• •
• •
•
• •
• •
• •
•
π⎞ ⎛ circuit ⎜∵φ = ⎟ . This current is refered as wattless ⎝ 2⎠ current. Power is dissipated only in resistor even circuit has RL, RC or LCR combination.
Resistance of an ideal coil is zero. LC Oscillation The oscillation of energy between capacitor (electric field energy) and inductor (magnetic field energy) is called LC-oscillation. 1 Frequency of oscillation υ = 2π LC If charge varies sinusoidally with time t as q = q0 cos ωt then current varies periodically with t π⎞ dq ⎛ as I = = q0ω cos ⎜ ωt + ⎟ 2⎠ dt ⎝ If initial charge on the capacitor is q0 then electrical
1 q02 2C If the capacitor is fully discharged, then total electrical energy is stored in the inductor in the form of magnetic energy. 1 U B = LI02 where I0 = Maximum current 2 Transformer A transformer is an electrical device which is used for changing alternating voltages. It is based on the phenomenon of mutual induction. If it is assumed that there is no loss of energy in the transformer then the power input = the power output, and since P = I × V then IPVP = ISVS Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. I P VS N S = = I S VP N P A transformer affects the voltage and current. We have energy stored in capacitor is U E =
•
T
1 (V I sin2 ωt cos φ − V0 I0 sin ωt cos ωt sin φ)dt T∫ 0 0 0 V I cos φ
= 0 0 ⇒ < P > = Vrms Irms cos φ 2 rms power Prms = Vrms Irms Average power Power factor (cos φ) = rms power R cos φ = Z Power dissipation is maximum in resistive circuit or at resonance in a LCR series circuit. No power is dissipated in purely inductive or capacitive circuit even a current is flowing in the
=
•
Choke Coil Circuit with a choke coil is a series L – R circuit. If resistance of choke coil = r (very small). Current in V the circuit, I = with Z = (R + r )2 + (ωL)2 Z It has a high inductance and negligible resistance coil. It is used to control current in ac circuit at negligible power loss. V r r ≈ →0 ∴ cos φ = = 2 2 2 Z ω L r +ω L
•
•
•
•
PHYSICS FOR YOU | OCTOBER ‘16
49
⎛N VS = ⎜ S ⎝ NP •
•
•
•
⎞ ⎛ NP ⎟VP and I S = ⎜ N ⎠ ⎝ S
⎞ ⎟ IP ⎠
Now, if NS > NP, the voltage is stepped up (VS > VP). This type of arrangement is called a stepup transformer. If NS < NP, we have a step-down transformer. In this case VS < VP and IS > IP. The voltage is stepped down and the current is increased. Displacement Current Maxwell assumed that a current also flows in the gap between the two plates of a capacitor, during the process of charging, known as displacement current ID. This displacement current originates due to time varying electric field between the plates of capacitor and is given by dφ I D = ε0 ε dt where φε is the electric flux linked with the space between the two plates of the capacitor. Using the concept of displacement current ID, Ampere’s circuital law can be modified as
∫ B ⋅ dl = μ0 (IC + I D ) •
f f
•
•
•
50
•
∫ B ⋅ dA = 0 (Gauss’s law for magnetism) dφ ∫ E ⋅ dl = − dtB (Faraday’s law)
•
•
•
dφ E dt (Ampere-Maxwell law)
Velocity of electromagnetic waves in free space is given by, 1 c= = 3 × 108 m s −1 μ0 ε0 The instantaneous magnitude of the electric and magnetic field vectors in electromagnetic wave are PHYSICS FOR YOU | OCTOBER ‘16
•
με μ 0 ε0
The energy is equally shared between electric field and magnetic field vectors of electromagnetic wave. Therefore the energy density of the electric field, 1 uE = ε0 E 2 and 2 1 B2 . the energy density of magnetic field, uB = 2 μ0 Average energy density of the electric field, 1 < uE > = ε0 E02 and average energy density of the 4 B2 1 magnetic field < uB > = 0 = ε0 E02 . 4μ 0 4 Average energy density of electromagnetic wave is =
∫ B ⋅ dl = μ0 IC + μ0ε0
Electromagnetic Waves An electromagnetic wave is a wave radiated by an accelerated charge which propagates through space as coupled electric and magnetic fields, oscillating perpendicular to each other and to the direction of propagation of the wave.
In a medium of refractive index n, the velocity v of an electromagnetic wave is given by 1 c 1 1 , Also v = v= = ⋅ n n μ0 ε0 με So that n =
•
Maxwell’s equations: q f ∫ E ⋅ dA = ε0 (Gauss’s law for electricity) f
related as |E| = c or E = Bc |B|
B2 1 ε0 E02 = 0 . 2 2μ0
Intensity of electromagnetic wave is defined as energy crossing per unit area per unit time perpendicular to the directions of propagation of electromagnetic wave. The intensity I is given by the relation 1 I = < u > c = ε0 E02c 2 The electromagnetic wave also carries linear momentum with it. The linear momentum carried by the portion of wave having energy U is given by p = U/c.
•
If the electromagnetic wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U/c to the surface.
•
If the incident wave is totally reflected from the surface, the momentum delivered to the surface is U/c – (–U/c) = 2U/c. It follows that the electromagnetic wave incident on a surface exerts a force on the surface.
1. Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I = I0 sin ωt, where I0 = 10 A and ω = 100 π rad s–1. The maximum value of emf in the second coil is (in V) (a) 2π (b) 5π (c) 6π (d) 12π 2. An express train takes 16 hours to cover the distance of 960 km between Patna and Gaziabad. The rails are separated by 130 cm and the vertical component of the earth’s magnetic field is 4.0 × 10–5 T. If the leakage resistance between the rails is 100 Ω, the retarding force on the train due to the magnetic field will be (a) 5 × 10–10 N (b) 8 × 10–10 N –5 (c) 15 × 10 N (d) 5 × 10–5 N 3. An inductor of inductance L = 400 H and resistors of resistances R1 = 2 Ω and R2 = 2 Ω are connected to a battery of emf 12 V as shown in figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is 12 –3t (a) 6e–5t V (b) e V t (c) 6(1 – e–t/0.2) V (d) 12e–5t V 4. A uniform magnetic field B exists in a direction perpendicular to the plane of a square frame made of copper wire. The wire has a diameter of 2 mm and a total length of 40 cm. The magnetic field changes with time at a steady rate dB/dt = 0.02 T s–1. What will be the current induced in the frame? (Resistivity of copper = 1.7 × 10–8 Ω m) (a) 0.1 A
(b) 0.2 A
(c) 0.3 A
(d) 0.4 A
5. In a uniform magnetic field of induction B, a wire in the form of semicircle of radius r rotates about the diameter of the circle with angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, then mean power generated per period of rotation is
Bπr 2 ω (Bπr 2 ω)2 (b) 2R 8R 2 (Bπr ω) (Bπr ω2 )2 (c) (d) 2R 8R 6. If a resistance of 100 Ω, an inductance of 0.5 H and a capacitance of 10 × 10–6 F are connected in series through 50 Hz ac supply, the impedance will be (a)
(a) 1.87 Ω (c) 18.7 Ω
(b) 101.3 Ω (d) 189.7 Ω
7. Two circular loops of equal radii are placed coaxially at some separation. The first loop is cut and a battery is inserted in between to drive a current in it. The current changes slightly because of the variation in resistance with temperature. During this period, the two loops (a) attract each other (b) repel each other (c) do not exert any force on each other (d) attract or repel each other depending on the sense of the current. 8. The instantaneous values of alternating current and voltage in a circuit are given as 1 1 I= sin (100πt) A and ε = sin (100πt + π/3) V 2 2 The average power in watts consumed in the circuit is 1 3 1 1 (b) (c) (d) 4 4 2 8 9. An electromagnetic wave of frequency 3 MHz passes from vacuum into a dielectric medium with permittivity εr = 4. Then (a) the wavelength and frequency both remain unchanged (b) the wavelength is doubled and the frequency remains unchanged (c) the wavelength is doubled and the frequency becomes half (d) the wavelength is halved and the frequency remains unchanged. (a)
PHYSICS FOR YOU | OCTOBER ‘16
51
10. The r.m.s. value of the electric field of the light coming from the sun is 720 N C–1. The average total energy density of the electromagnetic wave is (a) 3.3 × 10–3 J m–3 (b) 4.58 × 10–6 J m–3 (c) 6.37 × 10–9 J m–3 (d) 81.35 × 10–12 J m–3 11. The amplitude of the electric field in a parallel beam of light of intensity 2.0 W m–2 is (a) 38.8 N C–1 (b) 49.5 N C–1 –1 (c) 32.7 N C (d) 35.5 N C–1 12. A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving (a) along the electric field (b) along the magnetic field (c) along the direction of propagation of the wave (d) in a plane containing the magnetic field and the direction of propagation. 13. An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340t. The power loss in ac circuit is (a) 0.76 W (b) 0.89 W (c) 0.51 W (d) 0.67 W [NEET Phase I 2016] 14. A small signal voltage V(t) = V0 sinωt is applied across an ideal capacitor C (a) Current I(t) is in phase with voltage V(t) (b) Current I(t) leads voltage V(t) by 180° (c) Current I(t), lags voltage V(t) by 90° (d) Over a full cycle the capacitor C does not consume any energy from the voltage source. [NEET Phase I 2016] 15. A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10–3 Wb. The self-inductance of the solenoid is (a) 2 H (b) 1 H (c) 4 H (d) 3 H [NEET Phase I 2016] 16. Out of the following options which one can be used to produce a propagating electromagnetic wave? (a) A chargeless particle (b) An accelerating charge (c) A charge moving at constant velocity (d) A stationary charge [NEET Phase I 2016] 17. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (r.m.s.), 52
PHYSICS FOR YOU | OCTOBER ‘16
50 Hz ac supply, the series inductor needed for it to work is close to (a) 80 H (b) 0.08 H (c) 0.044 H (d) 0.065 H [JEE Main Offline 2016] 18. A series LR circuit is connected to a voltage source with V(t) = V0 sinωt. After very large time, current ⎛ ⎝
I(t) behaves as ⎜ t0 >>
L⎞ R ⎠⎟
(a)
(b)
(c)
(d) [JEE Main Online 2016]
19. A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which −
t
varies with time as B = B0e τ , where B0 and τ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time (t → ∞) is (a)
π 2r 4 B04 2 τR
(b)
π 2r 4 B02 2 τR
(c)
π 2r 4 B02 R τ
(d)
π 2r 4 B02 τR
[JEE Main Online 2016] 20. Microwave oven acts on the principle of (a) giving rotational energy to water molecules (b) giving translational energy to water molecules (c) giving vibrational energy to water molecules (d) transferring electrons from lower to higher energy levels in water molecule. [JEE Main Online 2016] SOLUTIONS d dI 1. (b) : As |ε| = M = M (I0 sin ωt) = MI0 ω cos ωt, dt dt εmax = MI0ω = (0.005)(10)(100π) V = 5π V 2. (a) : As the train moves in a magnetic field, a motional emf ε = vBl is produced across its width. Here B is the component of the magnetic field in a direction perpendicular to the plane of the motion, i.e., the vertical component.
960 km = 16.67 m s −1 16 h Thus, ε = (16.67 m s–1) (4.0 × 10–5 T)(1.3 m) = 8.6 × 10–4 V The leakage current is I = ε/R and the retarding force is The speed of the train is v =
F = IlB =
8.6 × 10−4 V × 1.3 m × 4.0 × 10−5 T 100 Ω
= 4.47 × 10–10 N 5 × 10−10 N 3. (d) : If I1 is the current through R1 and I2 is the ε current through L and R2, then I1 = and R1 I2 = I0(1 – e–t/), L 400 × 10 −3 = 0.2 s = Where τ = R2 2 ε 12 =6A and I0 = = R2 2 Thus, I2 = 6(1 – e–t/0.2) Potential drop across L, i.e., ε – R2I2 = 12 V – 2 × 6(1 – e–t/0.2) V = (12e–5t) V 4. (a) : Here, total length l = 40 cm = 40 × 10–2 m, Resistivity = 1.7 × 10–8 Ω m The area A of the loop ⎛ 40 cm ⎞ ⎛ 40 cm ⎞ 2 =⎜ ⎟⎜ ⎟ = 0.01 m ⎝ 4 ⎠⎝ 4 ⎠ If the magnetic field at an instant is B, the flux through the frame at that instant will be φ = BA. As the area remains constant, the magnitude of the emf induced will be dφ dB ε= =A dt dt = (0.01 m) (0.02 T s–1) = 2 × 10–4 V The resistance of the loop is (1.7 × 10−8 Ω m)(40 × 10−2 m) 3.14 × 1 × 10−6 m2
= 2.16 × 10−3 Ω
Hence, the current induced in the loop will be I=
2 × 10−4 V −3
2.16 × 10 Ω
= 9.3 × 10−2 A ≈ 0.1 A
⎛1 2⎞ 5. (b) : As φB = BA cos ωt = B ⎜ πr ⎟ cos ωt, ⎝2 ⎠ dφ B 1 ε=– = Bπr2ω sin ωt dt 2
Instantaneous power, P = T
Pav =
∫0
Pdt T
=
ε2 (Bπr 2 ω)2 = sin2 ωt R 4R
(Bπr 2 ω)2 (T / 2) 4R T T (as ∫0 sin2 ωt = T / 2)
(Bπr 2 ω)2 8R 6. (d) : As XL = 2πυL = 2π (50)(0.5) = 157.1 Ω, 1 1 = Ω = 318.4 Ω XC = 2πυC 2π(50)(10 −5 ) =
|XL – XC| = 161.3 Ω Z = R2 + ( X L − XC )2 = (100)2 + (161.3)2 Ω = 189.7 Ω 7. (a) 8. (d) : As εrms = I0
ε0
(1 / 2 )
=
2 (1 / 2 )
2
=
1 V, 2
1 A, 2 2 2 1 and cosφ = cos π/3 = 2 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ 1 Pav = εrmsIrms cosφ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = W ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 8 Irms =
=
=
9. (d) : Frequency remains unchanged with change of medium. c 1 / ε0 μ0 = εr μr Refractive index, n = = v 1 / εμ Since μr is very close to 1, n = εr = 4 = 2 λ λ Thus, λmedium = = n 2 1 1 2 10. (b) : uav = ε0 E02 = ε0 ( 2 Erms )2 = ε0 Erms 2 2 E (as Erms = 0 ) 2 = (8.85 × 10–12)(720)2 = 4.58 × 10–6 J m–3 11. (a) : The intensity of a plane electromagnetic wave is 1 I = uav c = ε0E02c 2 2I 2 × 2. 0 or E0 = = ε0 c 8.85 × 10 −12 × 3 × 108 = 38.8 N C –1 PHYSICS FOR YOU | OCTOBER ‘16
53
12. (a) 13. (c) : Here, L = 20 mH = 20 × 10–3 H, C = 50 μF = 50 × 10–6 F R = 40 Ω, V = 10 sin 340t = V0 sinωt ω = 340 rad s–1, V0 = 10 V XL = ωL = 340 × 20 × 10–3 = 6.8 Ω 1 1 104 XC = = = = 58.82 Ω ωC 340 × 50 × 10−6 34 × 5
Arc lamp will glow if I = 10 A, ∴
15. (b) : Here, N = 1000 , I = 4 A , φ0 = 4 × 10–3 Wb Total flux linked with the solenoid, φ = Nφ0 = 1000 × 4 × 10–3 Wb = 4 Wb Since, φ = LI ∴ Self-inductance of solenoid, φ 4 Wb L= = =1 H I 4A 16. (b) : An accelerating charge is used to produce oscillating electric and magnetic fields, hence the electromagnetic wave. 17. (d) : For a dc source I = 10 A, V = 80 V Resistance of the arc lamp, V 80 R= = =8 Ω I 10 For an ac source, εrms = 220 V υ = 50 Hz ω = 2π × 50 = 100 π rad s–1
54
PHYSICS FOR YOU | OCTOBER ‘16
εrms
ε or R2 + ω2 L2 = ⎜⎛ rms ⎟⎞ 2 2 2 ⎝ I ⎠ R +ω L
⎛ 220 ⎞ or 82 + (100 π)2 L2 = ⎜ ⎝ 10 ⎟⎠ 2
2
2
2
22 − 8 or L2 = (100 π)2
2 2 Z = R2 + (XC − X L )2 = (40) + (58.82 − 6.8)
= (40)2 + (52.02)2 = 65.62 Ω The peak current in the circuit is V 10 40 ⎞ R I0 = 0 = A , cos φ = = ⎜⎛ ⎟ Z 65.62 Z ⎝ 65.62 ⎠ Power loss in ac circuit, 1 = Vrms Irms cos φ = V0 I0 cos φ 2 1 10 40 = × 10 × × = 0.46 W 2 65.62 65.62 14. (d) : When an ideal capacitor is connected with an ac voltage source, current leads voltage by 90°. Since, energy stored in capacitor during charging is spent in maintaining charge on the capacitor during discharging. Hence over a full cycle, the capacitor does not consume any energy from the voltage source.
I=
∴
L=
30 × 14 = 0.065 H 100 π
18. (d) : Current in LR circuit is I=
π⎞ ⎛ sin ⎜ ωt − ⎟ 2⎠ ⎝ R2 + ω2 L2 V0
i.e., it is sinusoidal in nature. −
t
19. (b) : Here, B = B0e τ Area of the circular loop, A = πr2 Flux linked with the loop at any time, t, −
t
φ = BA = πr 2 B0e τ t
dφ 1 − Emf induced in the loop, ε = − = πr 2 B0 e τ dt τ Net heat generated in the loop ∞ 2
∞
0
0
2t
π2r 4 B02 − τ ε = ∫ dt = ∫ e dt R τ2 R =
π2r 4 B02
=
−π2r 4 B02
τ2 R
2τ2 R
×
1 2t ×⎡ − τ ⎢ ⎛ 2 ⎞ ⎣e ⎜− τ ⎟ ⎝ ⎠ × τ(0 − 1) =
∞
⎤ ⎥⎦ 0
π2r 4 B02 2τR
20. (a)
ANSWER KEY
MPP-4 CLASS XI 1. 6. 11. 16. 21. 26.
(d) (c) (b) (a) (b,c) (4)
2. 7. 12. 17. 22. 27.
(a) (c) (a) (d) (a,b,c) (a)
3. 8. 13. 18. 23. 28.
(c) (c) (a) (c) (b,d) (d)
4. 9. 14. 19. 24. 29.
(d) (d) (c) (a) (4) (b)
5. 10. 15. 20. 25. 30.
(a) (a) (b) (a,c) (2) (c)
CLASS XII Series 5
Optics
Time Allowed : 3 hours Maximum Marks : 70
GENERAL INSTRUCTIONS (i)
All questions are compulsory.
(ii)
Q. no. 1 to 5 are very short answer questions and carry 1 mark each.
(iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each. (iv) Q. no. 11 to 22 are also short answer questions and carry 3 marks each. (v)
Q. no. 23 is a value based question and carries 4 marks.
(vi) Q. no. 24 to 26 are long answer questions and carry 5 marks each. (vii) Use log tables if necessary, use of calculators is not allowed.
SECTION-A 1. For what angle of incidence, the lateral shift produced by a parallel sided glass slab is maximum? 2. Light of wavelength 6000 Å in air enters a medium of refractive index 1.5. What will be its frequency in the medium? 3. No interference pattern is detected when two coherent sources are infinitely close to one another. Why? 4. Differentiate a ray and a wavefront. 5. Why are mirrors used in searchlights (or car headlights) parabolic in shape and not concave spherical? SECTION-B
6. A light ray travels from medium 1 of refractive index μ1 to medium 2 of refractive index μ2, where μ2 < μ1. Write an expression for critical angle of incidence. 7. A convex lens made of material of refractive index μ2 56
PHYSICS FOR YOU | OCTOBER ‘16
Previous Years Analysis 2016 2015 2014 Delhi AI Delhi AI Delhi AI VSA SA-I SA-II VBQ LA
– 1
1 1
1 1
2 _
2 _
2 _
1
1
1
1 1
1 1
1 1
2
2 _
2 _
1
1
– 1
is held in a reference medium of refractive index μl. Trace the path of a parallel beam of light passing through the lens when (a) μl = μ2 , (b) μ1 < μ2 and (c) μl > μ2 . 8. What is the focal length of a combination of a convex lens of focal length 30 cm and a concave lens of focal length 20 cm ? Is the system a converging or diverging lens? Ignore thickness of the lenses. 9. In a double-slit interference experiment, the two coherent beams, have slightly different intensities I and (I + δI) where δI << I. Show that the resultant intensity at the maxima is nearly 4I, while at the minima is nearly
| δ I |2 4I
.
OR The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9:25. Find the ratio of the widths of the two slits.
10. A radar wave has frequency of 8.1 × 109 Hz. The reflected wave from an aeroplane shows a frequency difference of 2.7 × 103 Hz on the higher side. Deduce the velocity of aeroplane in the line of sight. SECTION-C
11. A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the magnifying power of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain. 1 1 1 = − for a concave f v u lens, using the necessary ray diagram.
12. Derive the lens formula,
13. Draw a graph to show the variation of the angle of deviation δ with the angle of incidence i for a monochromatic ray of light passing through a glass prism of refracting angle A. Hence deduce the relation ⎛δ + A⎞ sin ⎜ m ⎝ 2 ⎟⎠ . μ= ⎛A⎞ sin ⎜ ⎟ ⎝2⎠ 14. Define angular dispersion and dispersive power. Derive the expressions for these quantities in terms of refractive index. 15. A convex lens, of focal length 20 cm, is placed co-axially with a convex mirror of radius of curvature 20 cm. The two are kept 15 cm apart from each other. A point object is placed 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determined the nature and position of the image formed. 16. A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
17. State the condition for diffraction of light to occur. In a single slit diffraction pattern, how does the angular width of central maximum change, when (a) slit width is decreased, (b) distance between the slit and screen is increased and (c) light of smaller visible wavelength is used? Justify your answer in each case. 18. Two wavelengths of sodium light, 590 nm and 596 nm, are used in turn, to study the diffraction taking place at a single slit of aperture 2.0 × 10–4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maximum of the diffraction pattern obtained in the two cases. 19. Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids ? 20. Figure shows a modified Young's double slit λ experimental set up. Here SS2 − SS1 = . 4
(a) State the condition for constructive and destructive interference. (b) Obtain an expression for the fringe width. (c) Locate the position of the central fringe. OR What do you mean by coherent and incoherent sources of light? Why are coherent sources required to produce interference of light? State the conditions, which must be satisfied for two light sources to be coherent. 21. The absolute refractive index of air is 1.0003 and wavelength of yellow light in vacuum is 6000 Å. Find the thickness of air column which will contain one more wavelength of yellow light than in the same thickness of vacuum. PHYSICS FOR YOU | OCTOBER ‘16
57
22. Explain the phenomenon of total internal reflection. Under what conditions does it take place? How totally reflecting prism can be used to (a) deviate a ray through 90° (b) deviate a ray through 180°. SECTION-D
23. Rama was watching a programme on moon on the discovery channel. He came to know from the observation recorded on the surface of the moon that sunrise and sunset are abrupt there and the sky appears dark from there. He was surprised and determined to know the reason behind it. He discussed it with his Physics teacher next day, who explained him the reason behind it. (a) What were the values being displayed by Rama? (b) Why are sunrise and sunset abrupt on the surface of the moon? (c) Why does the sky appear dark from the moon? SECTION-E
24. Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices μ1 and μ2. Establish the relation between the distances of the object, the image and the radius of curvature from the central point of the spherical surface. Hence derive the expression of the lens maker’s formula. OR (a) How does one demonstrate, using a suitable diagram, that unpolarised light when passed through a polaroid gets polarised? (b) A beam of unpolarised light is incident on a glass-air interface. Show, using a suitable ray diagram, that light reflected from the interface is totally polarised, when μ = tan iB, where μ is the refractive index of glass with respect to air and iB is the Brewster's angle. 25. Draw a ray diagram for a compound microscope. Derive an expression for the magnifying power when the final image is formed at the least distance of distinct vision. State the expression for the magnifying power when the image is formed at infinity. Why is the focal length of the objective lens of a compound microscope kept quite small? OR Draw a ray diagram for the formation of image of a distant object at least distance of distinct vision by 58
PHYSICS FOR YOU | OCTOBER ‘16
an astronomical telescope. Deduce the expression for its magnifying power. Write two basic features which can distinguish between a telescope and a compound microscope. 26. (a) Use Huygen's geometrical construction to show how a plane wavefront at t = 0 propagates and produces a wavefront at a later time. (b) Verify, using Huygen's principle, Snell's law of refraction of a plane wave propagating from a rarer to a denser medium. (c) Illustrate with the help of diagrams the action of (i) convex lens and (ii) concave mirror, on a plane wavefront incident on it. OR In Young’s experiment, deduce the conditions for the constructive and destructive interference pattern observed on the screen by drawing the necessary diagram. Hence establish the relation between the fringe width, the wavelength of the monochromatic source, separation between the two slits and the distance between the screen and the plane of the slits. If the set-up were to be put up in a medium optically denser than air, what effect would there be on the observed fringe width? Give reason for your answer. SOLUTIONS
1. For i = 90°, lateral shift is maximum and is equal to the thickness of the slab. t sin(i − r ) d= cos r dmax =
t sin(90° − r ) cos r
=
t cos r cos r
=t
2. Frequency in air, υ=
c 3 × 108 = = 5 × 1014 Hz λ 6000 × 10−10
The frequency of light remains same as it travels from air to the given medium. ∴ Frequency in medium = 5 × 1014 Hz λD d 1 i.e., β ∝ , when d → 0, β → ∞ d Fringe width is very large. Even a single fringe may occupy the entire screen. The interference pattern cannot be observed.
3. Fringe width, β =
4. A wavefront is surface of constant phase but a ray is a line drawn normal to the wavefront and pointing in the direction of propagation of the wave or wavefront. 5. A spherical mirror of large aperture suffers with the spherical aberration. However, a parabolic reflector is free from spherical aberration. Therefore, when a source of light is placed at the focus point of a parabolic mirror, it produces a perfect parallel beam of light which is visible even from a long distance. 6. As shown in figure, when i = ic , r = 90° Using Snell's law of refraction, μ1 sin ic = μ2 sin90° μ sinic = 2 μ1 or ic = sin
= 2 I + δI + 2 I 1 +
δI ⎞ ⎛ = 4I ⎜ neglecting δI and ⎟ ⎝ I ⎠ I min = [ I + δI − I ]2 = I[(1 + δI / I )1/2 − 1]2 2
OR I 9 Given, min = I max 25
or
2 9 ⎛ r − 1⎞ ⎜⎝ ⎟⎠ = 25 r +1
r −1 3 or 5r − 5 = 3r + 3 = r +1 5 a r = 4 = 1 , the amplitude ratio a2
or
⎞ ⎜⎝ μ ⎟⎠ 1
2
⎡⎛ | δI |2 δI ⎞ ⎤ ⎛ δI ⎞ = I ⎢ ⎜1 + ⎟ − 1⎥ = I ⎜ ⎟ = ⎝ 2I ⎠ 2I ⎠ ⎦ 4I ⎣⎝
or −1 ⎛ μ2
δI I
∴
Width ratio of the slits, w1 I1 a12 16 = = = = 16 : 1 w2 I2 a22 1
7. The ray diagrams are shown below
10. Here υ = 8.1 × 109 Hz, Δυ = 2.7 × 103 Hz v As Δυ = . υ c ∴ v= 8. Given, focal length of convex lens, f1 = + 30 cm Focal length of concave lens, f2 = –20 cm Focal length of the combination is given by 1 1 1 1 1 1 = + = + =− f f1 f2 30 −20 60 or f = – 60 cm The negative value of f indicates that the combination behaves as a diverging lens. 9. For any interference pattern, I max = (a1 + a2 )2 = ( I1 + I 2 )2 2 and I min = (a1 − a2 )2 = ( I1 − I2 )2 (as I ∝ a )
Given, I1 = I and I2 = (I + δI), 2
∴ I max = ( I + I + δI )
= I + ( I + δI ) + 2 I ( I + δI )
Δυ 2.7 × 103 .c = × 3 × 108 = 100 m s −1 9 υ 8.1 × 10
Since the velocity of the aeroplane determined by radar waves is double of its actual velocity of approach, therefore, Actual velocity of the aeroplane = 50 m s–1 = 180 km h–1 11. (a) Here, area of each square = 1 mm2 u = –9 cm, f = + 10 cm 1 1 1 As − = v u f ∴
1 1 1 1 1 1 = + = − =− v f u 10 9 90
or v = – 90 cm Magnitude of magnification is v 90 = = 10 9 u Area of each square in the virtual image = (10)2 × 1 = 100 mm2 = 1 cm2 m=
PHYSICS FOR YOU | OCTOBER ‘16
59
25 D = = 2.8 |u| 9 (c) No, Magnification of an image by a lens and angular magnification (or magnifying power) of a optical instrument are two separate concepts. The latter is the ratio of the angular size of the object to the angular size of the object if placed at the near point (25 cm). Thus magnification v 25 and magnifying power is . magnitude is u |u| Only when the image is located at the near (b) Magnifying power, M =
point |v| = 25 cm, the two quantities are equal as will be seen. 12. Suppose O be the optical centre and F be the principal focus of concave lens of focal length f. AB is an object placed perpendicular to its principal axis. A virtual, erect and diminished image A′B′ is formed due to refraction through the lens.
ΔA′B′O and ΔABO are similar A′ B′ B′O ...(i) ∴ = AB BO Also, ΔA′B′F and ΔMOF are similar A′ B′ FB′ ∴ = MO FO But MO = AB, therefore A′ B′ FB′ ...(ii) = AB FO From (i) and (ii), we get B′O FB′ FO − B′O = = BO FO FO Using new cartesian sing convention, we get BO = –u, B′O = –v, FO = – f −v − f + v = ∴ −f −u or vf = uf – uv or uv = uf – vf Dividing both sides by uvf, we get As
60
PHYSICS FOR YOU | OCTOBER ‘16
1 1 1 = − f v u This is the thin lens formula for a concave lens. 13. For a given prism and for a given colour of light, the angle δ depends on i only. As i increases, the angle of δ first decreases and reaches a minimum value δm and then increases. Clearly, any given value of δ corresponds to two angles of incidence i and i ′. This fact is expected from the symmetry of i and i ′ in the equation : δ = i + i′ – A i.e., δ remains the same as i and i′ are interchanged. Physically, it means that the path of the ray in figure (a) can be traced back, resulting in the same angle of deviation. The minimum value of the angle of deviation suffered by a ray of light on passing through a prism is called the angle of minimum deviation and is denoted by δm.
When a prism is in the position of minimum deviation, ray of light passes symmetrically through the prism so that i = i′, r = r′, δ = δm As A + δ = i + i′ A + δm ∴ A + δm = i + i or i = 2 Also A = r + r′ = r + r = 2r A ∴ r= 2 From Snell's law, the refractive index of the material of the prism will be A + δm sin sin i 2 μ= or μ = A sin r sin 2 14. When a beam of white light passes through a prism, it gets dispersed into its constituent colours. Let δV, δR and δ be the angles of deviation for violet, red and yellow (mean) colours respectively, as shown in figure.
∴ O′I1 = distance of virtual object I1 from convex mirror = OI1 – OO′ = (30 – 15) cm = 15 cm For the convex mirror : u2 = + 15 cm and R = +20 cm Using mirror formula, we get 1 1 2 1 1 2 + = i.e., + = v2 u2 R v2 15 20 δV = (μV –1) A δR = (μR – 1) A δ = (μ – 1) A where μV, μR and μ are the refractive indices of the prism material for violet, red and yellow (mean) colours, respectively. The angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. ∴ Angular dispersion = δV – δR = (μV –1) A – (μR – 1) A = (μV – μR)A Clearly, the angular dispersion produced by a prism depends upon (i) angle of the prism and (ii) nature of the material of the prism. Dispersive power is the ability of the prism material to cause dispersion. It is defined as the ratio of the angular dispersion to the mean deviation. ∴ Dispersive power, Then
ω=
Angular dispersion
Mean deviation μ − μR or ω = V μ −1
δ − δR = V δ
15. The ray diagram, for the image formed by the combination, is shown in figure.
For the convex lens = u1 = –60 cm and f = +20 cm Using thin lens formula, we get 1 1 1 1 1 1 1 − = or = − = v1 − 60 20 v1 20 60 30 or v1 = 30 cm In the absence of the mirror, the lens would have formed the image of P at I1, which acts as a virtual object for the convex mirror.
or
1 1 1 1 = − = v2 10 15 30
∴ v2 = +30 cm Hence the final image I is a virtual image formed at a distance of 30 cm behind the convex mirror. 16. Image formed by concave mirror acts as a virtual object for convex mirror. Here parallel rays coming from infinity will focus at 110 mm an axis away from concave mirror.
Distance of virtual object for convex mirror = 110 – 20 = 90 mm For convex mirror 1 1 1 u = 90 mm, f = 70 mm ⇒ + = v u f 1 1 1 1 1 = − = − v f u 70 90 ⇒ v = 315 mm Hence image is formed at 315 mm from smaller convex mirror. 17. Diffraction of light occurs when the size of the obstacle or aperture is comparable to the wavelength of the light. 2D λ Linear width of central maximum, β0 = d β0 2 λ Angular width of central maximum = = D d (a) When slit width d decreases, angular width increases. (b) When distance D between the slit and screen is increased, angular width does not change. (c) When light of smaller wavelength λ is used, angular width decreases. PHYSICS FOR YOU | OCTOBER ‘16
61
18. As per question, slit width, a = 2.0 × 10–4 m; distance of screen from the slit, D = 1.5 m; λ1 = 590 nm = 5.9 × 10–7 m and λ2 = 596 nm = 5.96 × 10–7 m We know that for first diffraction maxima, 3λ a sin θ = 2 3λ If θ is small, then sin θ ≈ θ = 2a ∴ Linear distance of first maximum from central maximum, 3 λD x = Dθ = 2a 3λ D For wavelength λ1, x1 = 1 2a 3λ D and for wavelength λ2, x2 = 2 2a 3D ⇒ x2 − x1 = ( λ − λ1 ) 2a 2 3 × 1. 5 = [5.96 × 10−7 − 5.90 × 10−7 ] m −4 2 × 2.0 × 10 = 0.0675 mm 19. Let intensity of polarised light after passing through the first polaroid P1 be I. When polaroid P′ is rotated between two crossed polaroids P1 and P2, let at a certain instant, P′ is inclined at an angle θ with P1 and an angle (90° – θ) from P2. ∴ Intensity of light after passing through P′ = I cos2θ and final intensity of light transmitted even through P2 = I′ = (I cos2θ). cos2 (90° – θ) ⇒ I′ = I cos2 θ sin2 θ I I = . 4 cos2 θ sin2 θ = sin2 (2θ) 4 4 From this result, we concluded that intensity of transmitted light is maximum when 2θ = 90°, so that sin 2θ = 1. Thus, for maximum transmitted intensity, the polaroid sheet must be placed at an angle of 45° between two crossed polaroids and I then, I ′ = . 4 20.
Initial path difference between SS1 and SS2, 62
PHYSICS FOR YOU | OCTOBER ‘16
λ 4 Path difference between disturbances from S1 and S2 at point P, xd Δ= D Total path difference between the two disturbances at P, λ xd ΔT = Δ 0 + Δ = + 4 D ∴ In general for constructive interference, ⎛λ x d⎞ ΔT = ⎜ + n ⎟ = nλ ; n = 0, 1, 2,... ⎝4 D ⎠ Δ 0 = SS2 − SS1 =
xnd ⎛ 1⎞ = ⎜n − ⎟ λ ⎝ D 4⎠ For destructive interference, λ ⎛ λ x′ d ⎞ ΔT = ⎜ + n ⎟ = (2n − 1) ⎝4 2 D ⎠ x ′n d ⎛ 1⎞ λ or = ⎜ 2n − 1 − ⎟ ⎝ D 2⎠ 2 or
or
...(i)
x ′n d ⎛ 3⎞ λ = ⎜ 2n − ⎟ ⎝ D 2⎠ 2
λD d The position x0 of central fringe is obtained by putting n = 0 in equation (i). Therefore, λD ∴ x0 = − 4d The negative sign shows that the central fringe is obtained at a point O′ below the (central) point O. Fringe width, β = xn + 1 − xn =
OR Two sources of light which continuously emit light waves of same frequency (or wavelength) with a zero or constant phase difference between them, are called coherent sources. Two sources of light which do not emit light waves with a constant phase difference are called incoherent sources. When two monochromatic waves of intensity I1, I2 and phase difference φ meet at a point, the resultant intensity is given by I = I1 + I 2 + 2 I1I 2 cos φ The last term 2 I1 I 2 cos φ is called interference term. There are two possibilities:
(a) If cos φ remains constant with time, the total intensity at any point will be constant. The 2 intensity will be maximum ( I1 + I 2 ) at points, where cos φ = +1 and minimum
( I1 − I 2 )2 at points where cos φ = –1. The sources in this case are coherent. (b) If cos φ varies continuously with time assuming both positive and negative values, then the average value of cos φ will be zero. Then interference term averages to zero. There will be same intensity, I = I1 + I2 at every point. The two sources in this case are incoherent. Hence to observe interference, we need to have two sources with the same frequency and with a stable phase difference. Such a pair of sources is called coherent sources. Conditions for obtaining two coherent sources of light : (a) The two sources of light must be obtained from a single source by some method. Then the relative phase difference between the two light waves from the sources will remain constant with time. (b) The two sources must give monochromatic light. Otherwise, different colours will produce different interference patterns and fringes of different colours will overlap. 21. Wavelength of yellow light in vacuum, λ = 6000 Å Wavelength of yellow light in air,
is partly reflected back into the denser medium and partly refracted to the rarer medium. This reflection is called internal reflection. Under certain conditions, the whole of the incident light can be made to be reflected back into the denser medium. This gives rise to an interesting phenomenon called total internal reflection.
Necessary conditions for total internal reflection (i) Light must travel from an optically denser to an optically rarer medium. (ii) The angle of incidence in the denser medium must be greater than the critical angle for the two media. (a) To deviate a ray through 90° : A right angled prism shown in figure (a), as the light is incident normally on one of the faces containing right angle, it enters the prism without deviation. It is incident on the hypotenuse face at an angle of 45°, greater than the critical angle. The light is totally internally reflected. Having been deviated through 90°, the light passes through third face without any further deviation. Such prisms are used in periscopes.
6000 λ Å = μ 1.0003 Let a thickness t of vacuum contain n waves and the same thickness t of air contain n + 1 waves. t t ...(i) Then n = = λ 6000 Å λ′ =
and n + 1 =
1.0003 t t = λ ′ 6000 Å
...(ii)
From equations (i) and (ii), we get 1.003 t t +1 = or t + 6000 Å = 1.0003 t 6000 Å 6000 Å or t = 2 × 107 Å = 2 mm 22. If light passes from an optically denser medium to rarer medium, then at the interface, the light
(b) To invert an image with deviation of rays through 180° : As shown in figure (b), the light is incident normally on the hypotenuse face, it first suffers total internal reflection from one shorter face and then from the other shorter face. The final beam emerges through the hypotenuse face, parallel to the incident beam. The deviation is 180°. Such a prism is called a porroprism. PHYSICS FOR YOU | OCTOBER ‘16
63
23. (a) Keen observer and curiosity. (b) Moon has no atmosphere. There is no refraction of light. Sunlight reaches moon straight covering shortest distance. Hence sunrise and sunset are abrupt. (c) Moon has no atmosphere, so there is nothing to scatter sunlight towards the moon. No skylight reaches the moon surface, sky appears dark in the day time as it does at night. 24. Refer to point 6.5 (5 (i)), 6.6 (1) page no. 372, 374 (MTG Excel in Physics) OR (a) Refer to point 6.15 (7) page no. 453 (MTG Excel in Physics) (b) Refer to point 6.15 (9) page no. 455 (MTG Excel in Physics) 25. Refer to point 6.9 (1 (iv)) page no. 381 (MTG Excel in Physics) Angular magnification of objective will be large when u0 is slightly greater than f0. Now a compound
64
PHYSICS FOR YOU | OCTOBER ‘16
microscope is used for viewing very close objects, so u0 is small. Consequently, f0 has to be small. OR Refer to point 6.9 (2) page no. 382 (MTG Excel in Physics) The two important differences between a telescope and a compound microscope are (a) The aperture of the objective of a microscope is very small while that of the telescope is large. (b) Both the lenses of a compound microscope have short focal length while the objective of a telescope has large focal length. 26. (a) Refer to point 6.11 (1) page no. 444 (MTG Excel in Physics) (b) Refer to point 6.11 (5) page no. 445 (MTG Excel in Physics) (c) Refer to point 6.12 (1, 3) page no. 445 (MTG Excel in Physics) OR Refer to point 6.13 (6, 7, 8 (v)) page no. 447, 448 (MTG Excel in Physics)
CHAPTERWISE MCQs FOR PRACTICE Useful for All National and State Level Medical/Engg. Entrance Exams ATOMS
1. The total energy of an electron in the excited state corresponding to n = 3 state is E. What is its potential energy with proper sign? (a) –2E (b) 2E (c) –E (d) E 2. The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of wavelength λ is emitted for a transition 3 → 1. What will be the wavelength of emission for transition 2 → 1? λ 4λ 3λ (a) (d) 3λ (b) (c) 3 3 4 3. In Bohr model of hydrogen atom, the ratio of period of revolution of an electron in n = 2 and n = 1 orbit is (a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1 4. A hydrogen atom and a Li2+ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then (a) lH > lLi and EH > ELi (b) lH = lLi and EH < ELi (c) lH = lLi and EH > ELi (d) lH < lLi and EH < ELi 5. The ratio of the longest and shortest wavelengths in Brackett series of hydrogen spectra is 9 4 25 17 (a) (b) (c) (d) 5 3 9 6 6. The electric potential between a proton and an ⎛r⎞ electron is given by V = V0 ln ⎜ ⎟ , where r0 is a ⎝ r0 ⎠
constant. Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum number. 1 (a) rn ∝ n (b) rn ∝ n 1 (c) rn ∝ n2 (d) rn ∝ 2 n 7. A hydrogen atom is excited up to 9th level. The total number of possible spectral lines emitted by the hydrogen atom is (a) 36 (b) 35 (c) 37 (d) 38 8. The radius of the hydrogen atom in its ground state is a0. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged muon with mass 207 times that of an electron, is aμ equal to a0 a0 (c) (d) a0 207 (a) 207a0 (b) 207 207 9. Which energy state of doubly ionised lithium has the same energy as that of the ground state of hydrogen ? (Given Z for lithium = 3) (a) 4 (b) 3 (c) 2 (d) 1 10. When the electron in hydrogen atom is excited from the 4th stationary orbit to the 5th stationary orbit, the change in the angular momentum of the electron is (Planck’s constant, h = 6.63 × 10–34 J s) (a) 4.16 × 10–34 J s (b) 3.32 × 10–34 J s (c) 1.05 × 10–34 J s (d) 2.08 × 10–34 J s 257 11. If the atom 100 Fm follows the Bohr model and the 257 Fm is N times the Bohr radius of fifth orbit of 100 radius, then the value of N is (a) 100 (b) 200 (c) 4 (d) 1/4 PHYSICS FOR YOU | OCTOBER ‘16
65
12. The angular momentum of electron in 3d orbital of an atom is ⎛ k ⎞ (a) 2 ⎜ ⎟ (b) 3 ⎜⎛ k ⎞⎟ ⎝ 2π ⎠ ⎝ 2π ⎠ ⎛ k ⎞ 6⎜ ⎟ ⎝ 2π ⎠
⎛ k ⎞ 12 ⎜ ⎟ ⎝ 2π ⎠ 13. Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about (a) 53 pm (b) 27 pm (c) 18 pm (d) 13 pm (c)
(d)
14. An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of (a) 1 Å (b) 10–10 cm –12 (c) 10 cm (d) 10–15 cm 15. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is (a) 1215 Å (b) 1640 Å (c) 2430 Å (d) 4687 Å NUCLEI
16. In the following reaction, the energy released is 1 4 4 1H → 2 He + 2e+ + Energy Given
1 : Mass of 1H = 1.007825 u 4 Mass of 2 He = 4.002603 u +
Mass of e = 0.000548 u
(a) 12.33 MeV (c) 25.7 MeV
(b) 24.67 MeV (d) 49.34 MeV
17. A radioactive isotope A with a half life of 1.25 × 1010 years decays into B which is stable. A sample of rock from a planet is found to contain both A and B present in the ratio 1 : 16. The age of the rock is (a) 9.6 × 1010 years (b) 4.2 × 1010 years (c) 5 × 1010 years (d) 1.95 × 1010 years 18. A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 3 : 1. The ratio of radii of the fragments is (a) 1 : 31/3 (b) 31/3 : 4 (c) 4 : 1
(d) 2 : 1
19. A radioactive nucleus emits 3α-particles and 5β-particles. The ratio of number of neutrons to that of protons will be 66
PHYSICS FOR YOU | OCTOBER ‘16
A−Z A − Z − 12 (b) Z −1 Z −6 A − Z − 11 (d) (c) A − Z − 11 Z −1 Z −6 20. A radioactive sample has half life of 5 days. To decay from 8 microcurie to 1 microcurie, the number of days taken will be (a) 40 (b) 25 (c) 15 (d) 10 21. A free neutron decays spontaneously into (a) a proton, an electron and anti-neutrino (b) a proton, an electron and a neutrino (c) a proton and electron (d) a proton, and electron, a neutrino and an anti-neutrino. 22. The radius of germanium (Ge) nuclide is measured 9 to be twice the radius of 4Be. The number of nucleons in Ge are (a) 72 (b) 73 (c) 74 (d) 75 (a)
23. Fusion reaction takes place at high temperature because (a) nuclei break up at high temperature (b) atoms get ionised at high temperature (c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei (d) molecules break up at high temperature. 24. To generate power of 3.2 MW, the number of fissions of 235U per minute is (Energy released per fission = 200 MeV, 1 eV = 1.6 × 10–19 J) (a) 6 × 1018 (b) 6 × 1017 17 (c) 10 (d) 6 × 1016 232
25. The 90Th atom has successive alpha and beta decays 208 to the end product 82Pb. The numbers of alpha and beta particles emitted in the process respectively are (a) 4, 6 (b) 4, 4 (c) 6, 2 (d) 6, 4 26. An element X decays first by positron emission and then two α-particles are emitted in successive radioactive decay. If the product nucleus has mass number 227 and atomic number 89, the mass number and atomic number of element X are (a) (273, 93) (b) (235, 94) (c) (238, 93) (d) (237, 92) 27. A radioactive isotope has a decay constant λ and a molar mass M. Taking the Avogadro constant to be L, what is the activity of a sample of mass m of this isotope?
λmL M mλ mL (c) (d) ML λM 28. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because (a) they will break up (b) elastic collision of neutrons with heavy nuclei will not slow them down (c) the net weight of the reactor would be unbearably high (d) substances with heavy nuclei do not occur in liquid or gaseous state at room temperature. (a) λmML
(b)
29. The half life of radioactive radon is 3.8 days. The time at the end of which (1/20)th of radon sample will remain undecayed (given log10 2 = 0.30103) is (a) 3.8 days (b) 16.5 days (c) 33 days (d) 76 days 30. A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life time of one species is τ and that of the other is 5τ. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as function of time. Which of the following figures best represents the form of this plot?
(a)
(d)
SOLUTIONS 1. (b) 2. (d) :
E3 – E1 = 2E – E = E From figure, E2 – E1 = E2 – E3 + E3 – E1
Tn =
4ε02 h3n3 me 4
Tn ∝ n3 3
∴
T2 ⎛ 2 ⎞ 8 =⎜ ⎟ = T1 ⎝ 1 ⎠ 1
4. (b) : In second excited state, n = 3, ⎛ h ⎞ So, lH = lLi = 3 ⎜ ⎟ ⎝ 2π ⎠ 2 while E ∝ Z and ZH = 1, ZLi = 3 So, ELi = 9 EH or EH < ELi 5. (a) : For Brackett series, 1 1⎤ ⎡1 = R⎢ − ⎥ 2 λ ⎣ 4 n2 ⎦ where n = 5, 6, 7, 8, ........ For longest wavelength, n = 5 1 ⎡1 1⎤ ∴ = R⎢ − ⎥ λ Longest ⎣ 42 52 ⎦ .... (i)
.... (ii)
⎝ r0 ⎠
∴
hc =E λ
3. (c) : In Bohr model of hydrogen atom, the period of revolution of electron in nth orbit is given as
6. (a) : Given : V = V0 ln ⎛⎜ r ⎞⎟
(Using (i))
9 ⎡1 1⎤ = R⎢ − ⎥ = R ⎣16 25 ⎦ 400 For shortest wavelength, n = ∞ 1 1 ⎤ R ⎡1 ∴ = R⎢ − ⎥= 2 λShortest ⎣4 ∞ 2 ⎦ 16 Dividing (ii) by (i), we get λ Longest R 400 25 = × = λShortest 16 9R 9
(b)
(c)
4 = E − 2E + 2E − E 3 2 1 =− E+E= E 3 3 hc 1 hc = λ′ 3 λ λ′ = 3λ
...(i)
Potential energy, U = eV dU ⎛r⎞ ⎛r ⎞ 1 or U = eV0 ln ⎜ ⎟ ∴ = eV0 ⎜ 0 ⎟ ⎝ r ⎠ r0 dr ⎝ r0 ⎠ eV dU Force, F = − =− 0 dr r or
F =
=
eV0 r
eV0 r PHYSICS FOR YOU | OCTOBER ‘16
67
This force provides the necessary centripetal force. ∴
mv 2 r
=
eV0
or v =
r
eV0
m nh By Bohr’s postulate, mvr = 2π nh or v = 2πmr
...(i)
...(ii)
From equations (i) and (ii), we get eV0 nh = m 2πmr
or ∴
⎡ h ⎣⎢ 2π
r=⎢
or r = 1 meV0
nh m × 2πm eV0
⎤ ⎥×n ⎥⎦
rn ∝ n 7. (a) : Number of spectral lines emitted is n(n − 1) N= 2 9(9 − 1) 9 × 8 = = 36 Here n = 9 ∴ N = 2 2 h 2 ε0 a = 8. (b) : 0 πme 2 h 2 ε0 aμ = π(207m)e 2 Dividing (ii) by (i), we get aμ a 1 = or aμ = 0 a0 207 207
.... (i) .... (ii)
CZ 2
, where C is a constant. n2 For ground state of hydrogen atom, Z = 1, n = 1 ∴
C(1)2
=C (1)2 For nth state of Li2+ ion (Z = 3) En =
E1 =
C(3)2
=
9C
n n2 As En = E1 9C ∴ = C or n2 = 9 or n = 3 n2 10. (c) : According to Bohr’s quantisation condition nh Ln = 2π 4h 5h ∴ For n = 4, L4 = and for n = 5, L5 = 2π 2π ∴ Change in angular momentum when an electron is excited from n = 4 to n = 5 is 68
2
PHYSICS FOR YOU | OCTOBER ‘16
n2 a , where n is the orbit number. Z 0 For 257 100 Fm, Z = 100 25 1 ⎛ rs ⎞ a0 = a0 ⇒ r5 = ⎜⎝∵ a = N ⎟⎠ 100 4 0 r5 1 =N = ∴ a0 4
11. (d) : As rn =
12. (c) : The angular momentum is given by ⎛ k ⎞ L = l(l + 1) ⎜ ⎟ ⎝ 2π ⎠ For 3d orbital, l = 2 ⎛ k ⎞ ⎛ k ⎞ ∴ L = 2(3) ⎜ ⎟ = 6 ⎜ ⎟ ⎝ 2π ⎠ ⎝ 2π ⎠ 13. (c) : Radius of Li++ ion in its ground state, i.e.,
9. (b) : The energy of nth state of a hydrogen like atom is given as En =
5h 4h − 2π 2π h 6.63 × 10−34 J-s = 1.05 × 10–34 J s = = 2π 2 × 3.14 ΔL = L5 − L4 =
r0 =
ε0 h 2
πmZe 2 53 pm ε h2 a ≈ 18 pm As Bohr radius, a0 = 0 2 ⇒ r0 = 0 = πme
Z
3
(∵ for Li, Z = 3)
14. (c) : Gain in potential energy of α-particle = Loss in kinetic energy of α-particle 1 Ze × 2e 5 × 1.6 × 10−13 = ⋅ 4 πε0 r0 2 1 2Ze ⋅ r0 = 4 πε0 5 × 1.6 × 10−13 =
9 × 109 × 2 × 92 × (1.6 × 10−19 )2
5 × 1.6 × 10−13 = 5.3 × 10–14 m = 5.3 × 10–12 cm ≈ 10–12 cm 15. (a) : ∴
1 1⎤ ⎡1 = RZ 2 ⎢ − ⎥ 2 2 λ ⎣ n1 n2 ⎦ 1 ⎡1 1⎤ 5 = R(1)2 ⎢ − ⎥ = R λ1 ⎣ 22 32 ⎦ 36
1 1 ⎞ 3 ⎛ 1 = R(2)2 ⎜ − ⎟ = R 2 λ2 42 ⎠ 4 ⎝2 ∴ λ2 = 5 × 4 = 5 λ1 36 3 27 5 5 or λ2 = λ1 = × 6561 Å = 1215 Å 27 27 16. (c) : The given nuclear reaction is 1 4 4 1H → 2 He + 2e+ + Energy The energy released during the process is 1 4 Q = [4m(1H) – m( 2He) – 2(me+)]c2 = [4 × 1.007825 – 4.002603 – 2 × 0.000548]u × c2 = [4.0313 – 4.002603 – 0.001096]u × c2 = (0.027601 u)c2 = (0.027601)(931.5) MeV = 25.7 MeV 17. (c) : According to Rutherford and Soddy law for radioactive decay, Number of atoms remained undecayed after time t is N = N0e–λt N N e λt = 0 or λt = ln 0 N N 1 N0 t = ln λ N T1/2 N 0 1.25 × 1010 16 ln t= ln = ln 2 N ln 2 1 10 1.25 × 10 × 4 × ln 2 = 5 × 1010 years = ln 2 18. (a) : As the heavy nucleus at rest breaks, therefore according to law of conservation of momentum, we get m1v1 + m2v2 = 0 and
v1 m2 3 = = v2 m1 1 As nuclear density is same, 4 3 m1 ρ 3 πR1 R13 ∴ = = 3 4 m2 ρ πR23 R2 3 R13 m1 1 = = or R23 m2 3
or
....(i)
(Using (i))
∴ R1 : R2 = 1 : 31/3 19. (d) : During the emission of α-particle, the mass number and atomic number decreases by four and two respectively. During the emission of β-particle the mass number remains the same while the atomic number increases by 1. 3α 5β −12) ( A −12) ⎯⎯→ ( A (Z −6)Y ⎯⎯→ (Z −1)Y ′ Number of neutrons A − 12 − (Z − 1) ∴ = Number of protons Z −1 A − Z − 11 = Z −1 20. (c) : Here, Half life, T1/2 = 8 days Initial activity, R0 = 8 microcurie Final activity, R = 1 microcurie A ZX
n
R ⎛1⎞ = ⎜ ⎟ where n is the number of half lives As R0 ⎝ 2 ⎠ n
3
n
⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ = ⎜ ⎟ or ⎜ ⎟ = ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝8⎠ ⎝2⎠ or n = 3 t As n = or t = nT1/2 = (3)(5 days) = 15 days T1/2 ∴
21. (a) : A free neutron is unstable. It decays spontaneously into a proton, an electron and antineutrino. – n → p + e– + υ neutron
proton
electron
anti-neutrino
22. (a) : Nuclear radii, R = R0(A)1/3 (where R0 = 1.2 fm) or R ∝ (A)1/3 ∴ or
RBe (9)1/3 = RGe ( A)1/3 RBe (9)1/3 = 2RBe ( A)1/3
∴ (A)1/3 = 2 × (9)1/3 or A = 23 × 9 = 72 The number of nucleons in Ge is 72. PHYSICS FOR YOU | OCTOBER ‘16
69
23. (c) : For fusion to take place, high temperature is needed because at high temperature, the kinetic energy becomes large enough to overcome the coulomb repulsion between nuclei. 24. (a) : Here, power = 3.2 MW = 3.2 × 106 W Energy released per fission = 200 MeV = 200 × 106 eV = 200 × 106 × 1.6 × 10–19 J Number of fissions per minute 3.2 × 106 × 60 = = 6 × 1018 200 × 106 × 1.6 × 10−19 25. (d) : Let number of α particles emitted be x and number of β-particles emitted be y. Difference in mass number = 4x = 232 – 208 = 24 or x = 6 Difference in atomic number = 2x – y = 90 – 82 = 8 ⇒ 12 – y = 8 or y = 4 26. (b) : Let A and Z be mass number and atomic number of element X. β+
and M is the molar mass. λmL ⎛m⎞ ⇒ N = ⎜ ⎟ L or A = M ⎝M⎠ 28. (b) : During an elastic collision between two particles, the maximum kinetic energy is transferred from one particle to the other when they have the same mass. Consequently, a neutron loses all of its kinetic energy when it collides head-on with a proton, in analogy with the collision between a moving billiard ball and a stationary one. For this reason, materials which are abundant in hydrogen such as paraffin and water, are good moderators for neutrons. N 29. (b) : Here, T1/2 = 3.8 days, N = 0 20 ∴ Disintegration constant, 0.693 0.693 day −1 λ= = T1/2 3.8 As N = N0e–λt
2α
⎯⎯→ Z −A1Y ⎯⎯→ ZA−−58Y ′ Since the product nucleus has mass number 227 and atomic number 89. ∴ A – 8 = 227 or A = 235 and Z – 5 = 89 or Z = 94 A ZX
27. (b) : A = λN N = n⋅L where n is the number of moles and L is the Avogadro number. m Now, n = where m is the mass of the isotope, M
N0 = N 0e −λt or e λt = 20 20 Taking natural logarithm, λt = loge 20 or λt = 2.303 × log10 20 ∴
2.303 × 1.3010 2.303 × 1.3010 × 3.8 = days λ 0.693 =16.43 days 16.5 days or
t=
30. (d)
$
$
$
$
70
PHYSICS FOR YOU | OCTOBER ‘16
!
"#
s Google Glas e k li e ic v e d :A Coming soon gmented mind to read your ype `smart’ eyewear that can integrate suaurroundings.
out your le Glass-t information ab Imagine a Goog Wearable d feed you live an n e this possible. ow ak ur m yo ay m s er ch reality with ar se re less mentally m developed by mics, or make it A portable syste gnitive ergono co e technologies nc re ha fo en be o t Bu r displays can als searchers said. re s, sk ta way to monito in a rta ete ce , engineers need life ily taxing to compl da of rt s become a pa like Google Glas . stem affect the brain ey th d a portable sy w exactly ho the US develope in ty IRS) to rsi (fN ive y formation Un op m Drexel red spectrosc feed you live in Researchers fro tional near-infra with your own, nc ty fu ali for fNIRS are es re ns us d It te tio . en at ica th s said. The appl integrate augm n er that can do just ca ch It ar . se ity re , tiv n’s brain ac best, they said. operating room measure a perso disabilities learn be used in the ith en w s ev nt d de an s stu ng w undi dying ho about your surro controllers to stu ining air traffic endless -from tra
tter
ysterious ma
99% m A galaxy with spotted
ve t does not mo a th le b b u b A deal ve and it’s a big children who ha generations of same mass as y -roughly the t d a “ghost“ galax un fo onfly 44 is almos ve ag ha Dr ts r. Scientis of dark matte tly up os e m ad w m no ely r tir fo -and our own, but en the mysterious s never of dark matter, universe but ha e th entirely made up of % 27 up es ak m at least , at by th latively near theoretical -stuff the galaxy is re gh etely ou pl Th . m co en se ts tis actually been rk that scien iverse, it is so da sits in the It un . e ar th ye t of las ale d sc te in the finally spot n cades. But it was s from us. Whe missed it for de illion light year m 0 33 t ou a normal ab st r, ju t ste no clu y as w lax it Coma ga found that it at it further, they matter. Though scientists looked ade up of dark m t, os gh a one ly ad on ste , y in lax ut -b ga set of stars r own Milky Way rs, dust and me mass as ou al matter like sta has about the sa of up the norm e ad m is 1% hundredth of d us. dy knows what gas that surroun rk matter. Nobo ve made up of da % .99 galaxy could ha 99 a is w it , Rather out -or even ho ab al e rm m no ca e it m w so ho have exactly that is, agonfly 44 does rs than ed that way . Dr times more sta d re nd arisen that look hu a s ha y lax laxy Ga r ga t ou t os Bu the strange gh stars of its own. und out about that fo t s en er m em no ov tro -m are there. As galaxy’s stars e th of t en easures em m e mov esn’t by normal by looking at th matter that do Yale by d m ce fro en am flu te in seemed to be member of the a , um kk h Do uc n Pieter va tell you how m exist. Professor ions of the stars ot st “M : ju id ey sa , th US e matter is, University in th what form the ry They don’t care , stars move ve y lax ga matter there is. fly on e. “In the Drag times more er y th an it’s m d at th un u fo tell yo pancy . We as a huge discre e is mass in the fast. So there w e stars than ther th of ns io ot ng m by the ust be somethi mass indicated ow that there m kn ts er. But tis th ien ge Sc to y s.“ hold the galax stars themselve at is needed to th ity av e. gr er e th th ’t providing provide that isn ould normally the mass that w
rnt by around all over It is a lesson lea —they wobble les bb bu g in tists had fun blow . But now scien never last long ey th le d bb an bu e y ac the pl mobilise a tin managed to im lp he d ul co at th in France have hrough rprising breakt in water in a su ood clots. pushed doctors treat bl ill naturally be les in a liquid w bb bu edes in y im all ch rm Ar No ed by omenon describ d in a se er m im upwards, a phen y all t, wholly or parti jec ob ht of ny “A eig C. w 250B ual to the up by a force eq ician at m he at m fluid, is buoyed e th by the object,” ed lac stop sp to di ay id the flu d found a w now, no-one ha til s un er d ch An ar . se te re ro , w However m happening. eate cr d ul co this process fro ey th d Université foun h a tiny at Aix-Marseille electricity throug g in nn ru by es cy of the en qu microbubbl fre changing the By . er at w bubble in de electro could make the discovered they . electricity they ctrode nce from the ele ould stay a set dista h the water, it w ug ro th g in ris ly w slo ctrode, of ele ad So inste ey moved the sition. And if th po té ed rsi fix ive a Un in y ille sta x-Marse t with it. The Ai ea ilis ob m im the bubble wen d ul co monstrated they is as if the researchers de water electrolys m fro d te ea cr ally push it rm no microbubble ld t force that wou an oy bu ’ es Physics ed Archim ican Institute of exist. The Amer ’t uld dn di co … ds ar on w en up phenom and surprising dustry or in ar cle said: “This new nu e th ne, tions in medici lead to applica logy.” no ch te n io at ul micromanip
Courtesy : The Times of India
PHYSICS FOR YOU | OCTOBER ‘16
71
INDUCED ELECTRIC FIELD
This article expects a brief knowledge of induction;. i.e., when the magnetic flux through a closed conducting loop is changed, a current is induced within the loop such that it opposes the change that has produced it. According to Faraday's law, the induced emf in the loop is, dφ ε=− B dt where φB is magnetic flux through the loop. ∴ Induced current, dφ − B ε I= = dt R resistance of the loop Now, the question that should pop up in our head is how does current appear; i.e., what actually makes the free electrons of the loop push along the length of the wire? In current electricity chapter, you must have learnt that the terminals of the battery creates a potential difference across the length of the conductor which creates an electric field and hence the free electrons get pushed due to which current appears. So, it was the conservative electric field there since they were produced by static charges at the ends of the battery terminals. But how do we explain the generation of current without battery? An interesting observation found in nature is that a changing magnetic field leads to an induction of electric field in the surrounding and vice versa. So in time varying magnetic field, current is generated due to induced electric field which is very different from the conservative electric field. How do we know that its electric field only which is making the charges push along the length of conductor and not magnetic forces? Had it been magnetic forces, they would have been perpendicular to velocity [F = q(v × B)] and hence
perpendicular to length of wire too and hence charges could not have moved. Now, how do we find the strength of the induced electric field? We go back to the definition of emf of a battery which was the work done per unit charge by the nonconservative forces developed within the battery in moving the charges from one terminal to the other. Similarly, for induced emf, it will be the work done per unit charge by induced electric field in taking the charge through a closed loop once. Induced emf can be found out from Faraday's law as – dφ ε=− B dt where φB is the magnetic flux through the loop. Hence combining the two definitions, ε=
Winduced electric field q
∫ qEi ⋅ dl
dφ B q dt dφ ⇒ ∫ Ei ⋅ dl = − B dt where Ei is the induced electric field's value at an elemental length dl of the loop. To avoid the confusion of positive and negative in the above result, it would be convenient to use this law in two parts. 1. Calculation of magnitude using – dφ ∫ E ⋅ dl = dtB 2. Calculation of direction of induced electric field using Lenz's law, according to which if a conducting loop is placed in a region of time varying magnetic field (TVMF), current will be induced in it such that the induced magnetic field due to induced current is in opposite direction with respect to inducing ⇒
=−
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
72
PHYSICS FOR YOU | OCTOBER ‘16
magnetic field if the strength of inducing field increases with respect to time and in same direction if the strength decreases with respect to time. The direction of induced current would give us the direction of induced electric field. We will restrict our discussion to a cylindrical region of TVMF. Let us consider a cylindrical region of radius R which has a magnetic field uniform in space (along the length of cylinder) but variable in time and changing at the dB rate . dt We would be interested in finding out the induced electric field (Ei ) for internal as well as external points. At whichever point Ei is to be found out, we imagine that if we had placed a conducting loop in this cylindrical region symmetrically and passing through the given point, there would have been an induced current certainly and this induced current would give us the direction of Ei . dB >0 dt For internal points ( r < R): Current would have induced in anticlockwise sense, so would be the direction of Ei . Hence, note that Ei is forming closed circular loop. ∴ ∫ Ei ⋅ dl = ∫ Eidl cos θ
Let us assume
= Ei ∫ dl cos 0° = Ei (2πr) dB dφ B = πr 2 dt dt r ⎞ ⎛ dB ⎞ ⇒ Ei = ⎛⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ dt ⎠
∴ Ei (2πr) =
For external points (r ≥ R): dB ∴ Ei (2πr) = πR2 dt ⎛ 2 ⎞ dB ⇒ Ei = ⎜ R ⎟ ⎝ 2r ⎠ dt If the strength of Ei is plotted against r : (i) Ei ∝ r for r < R 1 (ii) Ei ∝ for r ≥ R r 74
PHYSICS FOR YOU | OCTOBER ‘16
Some of the important noteworthy points regarding Ei are These are non-conservative in nature and hence work done by such field in a closed path is non-zero. In fact, work done in closed path = charge × emf of loop. Induced electric field lines form closed loops. Potential energy cannot be associated with them. This comes not due to static charges but rather due to TVMF. Hence if a charge particle is placed at rest in a TVMF, it will experience a non-zero force due to induced electric field. Now, let us see some applications of it : Q1. A charge particle of charge q, mass m is placed at a distance 2R from the centre of a cylindrical region of radius R with TVMF where magnetic field varies ^ as B = (4t 2 − 2t + 6) k where t in sec gives B in tesla. Find the acceleration of the charge at t = 2 s. (Take R = 1 m) Sol.: dB = 8t − 2 dt ∴ At t = 2 s dB = 14 T s −1 dt which is greater than zero, so induced electric field must try to decrease the strength of inducing magnetic field, so must be circular loops in anticlockwise sense. 7 R2 dB 1 = (14) = N C −1 2 2(2R) dt 4 F qE 7q ∴ a= = i = m m 2m ∴ Ei =
Q2. A charge q, mass m is placed in a cylindrical region of uniform magnetic field of strength B and of radius R. It is attached to one end of a relaxed spring of stiffness
constant k as shown. If the magnetic field is suddenly switched off, find the maximum elongation or compression in the spring. Sol.: Switching off the magnetic field is decreasing the magnetic flux, to oppose which electric field would be induced in anticlockwise sense for the short duration in which the magnetic field changed its value from non-zero to zero value due to which the particle will experience an impulsive force and gain a kinetic energy using which the spring gets deformed. Clearly to increase the flux electric field gets induced in clockwise direction, hence spring will be elongated. The momentum gained by charge in switching duration Δp = mv − 0 = ∫ F ⋅ dt = ∫ (qEi ) dt = ∫ q ⋅
a dB dt 2 dt
qa q dB = aB ∫ 2 2 qaB ∴ v= 2m 1 2 1 ∴ kxmax = mv 2 2 2 =
m ⎛ qaB ⎞ ⇒ xmax = ⎟ ⎜ k ⎝ 2m ⎠
Q3. A conducting rod of length l is placed as a chord in a cylindrical region of TVMF of radius R where dB > 0. Find the potential difference across the dt ends of the rod. Sol.: Isn't there a contradiction in what has been taught in theory and numerical since we learnt that potential term cannot be associated with induced electric field. No, there isn't! Induced electric field pushes the free electrons of the conductor from one end to the other which creates a separation of charges across the ends of the rod and these separated charges create conservative electric field along the length of the rod in opposite direction to the component of induced electric field. We consider an arbitrary point on the rod at a radial distance r. The component of Ei along the length of rod,
E|| = Eicosθ ⎛ r dB ⎞ ⎛ y ⎞ =⎜ ⎝ 2 dt ⎟⎠ ⎜⎝ r ⎟⎠
y ⎞ ⎛ dB ⎞ = ⎜⎛ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ dt ⎠
= 2– 2
2
Do you notice something interesting here? The value is independent of r ! ∴ This E|| pushed the free electrons towards left. ∴ At equilibrium, if the drift of electron stops, conservative field strength, y dB Ec = E|| = 2 dt ∴ Potential difference across the ends y dB l= ΔV = Ecl = 2 dt
l2 4 dB l 2 dt
R2 −
Q4. Consider a rectangular conducting loop PQRS placed in a cylindrical region of radius R of dB TVMF where >0 dt is known. Find the induced emf across the ends QR. Sol.: Let us observe the distribution of induced electric field in the region once– There cannot be an induced emf in the section PS since Ei is perpendicular to length [Ei ⋅ dl = 0]. In the remaining 4 sections, PQ, QT, TR and RS, the induced emf will be identical as can be seen from symmetry. Hence, the induced emf of the loop = 4 times in QT ∴ ε=
dφ B πR2 dB = = 4εQT 2 dt dt
∴ εQT =
πR2 dB 8 dt
∴ εQR = 2εQT =
PHYSICS FOR YOU | OCTOBER ‘16
πR2 dB 4 dt 75
Class XII
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.
Magnetic Effect of Current and Magnetism Total Marks : 120
Time Taken : 60 min NEET / AIIMS / PMTs
Only One Option Correct Type
1. Consider a long, straight wire of cross-sectional area A carrying a current I. Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the I current with a speed v = and separated from πAe the wire by a distance r. The magnetic field seen by the observer is very nearly μ0 I 2μ 0 I μ I (a) 0 (b) zero (c) (d) πr πr 2 πr 2. Two short magnets of equal dipole moments M are fastened perpendicularly at their centres as shown in the figure. The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is μ 0 2M μ M (a) 0 3 (b) 4 π d3 4π d μ 0 2 2M μ0 2 M (c) 4 π (d) 3 4 π d3 d 3. An electron accelerated through a potential difference V passes through a uniform transverse magnetic field and experiences a force F. If the accelerating potential is increased to 2V, the electron in the same magnetic field will experience a force F (a) F (b) (c) 2 F (d) 2F 2 76
PHYSICS FOR YOU | OCTOBER ‘16
4. An electric current I enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged particle q moving along the axis of the circular wire passes through its centre at speed v. The magnetic force acting on the particle when it passes through the centre has a magnitude μ0 I μ0 I (a) qv (b) qv 2a 2 πa μ0 I (c) qv (d) zero a 5. A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpendicular to the beam. The protons and the electrons (a) will go undeviated (b) will be deviated by the same angle and will not separate (c) will be deviated by different angles and hence separate (d) will be deviated by the same angle but will separate. 6. A charged particle of mass 10–3 kg and charge 10–5 C enters a magnetic field of induction 1 T. If g = 10 m s–2, for what value of velocity will it pass straight through the field without deflection? (a) 10–3 m s–1 (b) 103 m s –1 6 –1 (c) 10 m s (d) 1 m s–1 7. Mark out the correct options. (a) Diamagnetism does not occur in all materials. (b) Diamagnetism results from the partial alignment of permanent magnetic moment.
(c) The magnetising field intensity is always zero in free space. (d) The magnetic field of induced magnetic moment is opposite to the applied field. 8. A particle is moving with velocity v = i + 3 j and it produces an electric field at a point given by E = 2k. It will produce magnetic field at that point equal to (all quantities are in SI units) (a) (6i − 2 j ) μ 0 ε0 (b) (6i + 2 j ) μ 0 ε0 (c) zero (d) cannot be determined from the given data 9. A point charge is moving in clockwise direction in a circle with constant speed. Consider the magnetic field produced by the charge at a fixed point P (not at the center of circle) on the axis of the circle. Then, (a) it is constant in magnitude only (b) it is constant in direction only (c) it is constant both in direction and magnitude (d) it is constant neither in magnitude nor in direction 10. A dip circle is so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is 40°. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30° with the magnetic meridian. In this position, the needle will dip by an angle (a) 40° (b) 30° (c) more than 40° (d) less than 40° 11. What should be the current in a circular coil of radius 5 cm to annul BH = 5 × 10–5 T? (a) 0.4 A (b) 4 A (c) 40 A (d) 1 A 12. A particle of mass 2 × 10–5 kg moves horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of 200 N C–1 acing upward. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both B and v . If g is 9.8 m s–2 and the charge on the particle is 10–6 C, then find the velocity of charge particle so that it continues to move horizontally. (a) 2 m s–1 (b) 20 m s–1 (c) 0.2 m s–1 (d) 100 m s–1
Assertion & Reason Type
Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 13. Assertion : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Reason : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. 14. Assertion : The true geographic north direction is found by using a compass needle. Reason : The magnetic meridian of the earth is along the axis of rotation of the earth. 15. Assertion : The net force on a closed circular current carrying loop placed in a uniform magnetic field is zero. Reason : The torque produced in a conducting circular ring is zero when it is placed in a uniform magnetic field such that the magnetic field is perpendicular to the plane of loop. JEE MAIN / JEE ADVANCED / PETs
Only One Option Correct Type
16. A charged particle of specific charge (charge/mass) α is released from origin at time t = 0 with velocity v = v0 (i + j ) in a uniform magnetic field B = B0 i. π Coordinates of the particle at time t = are ( B0α) ⎛ 2v0 −v0 ⎞ ⎞ ⎛ − v0 , 0, 0 ⎟ , (a) ⎜ 0, ⎟ (b) ⎜ ⎠ ⎝ 2 B0α ⎝ αB0 B0 α ⎠ −2v0 ⎞ ⎛ v0 π v π ⎞ ⎛ 2v , 0, (c) ⎜ 0, 0 , 0 ⎟ (d) ⎜ B0α ⎟⎠ ⎝ B0α ⎝ B0α 2 B0α ⎠ 17. A small block of mass m, having charge q, is placed on a frictionless inclined plane making an angle θ with the horizontal as shown in figure. There exists a uniform magnetic field B parallel to the inclined plane but perpendicular to the length of a spring. PHYSICS FOR YOU | OCTOBER ‘16
77
If m is slightly pulled on the incline in downward direction, the time period of oscillation will be (assume that the block does not leave contact with the plane) (a) 2π m / k (b) 2 π 2m / k
(c) 2π qB / K
(d) 2 π qB / 2K
18. A thin, plastic disk of radius R has a charge q uniformly distributed over its surface. If the disk rotates at an angular frequency ω about its axis, then magnetic dipole moment of the disk is ωqR 2 ωqR 2 (b) (c) ωqR2 (d) 2ωqR2 2 4 19. From a cylinder of radius R, a cylinder of radius R/2 is removed, as shown in the figure. Current flowing in the remaining cylinder is I. Then, magnetic field strength is (a)
(a) zero at point A (b) zero at point B μ I (c) 0 at point A 2 πR
(d)
μ0 I at point B 3 πR
More than One Options Correct Type 20. A conductor ABCDEF, shaded as shown, carries a current I. It is placed in the xy plane with the ends A and E on the x-axis. A uniform magnetic field of magnitude B exists in the region. The force acting on it will be
(a) zero, if B is in the x-direction (b) λBI in the z-direction, if B is in the y-direction (c) λBI in the negative y-direction, if B is in the z-direction (d) 2aBI, if B is in the x-direction 21. A microammeter has a resistance of 100 Ω and a full scale range of 50 μA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s) 78
PHYSICS FOR YOU | OCTOBER ‘16
(a) 50 V range with 10 kΩ resistance in series (b) 10 V range with 200 kΩ resistance in series (c) 5 mA range with 1 Ω resistance in parallel (d) 10 mA range with 1 Ω resistance in parallel 22. A direct current I flows along a long straight wire as shown in the figure. From point O the current spreads radially all over on infinite conducting plane perpendicular to the wire. Then (a) Magnetic field in region 1 is non-uniform (b) Magnetic field in region 2 in non-uniform (c) Magnetic field in region 3 is non-uniform (d) Magnetic field in region 3 is zero. 23. A current I flows in a long round uniform cylindrical wire made of paramagnetic material with susceptibility χ. Which of following statements are correct regarding the surface molecular current (Is) and the volume molecular current (Iv) ? (a) Both the currents Is and Iv have same magnitude. (b) Both the currents Is and Iv have different magnitude. (c) Both the currents Is and Iv flow in the same direction. (d) Both the currents Is and Iv flow in the opposite directions. Integer Answer Type 24. A current I = 10 A flows in a ring of radius r0 = 15 cm made of a very thin wire. The tensile strength of the wire is equal to T = 1.5 N. The ring is placed in a magnetic field, which is perpendicular to the plane of the ring so that the forces tend to break the ring. Find B (in T) at which the ring is broken. 25. An elevator carrying a charge of 0.5 C is moving down with a velocity of 5 × 103 m s–1. The elevator is 4 m from the bottom and 3 m horizontally from P as shown in figure. What magnetic field (in μT) does it produce at point P? 26. An iron of volume 10–4 m3 and relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.1 A is passed through the solenoid, find the magnetic moment of the rod.
Comprehension Type
Column I Column II (A) In the given situation, (P) Resultant force is acting along Pm (B) If loop is rotated (Q) Resultant force is acting opposite such that Pm is along positive z-direction to Pm (C) If loop is rotated (R) Fx = 0, Fy = 0 such that Pm is along negative z-direction (D) If loop is rotated (S) Fx = 0, Fz = 0 such that Pm is along positive y-direction A B C D (a) P P, R S Q (b) S P, R P, R Q, S (c) Q P, R P, R P, S (d) R R, S Q, R P, Q
A thin, uniform rod with negligible mass and length 0.200 m is attached to the floor by a frictionless hinge at point P as shown in the figure. A horizontal spring with force constant k = 4.80 N m–1 connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field B = 0.340 T directed into the plane of the proper. There is current I = 6.50 A in the rod, in the direction shown. 27. Calculate the torque due to the magnetic force on the rod, for an axis at P. (a) 0.0442 N m–1, clockwise (b) 0.0442 N m–1, anticlockwise (c) 0.022 N m–1, clockwise (d) 0.022 N m–1, anticlockwise 28. When the rod is in equilibrium and makes an angle of 53.0° with the floor, is the spring stretched or compressed? (a) 0.05765 m, stretched (b) 0.05765 m, compressed (c) 0.0242 m, stretched (d) 0.0242 m, compressed Matrix Match Type 29. An elementary current loop is placed in a nonuniform magnetic field as shown in the figure. Where, Pm is magnetic moment of loop.. In column I different orientations of loop are described and in column II, the corresponding forces experienced by the loop. Match the entries of column I with entries of column II.
30. Three wires are carrying same constant current I in different directions. Four loops enclosing the wires in different manners as shown in the figure. The direction of dl is shown in each loop. Match the entries of column I with entries of column II. Column I Column II (A) Along closed loop 1 (P) ∫ B ⋅ dl = μ 0 I
(B) Along closed loop 2 (Q)
∫ B ⋅ dl
(C) Along closed loop 3 (R)
∫ B ⋅ dl = 0
= −μ 0 I
(D Along closed loop 4 (S) Net work done by the magnetic force to move a unit charge along the loop is zero A B C D (a) P, Q Q, R Q, S R, S (b) Q, S P, S R, S R, S (c) P, Q P, R Q, S P, S (d) P, S Q, S R, S R, S 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.
PHYSICS FOR YOU | OCTOBER ‘16
79
P
hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five 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.
SINGLE OPTION CORRECT TYPE
1. A smooth cannon of mass M attached with an ideal spring of stiffness k fires a shell of mass m with muzzle velocity u. The recoil velocity of the cannon is
mu cos θ mu sin θ (b) M +m 2M + m mu cos θ 2mu cos θ (d) (c) M +m M +m 2. A uniform disc of mass m and radius R with a spring connected (spring constant K) to its centre stands in static equilibrium on a very rough inclined plane. Now a constant moment of couple M0, is applied at the centre as shown in the figure such that the disc rolls down the incline. The maximum distance the disc’s centre of mass will go down as measured from initial position is (a)
M0 2mg sin θ (a) + Rk 3k (c)
2 M0 Rk
(b)
2 M0 2mg sin θ + Rk k
(d)
2 M0 mg sin θ + Rk k
3. In a resonance tube experiment, a closed organ pipe of length 120 cm is used and tuned with a tuning fork of frequency 340 Hz. If water is poured into the pipe, then which of the following statements is incorrect? [Velocity of sound in air is 340 m s–1, neglect end correction] (a) Minimum length of water column to have the resonance is 45 cm. (b) The distance between two successive nodes is 50 cm. (c) The maximum length of water column to create the resonance is 95 cm. (d) The distance between two successive nodes is 25 cm. 4. A ball is hung vertically by a thread of length l from a point P of an inclined wall that makes an angle β with the vertical. The thread with ball is deviated through a small angle α(α > β) and set free. Assuming the wall to be perfectly elastic, the period of such pendulum is (a)
l g
⎡π −1 ⎛ β ⎞ ⎤ ⎢ 2 − sin ⎜⎝ α ⎟⎠ ⎥ ⎣ ⎦
(b)
l g
⎡ −1 ⎛ β ⎞ ⎤ ⎢ sin ⎜⎝ α ⎟⎠ ⎥ ⎣ ⎦
(c)
l g
⎡ −1 ⎛ β ⎞ ⎤ ⎢ cos ⎜⎝ α ⎟⎠ ⎥ ⎣ ⎦
(d) 2
l g
⎡ −1 ⎛ β ⎞ ⎤ ⎢ cos ⎜⎝ − α ⎟⎠ ⎥ ⎣ ⎦
By Akhil Tewari, Author Foundation of Physics for JEE Main & Advanced, Professor, IITians PACE, Mumbai.
80
PHYSICS FOR YOU | OCTOBER ‘16
COMPREHENSION TYPE
Direction for question 5 to 8 : The pressure of a monoatomic gas increases linearly from 4 × 105 N m–2 to 8 × 105 N m–2 when its volume increases from 0.2 m3 to 0.5 m3. Calculate 5. Work done by the gas (a) 2.8 × 105 J (b) 1.8 × 106 J 5 (c) 1.8 × 10 J (d) 1.8 × 102 J 6.
Increase in internal energy (a) 4.8 × 105 J (b) 4.8 × 104 J 5 (c) 6.8 × 10 J (d) 4.8 × 106 J
7.
Amount of heat supplied (a) 8.6 × 105 J (b) 12.6 × 105 J 5 (c) 6.6 × 10 J (d) 10.6 × 105 J
8. Molar heat capacity of the gas (R = 8.31 J mol–1 K–1) (a) 20.1 J mol–1 K–1 (b) 17.14 J mol–1 K–1 (c) 18.14 J mol–1 K–1 (d) 20.14 J mol–1 K–1 SUBJECTIVE TYPE
9. A body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity. Calculate the time taken by it to reach height h. 10. In Young's double slit experiment if the source consists of two wavelengths λ1 = 4000 Å and λ2 = 4002 Å. Find the distance from the centre where the fringes disappear, if d =1cm ; D =1 m.
PHYSICS FOR YOU | OCTOBER ‘16
81
Y U ASK
WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month.
Object is released from rest at y = 0, so u = 0 Equations of motion become at time t1 v1 = 0 – gt1 = – 9.8t m s–1 1 h = 0 − gt 2 = − 4.9 t 2 m 2 v12 = 0 − 2 gy = −19.6h m2s −2 Hence signs of u, v, a and h are negative as they are chosen along negative y-direction.
Q1. Are photons always moving? What about their rest mass? –Basavraj S. Awatiger, Hubballi (Karnataka)
Ans. Yes, photons always move with speed of light (c). In relativistic dynamics, rest mass of a particle, m0 = m 1 − v 2 /c 2 ; m = moving mass of particle,
(a) a – t graph (b) v – t graph (c) h – t graph
For photon, v = c ∴ m0 = m 1 − c 2 /c 2 = 0 Rest mass of photon is zero. Q2. Explain why Gauss’s law is not very useful in calculating the electric field of a charged disc. –Aquil Ahmed, Banda (U.P.)
Ans. Gauss’s law is useful as a calculational tool only in case of high symmetry, where one can produce a Gaussian surface on which the electric field is either constant or has no perpendicular component. It is not possible to do so in any simple way for the case of a charged disc. Gauss’s law still applies but it is just not particularly useful to find electric field. Q3. What are the sign used for initial velocity (u), final velocity (v), acceleration (a), and height (h) during free fall motion of an object and why? –Priyanka Soren
Ans. Assume motion of the object is in y-direction. Choose upward direction as positive so motion of an object under free fall is in negative y-direction. Since acceleration due to gravity is always downward so a = – g = – 9.8 m s–2 . 82
PHYSICS FOR YOU | OCTOBER ‘16
Q4. A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? –Inova Singh, Dispur (Assam)
Ans. There is only a uniform magnetic field so that a magnetised needle experiences no net force. An iron nail near a bar magnet is in the nonuniform magnetic field of the magnet. The nail is magnetised due to magnetic induction. This nail, which behaves like a small magnet, apart from experiencing a torque, does experience a force. I don’t know anything, but I do know that everything is interesting if you go into it deeply enough. – Richard Feynman
SOLUTION SET-38
1 2 2 2 1. (c) : Since, kinetic energy K = mω A cos ωt 2 1 2 2 2 and potential energy, U = mω A sin ωt 2 1 2 2 2 2 or K − U = mω A [cos ωt − sin ωt ] 2
∴ Angular frequency = 2ω, 2π π π × T ⎡ 2π ⎤ = = ∴ω = ⎥ time period, = ⎢ 2ω ω 2π T ⎦ ⎣ π×4 = =2 s 2π 2. (b) : Linear momentum is conserved. ∴ p1 = p2
where K2 is for α-particle and K1 is for nucleus. K2 = 54 K1
...(i)
Given K1 + K2 = 5.5 MeV From (i) and (ii), K1 + 54 K1 = 5.5 MeV 55 K1 = 5.5 MeV K1 =
1 10
or
K1 =
...(ii)
5.5 55
MeV 54
∴
K2 = 54K1 or K 2 =
or ∴
K2 = 5.4 MeV Kinetic energy of α-particle = 5.4 MeV.
10
MeV
⎛ dM ⎞ (L ) 3. (d) : Heat of radiator, P = ⎜ ⎝ dt ⎟⎠ Pt M L= ∴ L M t 4. (c) : Missing ones are dark fringes 1⎞ ⎛ d 2 + b2 − d = ⎜ n − ⎟ λ ∴ ⎝ 2⎠ ⎛ 1 b2 ⎞ ⎛ 1⎞ 1⎞ b2 ⎛ d ⎜1 + = ⎜n − ⎟ λ ⎟ − d = ⎜⎝ n − ⎟⎠ λ or 2 2 2 ⎝ d ⎠ 2d 2⎠ ⎝ or
P=
2 b2 3 = λ ⇒ λ=b 2d 2 3d
5. (c) : As υ =
nv n T = 2l 2l μ
Here, TI > TII, lI = lII , μI = μII and υI = υII ∴ nI < nII
Gm(dM ) x2
⎡ A + Bx 2 ⎤ = Gm ⎢ ⎥ dx ⎣ x2 ⎦
a+L ⎛ 1 ⎞ F = ∫ Gm ⎜ 2 ⎟ dx ( A + Bx 2 ) ⎝ x ⎠ a
and 2(216 m) K1 = 2(4 m) K 2
or
For n = 2,
dF =
But p = 2mK where K = kinetic energy.
216K1 = 4K2 or
b2 λ b2 ⇒ λ= = 2d 2 d
6. (b) : Mass per unit length of the rod = A + Bx2. So the mass of length dx is dM = dx(A + Bx2)
1 = mω 2 A2 cos 2ωt 2
or
For n = 1,
a+L ⎛A ⎞ A ⎡A ⎤ = ∫ Gm ⎜ 2 + B ⎟ dx = Gm ⎢ − + BL ⎥ ⎝ ⎠ ⎣a a+L ⎦ x a
7. Given, the pitch of screw gauge = 1 mm and total number of division on the circular scale = 50 1 mm = 0.02 mm Least Count, L.C = 50 The instruments has a positive zero error e = n × L.C = 6 × 0.02 = 0.12 mm i.e. zero correction = – 0.12 mm Linear scale reading (LSR) = 3 × 1 mm = 3 mm Circular scale reading (CSR) = 31 × (0.02 mm) = 0.62 mm Measured reading = LSR + CSR = 3 + 0.62 = 3.62 mm True reading = 3.62 – 0.12 = 3.50 mm
Solution Senders of Physics Musing SET-38
1. Suman Tiwari, Patna (Bihar) 2. Trideep Jyotida, New Delhi PHYSICS FOR YOU | OCTOBER '16
83
8. Consider dN number of turns of radius r and thickness dr. Let dε be the corresponding induced emf, then ⎛ dφ ⎞ dε = (dN ) ⎜ ⎟ ⎝ dt ⎠
L= =
⎛N ⎞ dε = ⎜ ⎟ dr (πr2ω × B0 cos ωt) ⎝a⎠
L=
⎛ N πωB0 cos ωt ⎞ 2 ε = ∫ dε = ⎜ ⎟⎠ ∫ r dr ⎝ a 0 a
N πω(B0 cos ωt )a N πωB0 cos ωta = 3a 3 3
=
εmax =
2
πNa 2 B0 ω
∴
t0
μ
1/2 ∫ (F0 − kt ) dt
0
t0 2 ⎡ (F0 − kt )3/2 ⎤⎦ 0 ⎣ μ (−3k)
1
2 −3k μ 2 3k μ k= =
3
1
∫ dx =
0
d × (πω2 × B0 sin ωt) dt
dε = dN
L
∴
[(F0 − kt 0 )3/2 − (F0 + k(0))3/2 ]
F03/2
(Using (i))
3 3×3×3 2 F0 2 = 3 L μ 3 × 1 3 × 10 −2
2×3 3 × 0. 1
= 20 N s −1
We get, n = 3 9. When outer surface is grounded charge –Q resides on the inner surface of sphere B. Now sphere A is connected to earth, potential on its surface becomes zero. Let the charge on the surface A becomes q kq kQ a − =0 ⇒ q= Q a b b In this position energy stored
SOLUTION OF SEPTEMBER 2016 CROSSWORD
2
U1 =
Q2 1 ⎡a ⎤ 1 ⎡a ⎤ Q Q (− Q ) + + ⎢ ⎥ 8 π ε0 a ⎣ b ⎦ 8 π ε0 b 4 π ε0 b ⎣⎢ b ⎥⎦
When S3 is closed, total charge will appear on the outer surface of shell B. In this position energy stored 2
U2 =
1 ⎛a ⎞ − 1⎟ Q 2 ⎠ 8 π ε0 b ⎜⎝ b
Heat produced = U1 – U2 =
Q 2 a (b − a) 8 π ε0 b3
= 1. 8 J
10. Time taken by the pulse to reach from P to Q, F ...(i) t0 = 0 k where F0 = 3 N dx F − kt T ⇒ = 0 Now, v = μ dt μ 84
PHYSICS FOR YOU | OCTOBER' 16
Winner (September 2016) s Preeti Dahiya, Sonipat (Haryana)
Solution senders (August 2016) s Maduree Shirimal, Sehore (Madhya Pradesh) s Nabanika Das, Lohit (Arunachal Pradesh)
Readers can send their responses at [email protected] or post us with complete address by 25th of every month to win exciting prizes. Winners' name with their valuable feedback will be published in next issue.
ACROSS 1. A diffraction grating made of parallel glass plates, each of which extends slightly beyond the next, used to examine extremely fine structures through interferometry. [7, 7] 2. A propulsion engine in which a fuel burns in air that has been compressed by the forward motion of the engine only. [6] 7. A thermometer designed for the measurement of very low temperatures. [9] 10. The removal of any unwanted a.c. components from a circuit or circuit element. [10] 11. The dissociation of molecules by nuclear radiation. [10] 14. A tradename for an alloy of iron, cobalt and nickel with an expansivity similar to that of glass. [5] 15. A mixture of two substances that solidifies as a whole when cooled, without change in composition. [8] 18. Absence of an electron in a semiconductor. [4] 21. A quantum mechanical description of an elementary vibrational motion in which a lattice of atoms or molecules uniformly oscillates at a single frequency. [6] 23. An instrument for measuring the total solar radiation intensity received on a horizontal surface. [11] 25. A non SI unit of luminance. [3] 27. Process of connecting a charged object to earth to remove object’s unbalanced charge. [9] 28. Point of maximum displacement of two superimposed waves. [8] 29. The study of the behaviour and characterstics of nucleons or atomic nuclei. [10] 30. A now-discarded hypothetical medium once thought to fill all space and to be responsible for carrying light waves and other electromagnetic waves. [6] 31. The process of preventing the plasma from coming into contact with the walls of the reaction vessel in a controlled thermonuclear reaction. [11] DOWN 3. The molecular attraction exerted between the surfaces of bodies in contact. [8] 4. A grid or pattern placed in the eyepiece of an optical instrument, used to establish scale or position. [7]
5. 6. 8. 9. 12. 13. 16. 17. 19. 20. 22. 24. 26.
The SI unit of dose equivalent. [7] Study of motion of particles acted on by forces. [8] An instrument for studying thin films on solid surfaces. [12] An apparatus for generating very high frequency currents at high potential. [5, 4] Roughly spherical ice particles, usually a few millimeters in radius, produced in very turbulent clouds. [4] Number of decays per second of a radioactive substance. [8] Two or more sounds that, when heard together, sound pleasant. [10] The amount of potential energy stored in an elastic substance by means of elastic deformation. [10] A hypothetical quantum of gravitational energy, regarded as a particle. [8] Defect of eye, in which distant objects focus in front of the retina. [6] An exposed ring surroundings a strongly illuminated spot on a photographic emulsion. [8] A portable insulation tester calibrated directly in mega ohms. [6] Point where disturbances caused by two or more waves result in no displacement. [4] PHYSICS FOR YOU | OCTOBER ‘16
85
86
PHYSICS FOR YOU | OCTOBER ‘16