

R C

center of the circle is (a) Ans:

3μ 0 I 2π R

(b)

μ 0 I

2 3π R

(c)

3μ 0 I π R

(d)

3 0 I 2π R

(c)

Solution: d = R cos60 0 =

3 2

R



fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ∵ B

=

μ 0 I

4π d

( sin θ 2 − sin θ 1 ) ⇒ B1 =

⇒ B = 6 B1 = 6 × Q23.

μ0 I

=

2 3π R

3μ0 I

μ0 I

4π d

=

3π R

μ0 I

2sin 300 =

4π

3 R 2

2sin 30 0 =

μ 0 I

2 3π R

3μ 0 I π R

An observer is located on a horizontal, circular turntable which rotates about a vertical axis passing through its center, with a uniform angular speed of 2rad/sec . A mass of 10 grams is sliding without friction on the turntable. At an instant when the mass is at a distance of 8 cm from the axis it is observed to move towards the center with a speed of 6 cm/sec. The net force on the mass, as seen by the observer at that instant, is (a) 0.0024 N

Ans:

(b) 0.0032 N

(c) 0.004 N

(d) 0.006 N

(c)

Solution: Two forces will act on the particle First coriolis force Fc = −2m(ω × v) = −240 × 10 −5 N (in tangential direction) radial direction) Another force is centrifugal force Fr = mω 2 r = 320 ×10 −5 N (in

Total force F = Q24.

Fc + Fc 2

2 r

= 0.04 N

Miller indicates of a plane in cubic structure that contains all the directions [100], [011] and

[111] are (a) (011) Ans:

(b) (101)

(c) (100 )

(a)

(d) (110 ) y

Solution: The name of the plane containing all the directions

[100] , [ 011] & [111] is ( 0 11)

[111]

or ( 01 1 )

The best suitable answer is option (a)

[ 011] x

[100] z




Q25.

Seven uniform disks, each of mass m and radius r , are inscribed inside a regular hexagon as shown. The moment of inertia of this system of seven disks, about an axis passing through the central disk and perpendicular to the plane of the disks, is 7

(a)

2 13

(c) Ans:

(b) 7mr 2

2

(d)

m r

55 2

2

m r

(d)

Solution: Q26.

2

2

mr

2 ⎛ mr 2 54mr 2 55mr 2 mr 2⎞ +6×⎜ + 4mr ⎟ = + = 2 2 2 2 ⎝ 2 ⎠

mr

2

A nucleus has a size of 10 −15 m . Consider an electron bound within a nucleus. The estimated energy of this electron is of the order of (b)10 2 MeV

(a)1 MeV Ans:

(c) 10 4 MeV

(d) 10 6 MeV

(d)

Solution: p =

∵ E

=

h

λ p

= 2

2me

⇒ E =

6.6 ×10 −34 10

=

−15

= 6.6 ×10−19 kgm / sec

44 × 10−38 2 × 9.1×10

2.4 ×10 −7 1.6 ×10

−19

−31

= 2.4 ×10−7 Joule

eV = 1.5 × 1012 eV = 1.5 ×106 MeV 

Q27.

Consider a vector field F = yiˆ + xz 3 ˆj − zykˆ . Let C be the circle x 2 + y 2 = 4 on the plane z = 2 , oriented counter-clockwise. The value of the contour integral (a) 28 π

Ans:

(c) − 4 π

(b) 4 π



∫ C



F ⋅ d r is

(d) − 28 π

(a)

Solution:

  ∵  F .d r

∫

C







= ∫ ( ∇ × F ).d a S



fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 

xˆ



yˆ

zˆ

∇ × F = ∂ / ∂x ∂ / ∂y ∂ / ∂z 3 − zy y xz ⎛ ∂ ( − yz ) ∂ ( xz 3 ) ⎞ ⎛ ∂ y ∂ ( − zy ) ⎞ ⎛ ∂ ( xz 3 ) ∂y ⎞ ⎟ + yˆ ⎜ − ⇒ ∇ × F = xˆ ⎜ − − ⎟ ⎟ + zˆ ⎜ ⎜ ∂ y ∂z ⎟ ⎝ ∂z ∂x ⎠ ⎜ ∂x ∂y ⎟ ⎝ ⎠ ⎝ ⎠   ⇒ ∇ × F = xˆ ( − z − 3xz 2 ) + yˆ ( 0 − 0 ) + zˆ ( z 3 − 1) 





∵ z



= 2 ⇒ ∇ × F = − ( 2 + 12 x ) xˆ + 7 zˆ

 ∵da







= rdrdφ zˆ ⇒ ( ∇ × F ) .d a = ⎡⎣ − ( 2 + 12 x ) xˆ + 7 zˆ ⎤⎦ .rdrdφ zˆ = 7 rdrdφ 





2

2π

0

0

⇒ ∫ ( ∇ × F ).d a = 7 ∫ rdr ∫ dφ = 28π S

Q28.

Consider the equation

dy dx

=

y

2

x

with the boundary condition y (1) = 1 . Out of the

following the range of x in which y is real and finite is (a) − ∞ ≤ x ≤ −3

(b) − 3 ≤ x ≤ 0

(c) 0 ≤ x ≤ 3

(d) 3 ≤ x ≤ ∞

Ans:

Solution of the differential equation is satisfied by options (c) and (d).

Q29.

The Fourier series for an arbitrary periodic function with period 2 L , is given by f ( x ) =

a0

2

+ ∑n =1 a n cos ∞

nπ x L

+ ∑n =1 bn sin ∞

shown in the figure the value of a 0 is

nπ x L

. For the particular periodic function

f ( x ) 1

1/ 2

−2 (a) 0 Ans:

(b) 0.5

−1

0

1

(c) 1

2

x

(d) 2

(c)

Solution: The wavefunction of the given function can be written as Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498 Website: www.physicsbyfiziks.com



0 < x <1

⎧ x f ( x ) = ⎨ ⎩ − x

−1 < x < 0

Coefficient a0 is defined as

∫

a0 = 1

0

−1

1

− x dx + 1∫ x dx 0

1 ω ⎡ ( −1) 2 ⎤ ⎡ (1) 2 ⎤ ⎡ x 2 ⎤ ⎡ x2 ⎤ 1 1 = − ⎢ ⎥ + ⎢ ⎥ = − ⎢0 − − 0⎥ = + + − 1 ⎥+⎢ 2 ⎥ ⎢ 2 2 2 ⎢⎣ ⎥⎦ ⎣ 2 ⎦ −1 ⎣ 2 ⎦ 0 ⎦ ⎣

∴ a0 = 1 Q30.

The phase of the complex number (1 + i ) i in the polar representation is (a)

Ans:

π

(b)

4

π

(c)

2

3π 4

(d)

5π 4

(c)

Solution: z = (1 + i ) i ⇒ z = ( −1 + i ) ⇒ z = x + iy tan θ =

y x

= −1 ⇒ θ = tan −1 ( −1) ⇒ θ =

3π 4




SECTION – B: MSQ Q1 – Q10 carry two marks each. Q1. For an electromagnetic wave traveling in free space, the electric field is given by  V 8 E = 100 cos 10 t + kx jˆ . Which of the following statements are true? m (a) The wavelength of the wave in meter is 6π

(

)

(b) The corresponding magnetic field is directed along the positive z direction (c) The Poynting vector is directed along the positive z direction (d) The wave is linearly polarized Ans:

(a) and (d) 

Solution: E = 100 cos (108 t + kx ) ˆj V / m Option (a) is true 2π c 2π × 3 ×108 8 ω = 10 ⇒ = 10 ⇒ λ = = 6π λ 108 8

Option (b) is wrong 

(



)

B ∝ kˆ × E ∝ ( − xˆ × yˆ ) ∝ − zˆ

Option (c) is wrong 

S ∝ kˆ ∝ − xˆ

Option (d) is true Q2.

In an ideal Op-Amp circuit shown below R1 = 3k Ω, R2 = 1k Ω and Vi = 0.5sin ω t (in Volt). Which of the following statements are true? V i

+

(a) The current through R1 = The current through R2

−

(b) The potential at P is V 0

R2 R1

(c) The amplitude of V 0 is 2V

V 0

P R 2

R1

(d) The output voltage V 0 is in phase with V i Ans:

(b), (c) and (d) Option (a) is wrong




Current through R1 is I 1 =

Vo − V i R1

and Current through R2 is I 2 =

V i R2

Option (b) is true The potential at P is V 0

R2 R1

(voltage divider rule)

Option (c) is true

⎛

V0 = ⎜ 1 +

⎝

R2 ⎞

⎛ 3⎞ ⎟ Vi = ⎜1 + ⎟ 0.5sin ωt = 2sin ω t ⇒ Vm = 2 V R1 ⎠ ⎝ 1⎠

Option (d) is true Q3.

A particle of mass m is moving in x − y plane. At any given time t , its position vector is  given by r ( t ) = A cos ωt i + B sin ω t ˆj where A, B and

are constants with A ≠ B .

Which of the following statements are true? (a) Orbit of the particle is an ellipse (b) Speed of the particle is constant (c) At any given time t the particle experiences a force towards origin (d) The angular momentum of the particle is mω ABk ˆ Ans:

(a), (c) and (d)

 x y = sin ω t Solution: (a) r ( t ) = A cos ωt iˆ + B sin ω t jˆ ⇒ x = A cos ωt , y = B sin ω t ⇒ = cos ωt , A B

⇒

x

2

A

+

2

y

2

B

2

= 1 (Ellipse)



(b)

d r dt

= − Aω sin ωt iˆ + Bω cos ωt ˆj 

Speed =

d r dt

= A2ω 2 sin 2 ωt + B 2ω 2 cos 2 ωt . Speed is function of time, so not constant.

2

(c)

d r

2

dt



= − Aω cos ωt iˆ − Bω 2 sin ωt ˆj = −ω 2 r . Force act towards origin. 2



fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES i ˆj ⎛ k ˆ⎞   ⎜ ⎟ B sin ω t 0 ⎟ ⇒ L = mω ABk (d) L = ( r × p ) = m ⎜ A cos ωt ⎜⎝ − Aω sin ωt Bω cos ωt 0⎟⎠ 

Q4.

A rod is hanging vertically from a pivot. A partic1e traveling in horizontal direction, collides with the rod as shown in the figure. For the rod-particle system, consider the linear momentum and the angular momentum about the pivot .Which of the following statements are NOT true? (a) Both linear momentum and angular momentum are conserved (b) Linear momentum is conserved but angular momentum is not (c) Linear momentum is not conserved but angular momentum is conserved (d) Neither linear momentum nor annular momentum are conserved

Ans:

(b), (c) and (d)

Q5.

A particle is moving in a two dimensional potential well V ( x, y ) = 0,

= ∞,

0 ≤ x ≤ L, 0 ≤ y ≤ 2 L elsewhere

which of the following statements about the ground state energy E 1 and ground state eigenfunction (a) E 1 = (c) ϕ 0 = Ans:



2

π 2 2

mL

0

are true? 52π 2 (b) E 1 = 8mL2

π x π y 2 sin sin L L 2L

(d) ϕ 0 =

π x π y 2 cos cos L L 2 L

(b) and (c)

Solution: E n =

2 π 2  2 ⎛ n x

⎜ 2m ⎜⎝ L2

+

2 n y ⎞

⎟ 4 L2 ⎟⎠

1 π 2 2 ⎛ 1 + 2 Ground state n x = 1, n y = 1 ⇒ E x = ⎜ 2 2m ⎝ L 4 L Wave function ψ =

2 L

⋅

2 2L

⋅

⎞ 5π 2 2 ⎟ = 8mL2 ⎠

sin π x sin π y L

2L




Q6.

Consider the circuit, consisting of an AC function generator V (t ) = V 0 sin 2π vt with V 0 = 5V an inductor L = 8.0mH , resistor R = 5Ω and a capacitor C = 100 F . Which of

the following statements are true if we vary the frequency? L C

R

(a) The current in the circuit would be maximum at ν = 178 Hz (b) The capacitive reactance increases with frequency (c) At resonance, the impedance of the circuit is equal to the resistance in the circuit (d) At resonance, the current in the circuit is out of phase with the source voltage Ans:

(a) and (c)

Solution: Option (a) is true ν =

1 2π LC

=

1 2 × 3.14

−3

−6

(8 ×10 )(100 ×10 )

= 178 Hz

Option (b) is wrong X C =

1 ⇒ X C ↓ as ω ↑ ω C

Option (c) is true Option (d) is wrong Q7.

Muons are elementary particles produced in the upper atmosphere. They have a life time of 2.2 s . Consider muons which are traveling vertically towards the earth’s surface at a speed of 0.998c . For an observer on earth, the height of the atmosphere above the surface of the earth is 10.4 km . Which of the following statements are true? (a) The muons can never reach earth’s surface (b) The apparent thickness of earth’s atmosphere in muon’s frame of reference is 0.96 km (c) The lifetime of muons in earth’s frame of reference is 34.8 s (d) Muons traveling at a speed greater than 0.998 c reach the earth’s surface

Ans:

(c) and (d)




Δt 0

Solution: Δt =

1−

v

2

c

2

⇒ Δt =

2.2 ×10−6 1 − ( 0.998 )

2

= 34.8 ×10 −6 sec

Now distance will be = Δt × v = 34.8 ×10 −6 × 0.998 × 3 ×108 = 10.4192 km Apparent thickness Δ X = Δt × v = 2.2 ×10 −6 × 0.998 × 3 ×108 = 0.658 km Q8.

As shown in the P − V diagram AB and CD are two isotherms at temperatures T 1 and T 2 , respectively (T 1 > T 2 ) . AC and BD are two reversible adiabats. In this Carnot cycle,

which of the following statements are true? (a)

Q1 T1

=

P

Q2

Q1

A

B

T 2

T 1

(b) The entropy of the source decreases C

(c) The entropy of the system increases

Q2 D

(d) Work done by the system W = Q1 − Q2

T 2 V

Ans:

(a), (b) and (d)

Q9.

The following figure shows a double slit Fraunhofer diffraction pattern produced by two slits, each of width a separated by a distance b, a < b . Secondary maxima

Primary maxima

Which of the following statements are correct? (a) Reducing a increases the separation between consecutive primary maxima (b) Reducing a increases the separation between consecutive secondary maxima (c) Reducing b increases the separation between consecutive primary maxima (d) Reducing b increases the separation between consecutive secondary maxima Ans:

(a) and (d)

Solution: The minima condition for double slit Fraunhofer diffraction is



fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES nλ

a sin θ = nλ ⇒ sin θ =

where a is the width of slit. a Reducing ‘ a ’ increases the separation between diffraction minima i.e. increases the separation between consecutive primary maxima. The condition of interference maxima is mλ where b is the separation between slits. b sin θ = mλ ⇒ sin θ = b The position of interference maxima gives the separation between secondary maxima. Reducing ‘ b ’ increases the separation between consecutive secondary maxima. The correct answer is option (a) and (d). 

Q10.

A unit cube made of a dielectric material has a polarization P = 3iˆ + 4 jˆ units. The edges of the cube are parallel to the Cartesian axes. Which of the following statements are true? (a) The cube carries a volume bound charge of magnitude 5 units (b) There is a charge of magnitude 3 units on both the surfaces parallel to the y − z plane (c) There is a charge of magnitude 4 units on both the surfaces parallel to the x − z plane (d) There is a net non-zero induced charge on the cube

Ans:

(b) and (c)

Solution:

 ∵P

= 3iˆ + 4 ˆj

Option (a) is wrong  

ρ b = −∇.P = 0

Option (b) is true 

(

)( )

(

)( )

At x = 0 , σ b = P.nˆ = 3iˆ + 4 ˆj . −iˆ = −3 

At x = 1 , σ b = P.nˆ = 3iˆ + 4 ˆj . iˆ = 3 Option (c) is true 

(

)( )

At y = 0 , σ b = P.nˆ = 3iˆ + 4 ˆj . − ˆj = −4 

(

)( )

At y = 1 , σ b = P.nˆ = 3iˆ + 4 ˆj . ˆj = 4 Option (d) is wrong Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐16 Phone: 011‐26865455/+91‐9871145498 Website: www.physicsbyfiziks.com


fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES SECTION – C: NAT Q1 – Q10 carry one mark each.

Q1.

The power radiated by sun is 3.8 × 10 26 W and its radius is 7 × 10 5 km . The magnitude of the Poynting vector (in

Ans:

cm

2

) at the surface of the sun is _________________

6174

Solution: I = Q2.

W

P

3.8 ×1026

=

A

4π × ( 7 ×10

10

)

W / cm2 = 6174 W / cm 2

A particle is in a state which is a superposition of the ground state excited state

Φ=

1 5

1

ϕ 0 +

0

and the first

of a one-dimensional quantum harmonic oscillator. The state is given by

2 5

ϕ 1 . The expectation value of the energy of the particle in this state (in

units of  , being the frequency of the oscillator) is ________ Ans:

1.3

⎛ ⎝

Solution: ∵ En = ⎜ n +

⇒ E = Q3.

ω

2

⎛ 3ω ⎞ 4 ⎛ ω ⎞ 1 ω and P ⎜ = , P⎜ = ⎟ ⎟ ⎝ 2 ⎟⎠ 5 2⎠ ⎝ 2 ⎠ 5

1⎞

1

3ω

5

2

× +

4

13ω

5

10

× =

= 1.3ω

In an experiment on charging of an initially uncharged capacitor, an RC circuit is made with the resistance R = 10k Ω and the capacitor C = 1000 F along with a voltage source of 6V . The magnitude of the displacement current through the capacitor (in

A ),

5 seconds after the charging has started, is _______________ Ans: Solution: I =

364 V R

e

− t / RC

=

6 10 ×10

3

e

−5 /10×103 ×1000×10−6

=

6 10

4

e

−5 /10

=

6 e ×104


=

6 1.65 ×10 4

= 364 μ A



Q4.

In the given circuit VCC = 10V and β = 100 for n − p − n transistor. The collector voltage V CC

V C (in volts) is __________.

1 K 100 K

V C

+ 5V

Ans:

−

5.7

Solution: I B =

5 − 0.7 100 ×10

3

= 4.3 ×10−5 A ⇒ I C = β I B = 4.3 mA

⇒ VC = VCC − I C RC = 10 − 4.3 = 5.7 V Q5.

Unpolarized light is incident on a calcite plate at an angle of incidence 50o as shown in the figure. Take n0 = 1.6584 and ne = 1.4864 for calcite. The angular separation ( in degrees) between the two emerging rays within the plate is Air

Optic axis

Ans:

50 0

Calcite

3.51

Solution: Inside the crystal incident light split into two components, ordinary ray and extraordinary ray According to Snell’s law

i = 500

sin i sin r

=n

For ordinary ray i = 500 , no = 1.6584

⎛ sin i ⎞ sin i ∴ sin ro = ⇒ r o = sin −1 ⎜ no ⎝ no ⎟⎠

r o r e e- ray o- ray




⎡ sin 500 ⎤ ⎡ 0.766 ⎤ ⇒ r o = sin ⎢ = sin −1 ⎢ = sin −1 [ 0.462 ] ⇒ r 0 = 27.510 ⎥ ⎥ ⎣1.6584 ⎦ ⎣ no ⎦ −1

For extra-ordinary ray i = 500 , ne = 1.4864

∴ sin re =

⎛ sin i ⎞ sin i ⇒ r e = sin −1 ⎜ ne ⎝ ne ⎟⎠

⎡ sin 500 ⎤ −1 ⎡ 0.766 ⎤ ⇒ r e = sin ⎢ = sin −1 [ 0.515] ⇒ r e = 31.02 0 ⎥ = sin ⎢ ⎥ ⎣1.4864 ⎦ ⎣ ne ⎦ −1

Thus, the angular separation between the o - ray and e - ray is θ = re − r o = 3.510 Q6.

In the hydrogen atom spectrum. the ratio of the longest wavelength in the Lyman series (final state n = 1 ) to that in the Balmer series (final State n = 2 ) is ____________

Ans:

0.185

Solution: According to Bohr Theory

⎛ 1 1⎞ = R ⎜ 2 − 2 ⎟ λ L ⎝ n f ni ⎠ 1

n=3 n = 2

The longest wavelength in the Lyman series is

⇒

1 ⎛1 1 = R ⎜ − 2 λ L ⎝1 2

4 ⎞ ⎛3⎞ = ⇒ = R λ L ⎟ ⎜4⎟ 3 R ⎠ ⎝ ⎠

n = 1

H α

Lα

The longest wavelength in the Balmer series is

Q7.

⇒

1 ⎛1 1 = R ⎜ 2 − 2 λ B ⎝2 3

⇒

λ L λ B

=

4 3 R

×

5 R 36

=

1 36 ⎞ ⎛1 1⎞ ⎛9−4 ⎞ ⎛ 5 ⎞ R R R λ = − = ⇒ = ⇒ = B ⎟ ⎜4 9⎟ ⎜ 36 ⎟ λ ⎜ 36 ⎟ 5 R ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ B 5 27

= 0.185

A rod is moving with a speed of 0.8c in a direction at 60o to its own length. The percentage contraction in the length of the rod is __________

Ans:

9

Solution: l x = l0 x 1 −

v

2

c

2

2

1

l0 3

2

2

= l0 cos θ 1 − ( 0.8) ⇒ l x = l0 × × 0.6 = 0.3l0 and l y = l0 sin θ =



fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2

⎛ 3l0 ⎞ 3 = New length l = ( 0.3l0 ) + ⎜ l0 0.09 + = 0.916 l0 ⎟ ⎜ 2 ⎟ 4 ⎝ ⎠ (1 − 0.91) l0 × 100 = 0.09 × 100 = 9% % change in length 2

l0

Q8.

X − rays of wavelength 0.24 nm are Compton scattered and the scattered beam is

observed at an angle of 60 o relative to the incident beam. The Compton wavelength of the electron is 0.00243 nm . The kinetic energy of scattered elections in eV is___________ Ans:

25

Solution: λ = 0.24 nm, λ C = 0.00243 and θ = 600 ∵ λ′ − λ

= λC (1 − cos θ ) ⇒ λ ′ = λ + λC (1 − cosθ )

1 ⎛ 1⎞ ⇒ λ ′ = 0.24 + 0.00243 ⎜1 − ⎟ = 0.24 + 0.00243 × = 0.24 + 0.00121 = 0.2412 nm 2 ⎝ 2⎠ Kinetic Energy of scattered electron K .E. =

⇒ K .E. = ⇒ K .E. = Q9.

19.8 ×10−26 10

−9

396 ×10 −20 1.6 ×10−19

hc

λ

−

1 ⎞ 1 ⎛ 1 = 6.6 ×10−34 × 3 ×108 ⎜ − ⎟ × −9 Joules λ ′ 0.24 0.2412 ⎝ ⎠ 10

hc

( 4.17 − 4.15 ) = eV = 24.75 eV

19.8 ×10 −26 10

−9

× 0.02 = 396 ×10 −20 Joules

A diode at room temperature (kT = 0.025 eV ) with a current of 1 A has a forward bias voltage V F = 0.4V . For V F = 0.5V , the value of the diode current (in A ) is _________

Ans:

54.5

Solution: I = I 0 ( e

V / V T

Q10.

− 1) ⇒

I 2 I 1

(e = (e

V2 / V T V1 / V T

− 1)

(e = − 1) ( e

0.5 / 0.025 0.4 / 0.025

− 1)

(e = − 1) ( e

20

− 1)

16

− 1)

= 54.5 ⇒ I 2 = 54.5 μ A

GaAs has a diamond structure. The number of Ga-As bonds per atom which have to be broken to fracture the crystal in the (001) plane is _______

Ans:

4




Solution: Diamond structure has tetrahedral bond. To fracture the diamond structure along

( 0 0 1)

plane, four bonds need to be broken.

Q.11 – Q.20 Carry two marks each.

Q11.

In the thermodynamic cycle shown in the figure, one mole of a monatomic ideal gas is taken through a cycle. AB is a reversible isothermal expansion at a temperature of 800 K in which the volume of the gas is doubled. BC is an isobaric

the pressure and temperature return to their initial values. The

e r u s s e r P P1

net amount of heat (in Joules) absorbed by the gas in one

P2

contraction to the original volume in which the temperature is reduced to 300 K . CA is a constant volume process in which

complete cycle is _____________ Ans:

A C V

B

2V Volume

452

Solution: Process A → B is isothermal expansion T A = 800 K , VA , PA and T B = 800 K , VB = 2V A , PB =

P A

2

Process B → C is isobaric PC = PB =

P A

2

, VC = V A , TC = 300 K

C → A is Isochoric

⎛ V ⎞ ΔQ1 = nRT A ln ⎜ B ⎟ = 4602 J ⎝ V A ⎠ ΔQ2 = nCP ΔT = ΔQ3 =

R

⎛ γ ⎞ =⎜ R ( 300 − 800 ) = −10344 J (γ − 1) ⎝ γ − 1⎟⎠

nγ RΔT

R

× 500 = 6194 J (800 − 300 ) = (γ − 1) ( γ − 1)

Total heat exchange is Q1 + Q2 + Q3 = 452




Q12.

In a region of space, a time dependent magnetic field B (t ) = 0.4t Tesla points vertically upwards. Consider a horizontal, circular loop of radius 2 cm in this region. The magnitude of the electric field (in mV / m ) induced in the loop is ___________.

Ans:

4

−  ∂ B r ∂B 2 ×10 2 2 × π r ⇒ E = = Solution: E × 2π r = − 0.4 = 4 mV / m ∂t 2 ∂t 2 Q13. A plane electromagnetic wave of frequency 5 × 1014 Hz and amplitude 103 V / m traveling



in a homogeneous dielectric medium of dielectric constant 1.69, is incident normally at the interface with a second dielectric medium of dielectric constant 2.25. The ratio of the amplitude of the transmitted wave to that of the incident wave is __________. Ans:

0.93

2 ε r 1 ⎛ 2n1 ⎞ E 0T ⎛ ⇒ =⎜ E ⎟ 0 I E 0 I ⎜⎝ ε r1 + ε r 2 ⎝ n1 + n2 ⎠

Solution: E0T = ⎜ Q14.

⎞ ⎛ ⎞ 2 1.69 ⎟⎟ = ⎜⎜ ⎟⎟ = 0.93 ⎠ ⎝ 1.69 + 2.25 ⎠

For the arrangement given in the following figure, the coherent light sources A, B and C have individual intensities of 2 mW / m 2, 2 mW / m 2 and 5 mW / m 2 respectively at point P . The wavelength of each of the sources is 600 nm . The resultant intensity at point P P

(in mW / m 2 ) is ___________.

15 mm A

3.22 mm 2.04 mm

B

1m

C

Ans:

9.23 mw/ m

2

p

Solution: The electric field on the screen is the sum of the fields produced by the slits individually. A

E = E1 + E2 + E 3 iδ

= A + Ae + Be where δ =

2πd

λ

y d

iaδ

B

D

O

ad C

sin θ




The total intensity at θ is I = EE = 2 A + B + 2 A cos δ + 2 AB ⎡⎣cos ( aδ ) + cos (1 − a ) δ ⎤⎦ *

2

2

2

2π d y 3.22 ×10−3 15 ×10 −3 × = 2π × × = 505.7 where δ = sin θ ≅ θ = 6 ×10−7 1 λ λ λ D 2π d

2π d

δ = 145.80

given, A2 = 2 mw / m , B 2 = 5 mw / m 2 , d = 3.22 mm, ad = 2.04 mm, a = 0.6335 mm

∴ I = 2 × 2 × 10−3 + 5 ×10 −3 + 2 × 2 ×10 −3 cos (δ ) + 2 2 5 ×10 −3 ⎡⎣cos aδ + cos (1 − a ) δ ⎤⎦ = 9.23 × 10 −3 w / m2 I = 9.23 mw / m 2

Q15.

One gram of ice at 0 o C is melted and heated to water at 39 o C . Assume that the specific heat remains constant over the entire process. The latent heat of fusion of ice is 80 Calories/gm. The entropy change in the process (in Calories per degree) is _________.

Ans:

0.39

Solution: ΔS 1 =

ML T

=

1× 80 273

⇒ ΔS = ΔS1 + ΔS 2 ⇒ ΔS = Q16.

302

dT

273

T

∫

, ΔS2 = MC 80 273

+ 1.1ln

⇒ Δ S 2 = 1.1ln

302 273

302 273

= 0.29 + 0.1 = 0.39

A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 o to the horizontal. The linear acceleration of the disk (in m / sec 2 ) is _____________.

Ans:

3.266

Solution: Equation of Motion mg sin θ − f = ma Torque = fR = I α 2 ⎛ a mR ⎞ mg sin θ − = ma , ⎜ α = , I = ⎟ R R 2 ⎠ ⎝

I α

a =

g

3

= 3.266




Q17.

A nozzle is in the shape of a truncated cone, as shown in the figure. The area at the wide end is 25 cm 2 and the narrow end has an area of 1 cm 2 . Water enters the wider end at a rate of 500 gm / sec . The

50 cm

height of the nozzle is 50 cm and it is kept vertical with the wider end at the bottom. The magnitude of the pressure difference in kPa ( 1 kPa = 10 3 N / m 2 ) between the two ends of the nozzle is __________ Ans:

17.5

Solution: According to Bernoulli’s equation Pb + ρ ghb +

1 1 ρVb2 = Pt + ρ ght + ρ Vt 2 2 2

Pt , Vt , At

1

⇒ Pb − Pt = ρ g ( ht − hb ) + ρ (Vt 2 − Vb2 ) 2

ht − hb = 50 cm

Now given ρ AV t t = 500 gm / sec

⇒ V t =

500 × 10 −3 kg / sec ρ × At

=

500 × 10 −3 kg / sec 1000 kg / m3 × 10 −4 m 2

Pb , Vb , Ab

⇒ Vt = 5 m / sec According to equation of continuity AV t t = AbVb ⇒ Vb =

At Ab

V t ⇒ Vb =

1 cm 2 25 cm 2

∴ ΔP = Pb − Pt = 1000 × 10 × 50 × 10 −2 +

× 5 m / sec = 0.2 m / sec 1 2

2 × 1000 × ⎡⎣5 2 − (0.2 ) ⎤⎦

⇒ ΔP = 5000 + 500 ( 25 − 0.04 ) = 5000 + 12480 = 17480 N / m2 ⇒ ΔP = 17.5 kPa Q18.

A block of mass 2 kg is at rest on a horizontal table The coefficient of friction between the block and the table is 0.1. A horizontal force 3 N is applied to the block The speed of the block (in m/s) after it has moved a distance 10 m is ________________.

Ans:

3.225

Solution: f r = μ N = 0.1 × 2 × 10 = 2 N

∵m =

2 kg



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