PRACTICE PAPER 2016
OR Using Kirchhoff’s rules
determine the value of unknown resistance R in the circuit so that no current flows through 4 W resistance. Also find the potential difference between A and D.
17. Name the three different modes of propagation of electromagnetic waves. Explain the mode of propagation used in the frequency range from a few MHz to 40 MHz using a proper diagram.
18. In a diode AM detector, the output circuit consists of R = 1 kW and C = 10 pF. A carrier signal of 100 kHz is to be detected. Is it good? If yes, then explain why ? If not, what value of C would you suggest?
19. Show that the first few frequencies of light that is emitted when electrons fall to the nth level from levels higher than n, are approximate harmonics (i.e. in the ratio 1 : 2 : 3…) when n >> 1.
20. Define the term electric potential due to a point charge. Find the electric potential at the centre of a square of side 2 m, having charges 100 mC, –50 mC, 20 mC and – 60 mC at the four corners of the square.
21. A long solenoid S has n turns per meter, with diameter a. At the centre of this coil we place a smaller coil of N turns and diameter b (where b < a). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of mt2 + C.
22. In an intrinsic semiconductor, the energy gap Eg is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature.
What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration ni is
given by n n E
i k Tg
= − B
0exp 2 , where n0 is constant.
kB = 8.62 × 10–5 eVK–1. section-D
23. Mr. Sharma purchase heater marked with 80V – 800 W. He wanted to operate it on 100 V – 50 Hz ac supply. He calculated inductance
of the choke required for operating that heater.
(a) Specify the nature of Mr. Sharma. (b) How Mr. Sharma could have calculated the value of inductance? Explain.
section-e
24. Describe an astronomical telescope. Derive expression for its magnifying power when final image is
(i) at infinity
(ii) at least distance of distinct vision.
State Huygen’s principle. Use Huygen’s construction OR to explain refraction of plane wavefront at a plane surface. Draw diagrams to show the behaviour of a (i) convex lens, (ii) concave mirror when a plane wavefront falls on it.
25. Derive an expression for a potential at a point due to an electric dipole.
Obtain the expression for the capacitance of a OR parallel plate capacitor. Three capacitors of capacitances C1, C2 and C3 are connected in parallel. Derive an expression for the equivalent capacitance.
26. State Biot-Savart’s law. Using this law derive an expression for the magnetic field at a point situated at a distance of x metre from the centre of a circular coil of N turns and radius r carrying a current of I A.
Explain the difference between diamagnetic, OR paramagnetic and ferromagnetic substances.
solutions
1. Total charge within a surface S = +2q + (–q) = +q According to Gauss’s law,
Electric flux φ
= qe
2. The current in an ac circuit can be reduced by using 0
an inductor coil or a capacitor in the circuit, as the power dissipation in these elements is negligible.
3. In AM, the carrier wave instantaneous voltage is varied by modulating wave voltage. On transmission, noise signals can also be added and receiver assumes noise as a part of the modulating signal.
However in FM, the carrier wave frequency is changed as per modulating wave instantaneous voltage. This can only be done at the mixing/
modulating stage and not while signal is transmitting
in channel. Hence, noise doesn’t affect FM signal.
4. No, it is true only for metallic conductors.
Commercially available resistors are carbon resistors and wire bound resistors.
5. According to lens maker’s formula,
1 1 1 1
i.e., focal length of lens decreases.
6. Resistance of conductor A,
R l
A =
× −
ρ π( .0 5 10 3 2) Resistance of conductor B,
R l
7. Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle.
i.e., tanqB = m
m21 where m2 < m1.
When light travels in such a medium, the critical angle is sinqC = m
m21 where m2 < m1.
As |tanqB| > |sinqC| for large angles, qB < qC. Thus, polarisation by reflection shall definitely occur.
10. Consider the a-decay of nucleus.
Before decay, neutron to proton ratio, n p/ =238 92− =
92 1 58.
After decay, neutron to proton ratio, n p/ =234 90− =
90 1 60. Thus ratio increases.
OR
23892U → 23490Th + 42He + Q Energy released Q = Dmc2
= (mU – mTh – mHe)c2
= (238.05079 – 234.043630 – 4.002600)c2
= 0.00456 × 931.5 MeV Q = 4.25 MeV
11. Let m1 be the refractive index of lens m2 be the refractive index of medium
1 accelerated under a potential difference V, let v be the velocity acquired by particle. Then
qV=1mv mv= qVm wavelength is associated with electron as compared to proton. electron as compared to proton.
13. Here, d = 0.15 mm = 0.15 × 10–3 m = 15 × 10–5 m, l = 450 nm = 450 × 10–9 m = 4.5 × 10–7 m, D = 1.0 m (a) (i) Distance of the second bright fringe,
x D
(ii) Distance of the second dark fringe,
x D
When screen is moved away, D increases, therefore width of the fringes increases but the angular separation (l/d) remains the same.
14. A carrier wave is an electromagnetic wave of high frequency and of constant amplitude, which is employed to carry the audio signals from one location to other on the surface of earth.
For the transmission of audio signals, the high frequency carrier waves are used, because these high frequency carrier waves travel through space or medium with the speed of light and they are not obstructed by earth’s atmosphere.
15. At the equator, the value of vertical component of earth’s magnetic field is zero.
Given : BH = 0.4 × 10–4 Wb m–2, d = 30°
(i) The vertical component of earth’s magnetic field BV = BHtand = 0.4 × 10–4 × tan30°
= 0.23 × 10–4 Wb m–2
(ii) Total intensity of earth’s magnetic field, B= BH = ×
Let f be potential gradient of the wire.
Applying Kirchhoff’s loop rule to the closed loop ACA, we get
f(120) = e1 – e2 ...(i)
Again, applying Kirchhoff’s loop rule to the closed loop ADA, we get
f(300) = e1 + e2 ...(ii) increasing the length of the potentiometer wire.
OR As no current
flows through 4 W, the current in various branches as shown in the figure.
Applying Kirchhoff’s loop rule to the closed loop AFEBA, we get – I – I – 4 × 0 – 6 + 9 = 0
or 9 – 6 – 2I = 0 or 2I = 3 or I= 3
2 ...(i)
Again, applying Kirchhoff’s loop rule to the closed loop BEDCB, we get 6 + 4 × 0 – IR – 3 = 0 or IR = 3
R= = × =I3 3 2
3 2W (Using (i))
Potential difference between A and D = Potential difference between A and E
\ VAD =2I= × =2 3 2 3 V
17. Three modes of propagation of electromagnetic waves are :
(i) Ground waves (ii) Sky waves (iii) Space waves
In the frequency range from a few MHz to 40 MHz, the sky wave propagation is used. In the ionosphere of the earth’s atmosphere, there are a large number of ions or charged particles. It extends from a height of about 65 km to about 400 km above the earth’s surface. Ionisation occurs due to the absorption of the ultraviolet and other high-energy radiation coming from the sun by air molecules.
The ionosphere is further subdivided into several layers, D, E, F1 and F2. The degree of ionisation varies with the height. The density of atmosphere decreases with height. At great heights, the solar radiation is intense but there are few molecules to be ionised. Close to the earth, even though the molecular concentration is very high, the radiation intensity is low so that the ionisation is again low.
However, at some intermediate heights, there occurs a peak of ionisation density. The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz). Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape. These phenomena are shown in the figure.
18. Here, 1 1
Whereas for demodulation, u1
c << RC. So it is not
umn cRZ bring a unit positive charge from infinity to that point against the electrostatic forces.
AC BD= =
( )
2 2+( )
2 2=2m\ AO = OC = BO = OD = 1 m
Potential at the centre of the square O is
V q 21. Magnetic field due to a solenoid S,
B = m0nI
Negative sign signifies opposite nature of induced emf. The magnitude of emf varies with time as shown in the figure.
22. Conductivity is given by s = e(neme + nh mh)
For intrinsic semiconductor, ne = nh = ni
Also, mobility of holes (mh) << mobility of electrons (me)
So, conductivity s = eneme
Temperature dependence of intrinsic carrier
concentration, n n ei
E
Conductivity at 600 K, s s1 0 Conductivity at 300 K,
s2 s0
23. (a) He seems to be a techno friendly person, having confidence and knowledge of electrical gadgets.
(b) Current required, I P
= =V 800=
Refer point 6.10(6), 6.11 and 6.12 page no. 443, 444 OR and 445 (MTG Excel in Physics)
25. Refer point 1.5(10) page no.9 (MTG Excel in Physics)
Refer point 1.11 (4, 8) page no. 15, 16 (MTG Excel OR in Physics).
26. Refer point 3.1(1) page no. 169 and point 3.1(3 (vi)) page no. 170 (MTG Excel in Physics).
Refer point 3.8(8) page no. 180, 181 (MTG Excel in OR
Physics). nn
oscillations
1. The displacement of two particles executing SHM are represented by equations y1 = 2sin(10t + q), y2 = 3cos 10t. The phase difference between the velocity of these particles is
(a) q (b) –q
(c) q p+
2 (d) q p
-2. The bob of a simple pendulum of length L is released 2 at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by
(a) X a L
g t
= sin2p × (b) X a g L t
= cos2p × (c) X a g
L t
= sin × (d) X a g L t
= cos × 3. A horizontal platform with an object placed on
it, is executing SHM in the vertical direction. The amplitude of oscillation is 3.92 × 10–3 m. What must be the least period of these oscillations, so that the object is not detached from the platform?
(a) 0.1256 s (b) 0.1356 s (c) 0.1456 s (d) 0.1556 s
4. A particle of mass m is executing oscillations about the origin on the x-axis with amplitude A.
Its potential energy is given as U(x) = ax4, where a is positive constant. The x-coordinate of motion where potential energy is one-third of the kinetic energy of particle is
(a) ± A
3 (b) ± A
2 (c) ± A
3 (d) ± A 2
5. On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm support. If each spring has force constant k, the period of oscillation of the body (assuming the springs as massless) is
M
(a) 2p[M/2k]1/2 (b) 2p[2M/k]1/2 (c) 2p[Mg sin q/2k]1/2 (d) 2p[2Mg/k]1/2 6. In damped oscillations, the amplitude of oscillations
is reduced to one-third of its initial value a0 at the end of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be (a) a0/2 (b) a0/6
(c) a0/9 (d) a0/4
7. The x-t graph of a particle undergoing SHM is as shown in figure. The acceleration of the particle at t = 4/3 s is
(a) 3
32p cm s2 -2 (b) -p2 -2 32cm s (c) p2 2
32cm s- (d) - 3 -32p cm s2 2