Physics Musing Problem Set 34 8
JEE Advanced Practice Paper 10
JEE Main Solved Paper 2016 22
PMT Practice Paper 31
BITSAT Practice Paper 39
Brain Map 46
Exam Prep 2016 51
AIIMS Practice Paper 56
Olympiad Problems 65
JIPMER Practice Paper 69
Core Concept 74
Live Physics 81
Physics Musing Solution Set 33 82
You Ask We Answer 84
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1. A particle is projected from ground in vertical direction at t = 0. At t = 0.8 s, it reaches h = 14 m. It will again come to same height at t = (g = 10 m s–2) (a) 2 s (b) 145 s (c) 3 s (d) 7
2 s
2. In the figure shown, the blocks A and C are pulled down with constant velocities u . Acceleration of block B is b u B A C u b (a) u b 2 2 tan secθ θ (b) u b 2 3 tan θ (c) u b 2 2
sec tanθ θ (d) zero
3. Two particles A and B each of mass m are attached by a light inextensible string of length 2l. The whole system lies on a smooth horizontal table with B initially at a distance l from A. The particle at end B is projected across the table with speed u perpendicular to AB. Velocity of ball A just after the jerk is
(a) u 3
4 (b) u 3
(c) u 3
2 (d) u2
4. A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r as shown in the figure. The particle looses its contact with hemisphere at point B. C is centre of the hemisphere. The equation relating a and b is
C A
B
r
(a) 3 sin a = 2 cos b (b) 2 sin a = 3 cos b (c) 3 sin b = 2 cos a (d) 2 sin b = 3 cos a
5. The intensity of radiation emitted by the Sun has its maximum value at a wavelength of 510 nm and that emitted by the North star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperatures of the Sun and the North star is
(a) 1.46 (b) 0.69 (c) 1.21 (d) 0.83
6. N(< 100) molecules of a gas have velocities 1, 2, 3... N km s–1 respectively. Then
(a) rms speed and average speed of molecules is same
(b) ratio of rms speed to average speed of molecules is (2N+1)(N+1 6)/ N
(c) ratio of rms speed to average speed of molecules is (2N+1)(N+1 6)/
(d) ratio of rms speed to average speed of molecules is 2 2 1 6 1 ( ) ( ) N N + +
7. A thermodynamic process of one mole ideal gas is shown in the
figure. The efficiency of cyclic process ABCA will be (a) 25% (b) 12.5% (c) 50% (d) 7.7% V A B C P 2V0 V0 T 0 2T0 4T0 P0 2P0
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.
PHYSICS
PHYSICS
MUSING
MUSING
Section 1 (MaxiMuM MarkS : 32)
This section contains EIGHT questions •
The answer to each question is a SINGLE DIGIT •
INTEGER ranging from 0 to 9, both inclusive
1. A current I flows in a rectangularly shaped wire whose center lies at (x0, 0, 0) and whose vertices are located at the points A(x0 + d, –a, –b), B(x0 – d, a, –b), C(x0 – d, a, b), and D(x0 + d, –a, b) respectively. Assume that a, b, d << x0. Find the magnitude of magnetic dipole moment vector of the rectangular wire frame in J T–1. (Given: b = 10 m, d = 4 m, a = 3 m, I = 0.01 A)
2. A very long, straight, thin wire carries –3.60 nC m–1 of fixed negative charge. The wire is to be surrounded by a uniform cylinder of positive charge, radius 1.50 cm, coaxial with the wire. The volume charge density r of the cylinder is to be selected so that the net electric field outside the cylinder is zero. Calculate the required positive charge density r (in mC m–3).
3. A long coaxial cable consists of two thin-walled conducting cylinders with inner radius 2 cm and outer radius 8 cm. The inner cylinder carries a steady current 0.1 A, and the outer cylinder provides the return path for that current. The current produces a magnetic field between the two cylinders. Find the energy stored in the magnetic field for length 5 m of the cable. Express answer in nJ (use ln 2 = 0.7).
4. A 0.5 kg block slides from the point A on a horizontal track with an initial speed 3 ms–1 towards a weightless horizontal spring of length 1 m and force constant 2 Nm–1. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, find the
total distance (nearest integer) in m through which the block moves before it comes to rest completely. [g = 10 m s–2]
A B D C
5. An open organ pipe containing air resonates in fundamental mode due to a tuning fork. The measured values of length l (in cm) of the pipe and radius r (in cm) of the pipe are l = 94 ± 0.1, r = 5 ± 0.05. The velocity of the sound in air is accurately known. The maximum percentage error in the measurement of the frequency of that tuning fork by this experiment is given by a2%. Find the value of 10 a.
6. An initially uncharged capacitor C is fully charged by a constant emf e in series with a resistor R. Rate of energy dissipation in the resistor is equal to rate of energy stored in capacitor at time CR ln k. Find the value of k.
7. You are at a distance of R = 1.5 × 106 m from the centre of an unknown planet. You notice that if you throw a ball horizontally it goes completely around the planet hitting you in the back 90,000 s later with exactly the same speed that you originally threw it. The length of semi major axis of the motion of ball is 2R. The mass of the planet in scientific notation is a × 1021 kg. Find a.
8. In a certain polytropic process, the volume of argon was increased four times. Simultaneously, the pressure decreased eight times. Find the molar heat capacity (in SI unit) of argon in this process, assuming the gas to be ideal.
PaPer-1
JEE
Advanced
JEE
Advanced
PRACTICE PAPER 2016
exam on 22nd May 2016JEE
Advanced
JEE
Advanced
PRACTICE PAPER 2016
Section 2 (Maximum Marks : 40)
This section contains TEN questions •
Each question has FOUR options (a), (b), (c) and (d). •
ONE OR MORE THAN ONE of these four option(s) is(are) correct
9. For a certain radioactive substance, it is observed that after 4 h, only 6.25% of the original sample is left undecayed. It follows that
(a) the half-life of the sample is 1 h (b) the mean life of the sample is 1
2 ln h
(c) the decay constant of the sample is ln(2) h–1 (d) after a further 4 h, the amount of the substance
left over would by only 0.39% of the original amount.
10. A system consists of two identical m
m cubes, each of mass m, linked
together by the compressed weightless spring of stiffness k. The cubes are also connected by a thread which is burned through at a certain moment.
(a) The lower cube will bounce up after the thread has been burned through when the initial compression of the spring is 2mg
k .
(b) If the initial compression of the spring is 7 mg/k, then centre of gravity of this system will rise to height 8mg
k .
(c) The lower cube will bounce up after the thread has been burned through when the initial compression of the spring is 5mg
k . (d) All are correct.
11. It is desired to make a long cylindrical conductor whose temperature coefficient of resistivity at 20°C will be close to zero. If such a conductor is made by assembling alternate disks of iron and carbon, find the ratio of the thickness of a carbon disk to that an iron disk.
(For carbon, r = 3500 × 10–8 W m and a = –0.50 × 10–3 °C–1 for iron, r = 9.68 × 10–8 W m and a = 6.5 × 10–3 °C–1)
(a) 0.36 (b) 0.036 (c) 1.0 (d) 2.0
12. A rocket set for vertical firing weighs 50 kg and contains 450 kg of fuel. It can have a maximum exhaust velocity of 2 km s–1.
(a) Minimum rate of fuel consumption to just lift it off the launching pad is 2.45 kg s–1.
(b) Minimum rate of fuel consumption to give it an acceleration of 20 m s–2 is 3.5 kg s–1.
(c) The speed of the rocket is 4.2 kms–1 when the rate of consumption of fuel is 10 kgs–1 after whole of the fuel is consumed.
(d) All are correct.
13. One mole of a diatomic ideal gas (γ = 1.4) is taken through a cyclic process starting from point A. The process A → B is an adiabatic compression, B → C is isobaric expansion, C → D is an adiabatic expansion, and D → A is isochoric. The volume ratios are VA/VB = 16 and VC/VB = 2 and the
temperature at A is TA = 300 K.
(a) Temperature of the gas at B is 909 K. (b) Temperature of the gas at D is 791 K. (c) The efficiency of the cycle is 61.4%. (d) The efficiency of the cycle is 38.6%.
14. A point charge q is located
a b q O at centre O of a spherical uncharged conducting layer provided with a small orifice as shown in the figure. The inside and outside radii of the layer are equal to a and b
respectively. What amount of work has to be performed to slowly transfer the charge q from the point O through the orifice and into infinity ? (a) q a b 2 0 8pe 1 1− (b) q b a 2 0 8pe 1 1− (c) q a b 2 0 4pe 1 1− (d) q b a 2 0 4pe 1 1−
15. n drops of a liquid each with surface energy E join to form a single drop. Then
(a) some energy will be released in the process (b) some energy will be absorbed in the process (c) the energy released will be E(n – n2/3) (d) the energy absorbed will be nE(22/3 – 1)
16. A 30 cm violin string with linear mass density 0.652 g m–1 is placed near a loudspeaker that is fed by an audio oscillator variable frequency. It is found that the string is set into oscillation only at the frequencies 880 Hz and 1320 Hz as the frequency of the oscillator is varied continuously over the range 500-1500 Hz. What is the tension in the string ? (a) 120 N (b) 60 N (c) 90.8 N (d) 45.4 N
17. A particle of mass m moves in a certain plane due to a force F whose vector rotates in that place with a constant angular velocity w. Assuming the particle to be stationary at the moment t = 0, then
(a) its velocity as a function of time is F m t w w sin 2
(b) its velocity as a function of time is 2 2 F m t w w sin
(c) the distance covered by the particle between two successive stops is 8F2
mw
(d) the mean velocity over two successive stops is 4F
m p w.
18. A massless rope is tossed over a wooden dowel of radius r in order to lift a heavy object of weight W off the floor, as shown in the figure. The coefficient of sliding friction between the rope and the dowel is m. Which of the following relation is correct for minimum downward pull (Fmin) on the rope necessary to lift the object ?
W r
F (a) Fmin = We–pµ (b) Fmin = We+pµ
(c) Fmin = –Wepµ (d) Fmin = We–p Section 3 (Maximum Marks : 16)
This section contains TWO questions •
Each question contains two columns, Column I and •
Column II
Match the entries in Column I with the entries in •
Column II
One or more entries in Column I may match with one •
or more entries in Column II
19. In each situation of Column I, a physical quantity related to orbiting electron in hydrogen-like atom is given. The terms Z and n given in Column II have usual meaning in Bohr's theory. Match the quantities in Column I with the terms they depend on in Column II.
Column-I Column-II
(A) Frequency of orbiting
electron (P) is proportional directly to Z2
(B) Angular momentum
of orbiting electron (Q) is proportional directly to n
(C) Magnetic moment of
orbiting electron (R) is proportional inversely to n3
(D) The average current due to orbiting of electron
(S) is independent of Z
20. In Young's double-slit experiment, the point source S is placed slightly off the central axis as shown in the figure. If l = 500 nm, then match the following.
P O S 2 mm 20 mm 1 m 2 m y = 10 mm S1 S2 Column-I Column-II
(A) Nature and order of interference at point P, OP = 10 mm
(P) Bright fringe of order 80 (B) Nature and order of
interference at point O (Q) Bright fringe of order 262 (C) If a transparent paper
(refractive index µ = 1.45) of thickness t = 0.02 mm is pasted on S1, i.e., one of the slits, the nature and order of the interference at P
(R) Bright fringe of order 62
(D) After inserting the transparent paper in front of slit S1, the nature and order of interference at O
(S) Bright fringe of order 280
PaPer-2 Section 1 (MaxiMuM MarkS : 32)
This section contains EIGHT questions •
The answer to each question is a SINGLE DIGIT •
INTEGER ranging from 0 to 9, both inclusive
1. For the given circuit in the steady state condition, charge on the capacitor is q0 = 16 µC. If now the battery is removed and the nodes A and C are shorted. The time during which charge on the capacitor becomes 4 µC is t(µs) and emf of battery is e(V). Find the value of 3
2 t eln . 2 3 3 1 4 4 + – + – C A D B C = 4 µF 24 V
2. A glass of refractive index 1.5 is coated with a thin layer of thickness t of refractive index 1.8. Light of wavelength l travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. If l = 648 nm, obtain the least value of t(in 10–8 m) for which the rays interfere constructively.
3. A charged particle enters a uniform magnetic field with velocity v0 = 4 m s–1 perpendicular to it, the length of magnetic field is x = 3
2 R, where R is the radius of the circular path of the particle in the field. Find the magnitude of change in velocity (in m s–1) of the particle when it comes out of the field.
4. Two tuning forks A and B each
O
A B
of natural frequency 85 Hz move with velocity 10 m s–1 relative to stationary observer O. Fork A moves away from the observer while the fork B moves towards him as shown in the figure. A wind
with a speed 10 m s–1 is blowing in the direction of motion of fork A. Find the beat frequency measured by the observer in Hz. [Take speed of sound in air as 340 m s–1]
5. In two calorimeters, we poured 200 g of water each at temperatures of +30 °C and +40 °C. From the hot calorimeter 50 g of water, is poured into cold calorimeter and stirred. Then from cold calorimeter 50 g of water is poured in hot and again stirred. How many times do you have to pour the same portion of water back and forth so that the temperature difference between water in the calorimeters becomes less than 3 °C ? Heat loss during the transfer and head capacity of calorimeters is neglected.
6. The tap in the garden was
3 cm
d1= 6 mm
d2= 3 mm
closed inappropriately resulting in the water flowing freely out of it which forms a downward narrowing beam. The beam of water has a circular cross-section, the diameter
of the circle is 6 mm at one point and 3 cm below it is only 3 mm as shown in figure. If the rate of water wasted is 55.65 × 10–n L s–1 then find the value of n. (Neglect the effect of viscosity and surface tension of the flowing water.)
7. Nuclei A and B convert into a stable nucleus C. Nucleus A is converted into C by emitting 2 a-particles and 3 b-particles. Nucleus B is converted into C by emitting one a-particle and 5 b-particles. At time t = 0, nuclei of A are 4N0 and nuclei of B are N0. Initially, number of nuclei of C are zero. Half-life of A (into conversion of C) is 1 min and that of B is 2 min. Find the time (in minutes) at which rate of disintegration of A and B are equal.
8. A block of mass m is being pulled up the rough incline, inclined at an angle 37° with horizontal by an agent delivering constant power P. The coefficient of friction between the block and the incline is µ. Find the maximum speed (in m s–1) of the block during the course of ascent.
[Take: P = 60 W, m = 1 kg, m = 0.5]
Section 2 (Maximum Marks : 32)
This section contains EIGHT questions •
Each question has FOUR options (a), (b), (c) and (d). •
ONE OR MORE THAN ONE of these four option(s) is(are) correct
9. A thin uniform bar lies on a
10 m s–1
6 m s–1
A B frictionless horizontal surface
and is free to move in any way on the surface. Its mass is 0.16 kg and length 3 m.
Two particles, each of mass 0.08 kg, are moving on the same surface and towards the bar in a direction perpendicular to the bar, one with a velocity of 10 m s–1, and other with 6 m s–1 as shown in the figure. The first particle strikes the bar at point A and the other at point B. Points A and B are at a distance of 0.5 m from the centre of the bar. The particles strike the bar at the same instant of time and stick to the bar on collision. The loss of the kinetic energy of the system in the collision process is
(a) 2 J (b) 4 J (c) 2.72 J (d) 5.44 J
10. Consider the earth as a uniform sphere of mass M and radius R. Imagine a straight smooth tunnel made through the earth which connects any two points on its surface. The time taken by a particle to go from one end to other through the tunnel is (a) 2p R GM 3 (b) p R GM 3 (c) R GM 3 (d) R GM 3 2
11. A linear object of size 1.5 cm is placed at 10 cm from a lens of focal length 20 cm. The optic centre of lens and the object are displaced a distance D. The magnification of the image formed is m. (Take optic centre as origin). The coordinates of image of A and B are (x1, y1) and (x2, y2) respectively. Then
(a) (x1, y1) = (–20 cm, –1 cm) (b) (x2, y2) = (–20 cm, 2 cm) (c) m = 3 (d) m = 2
12. A conducting wire of length l and mass m is placed on two inclined rails as shown in the figure. A current I is flowing in the wire in the direction shown. When no magnetic field is present in the region, the wire is just on the verge of sliding. When a vertically upward magnetic field is switched on, the wire starts moving up the incline. The distance travelled by the wire as a function of time t will be
C x l I z y (a) 1 2 2 2 IBl m − g t (b) 1 2 1 2 2 IBl m × − g t cosq sinq (c) 1 2 2 2 IBl m − g t sinq (d) 1 2 2 2 2 IBl m g t cos cos sin q q − q
13. For a certain metal, the K absorption edge is at 0.172 Å. The wavelength of Ka, Kb and Kg lines of K series are 0.210 Å, 0.192 Å, and 0.180 Å, respectively. The energies of K, L and M orbits are EK, EL and EM,
respectively. Then
(a) EK = –13.07 keV (b) EL = –7.52 keV (c) EM = –3.21 keV (d) EK = 13.04 keV
14. Two light springs of force constants k1 and k2 and a block of mass m are in one line AB on a smooth horizontal table such that one ends of each spring is fixed on rigid supports and the other end is free as shown in the figure.
A C D B
60cm
k1 v k2
The distance CD between the free ends of the springs is 60 cm. If the block moves along AB with a velocity 120 cm s–1 in between the springs, calculate the period of oscillation of the block.
(k1 = 1.8 N m–1, k2 = 3.2 N m–1, m = 200 g) (a) 3 s (b) 4 s
15. PQR is an equilateral triangular frame of mass m and side r. It is at rest under the action of horizontal magnetic field B as shown in the figure and the gravitational field.
r 3
4
(a) The frame remains at rest if the current in the frame is 2mg
rB .
(b) The frame remains at rest if the current in the frame is 2
3 mg rB .
(c) The frame is in simple harmonic motion when frame is slightly displaced in its plane perpendicular to PQ. The period of oscillation is p r g 3 1 2 /
(d) For same as in above option, the period of oscillation is p 3 2 1 2 r g /
16. Two identical cylindrical vessels with their bases at the same level each contain a liquid of density r. The height of the liquid in one vessel is h1 and in other vessel is h2. The area of either base is A. What is the work done by gravity in equalizing the levels when the two vessels are connected ?
(a) rAg 4 (h1 – h2) 2 (b) rAg 4 (h1 + h2)2 (c) rAg 2 (h1 – h2) 2 (d) rAg 2 (h1 + h2)2 Section 3 (Maximum Marks : 16)
This section contains TWO paragraphs •
Based on each paragraph, there will be TWO questions •
Each question has FOUR options (a), (b), (c) and (d). •
ONE OR MORE THAN ONE of these four options(s) is(are) correct
PARAGRAPH 1
The figure shows a cross-section of a double glass unit of a window on a vertical wall. A graph of the temperatures at different points within the unit is shown next to it.
The temperature difference across the unit is 13 K. It has a cross-sectional area of 1.3 m2 and the rate of heat flow through it is 65 W. Glass has a thermal conductivity of 1 W m–1 K–1.
17. Select the correct statement.
(a) The unit is in steady state and in thermal equilibrium.
(b) The unit is in steady state but not in thermal equilibrium.
(c) The unit is not in steady state but is in thermal equilibrium.
(d) The unit neither in steady state nor in thermal equilibrium.
18. The thermal conductivity of air is (a) 1
10W m–1K–1 (b) 112W m–1K–1 (c) 1
14W m–1K–1 (d) 9130W m–1K–1 PARAGRAPH 2
In the given setup, the parallel plate capacitor AB has vertical plates with separation, d = 50 mm and capacitance C0. From the plate A a small conducting ball hangs on a non-conducting silk thread of length L = 100 mm. The mass of the ball is m and capacitance is C1. It initially touches the plate A, as shown in the figure. The plate B is grounded while plate A is connected to a power supply of potential V0 for a short time by closing the switch S and then opening it again.
The motion of the conducting ball was observed. It was found that due to the charge deposited on the plate and the ball, the ball swings across, touches the plate B, swings back, touches A and finally swings out again such that it almost touches plate B. Take g = 10 m s–2.
Energy in volume of element (length l) dUB = uB dV = mp0 2 2 2 8 i r (2prl)dr = m p 02 4 i l dr r UB = m0p mp 2 0 2 4 4 i l dr r i l b a a b =
∫
lnUsing values, we get UB = 7 nJ
4. (4) : As the track AB is frictionless, the block moves this distance without loss in its initial KE = 1
2mv2 = 12 × 0.5 × 32 = 2.25 J.
In the path BD as friction is present, so work done against friction
= µkmg × (BD) = 0.2 × 0.5 × 10 × 2.14 = 2.14 J
So, at D the KE of the block = 2.25 – 2.14 = 0.11 J. Now, if the spring is compressed by x
0.11 = 12 × k × x2 + µk mgx
or 0.11 = 12 × 2 × x2 + 0.2 × 0.5 × 10x or x2 + x – 0.11 = 0
On solving, x = 0.1 m or –1.1 m x ≠ –1.1 m, so, x = 0.1 m
After moving the distance x = 0.1 m the block comes to rest. Now the compressed spring exerts a force:
F = kx = 2 × 0.1 = 0.2 N
on the block while limiting frictional force between block and track is fL = µs mg = 0.22 × 0.5 × 10 = 1.1 N. Since, F < fL. The block will not move back. So, the
total distance moved by the block
= AB + BD + 0.1 = 2 + 2.14 + 0.1 = 4.24 m 5. (4) : u = v l e 2( +2 ) \ u =2(l+ ×2 0 6v . )r =2(l v1 2. )r + \ Du D D D D D u = − + + = − + + v v l r l r v v l r l r ( . ) . . . 1 2 1 2 1 2 1 2 Here Dv v = 0 ; Du D D u ×100= − ++ × 1 2 1 2 100 l r l r . . %
For maximum % error: Dl = 0.1 cm, Dr = 0.05 cm Du u × 100 = ++ ×× × 0 1 1 2 0 05 94 1 2 5 100 % . .. . % max = 0.16% = a2% \ a = 0.4. Hence, 10a = 4.
6. (2) : The capacitor charge as a function of time is given by
q = Ce(1 – e–t/RC),
19. The ratio of the charge carried by the plates and the ball finally is
(a) C1 : C0 (b) C12 : C02 (c) C02 : C12 (d) C0 : C1
20. Based on the description given in the passage, the required power supply voltage (V0) is given by (a) V0 = 1 1 2 3 0 1 + C C mC (b) V0 = 1 2 3 0 1 1 + C C m C (c) V0 = 1 5 3 0 1 1 + C C m C (d) V0 = 1 5 3 1 0 1 + C C m C solutions paper-1
1. (2) : Magnetic moment of a current carrying loop, m= IS
Area of the loop, S AB BC = × Here, AB = −2d i+2a j BC , =2b k
\ S= −( 2d i+2a j) (× 2b k)=4bd j+4ab i \ | | | |m =I S =4Ib a d2+ 2
= 4 × 0.01 × 10 × 3 42+ 2
= 0.4 × 5 = 2 J T–1
2. (5) : We don't really need to write an integral, we just need the charge per unit length in the cylinder to be equal to zero. This means that the positive charge in cylinder must be +3.60 nC m–1. This positive charge is uniformly distributed in a circle of radius R = 1.50 cm, so r = 3 60 3 60 0 015 1 2 1 2 . . ( . ) nC m nC m m − − = pR p ≈ 5 µC m–3.
3. (7) : The magnetic field inside is only due to the current of the inner cylinder. B = m
p0 2
i r Magnetic field energy density is not uniform in the space between the cylinders. At a distance r from the centre
uB = B i r 2 0 0 2 2 2 2m 8 m p = r
While the current through the circuit (and the resistor) is given by i = e Re t RC − /
The energy stored in the capacitor is given by U = 2qC2
So the rate that energy is being stored in the capacitor is PC = dU dt q C dq dt q Ci = = .
The rate of energy dissipation in the resistor is PR = i2R
So the time at which the rate of energy dissipation in the resistor is equal to the rate of energy storage in the capacitor can be found by solving
PC =PR or, i2R = qCi or, iRC = q,
eCe–t/RC = Ce(1 – e–t/RC), e–t/RC = 1/2 or, t = RC ln 2.
7. (2) : For the ball, centripetal force = gravitational force mw2(2R) = GMm R ( )2 2 or M = 8w2 3R 32p22 3 G R T G = w= p 2 T \ M = 32 3 149000022 6 67 101 5 106 311 × × × × × − ( . ) ( . ) ( ) . = 1.97 × 1021 ≈ 2 × 1021 kg
8. (4) : Let the process is polytropic. According to the law pVn = constant Thus, pfVnf = piVni or, VVf pp i n i f = Given V V p p f i i f = =a 4and = =b 8 So, an = b or ln b = n ln a or n = lnlnba
In the polytropic process, molar heat capacity is given by Cn = R n n R R n ( ) ( )( ) − − − = − − − γ γ γ 1 1 1 1 = R R γ a b a −1− − ln ln ln So, Cn = 1 4 18 314.. − −ln8 314 4.8−lnln4 = 4.16 ≈ 4 J mol–1K–1 9. (a, b, c, d) : We have, 6.25% = 6 25 100 1 16 1 24 . = =
The given time of 4 h thus equals 4 half-lives so the half-life is 1 h.
Since half-life = decay constantln2 and mean life =decay constant1 decay constant, l = ln2 h–1 mean life, t = 1
2 ln h
After further 4 h, the amount left over would be 1 2 1 2 4× 4 , i.e., 1256 100 256 or % or 0.39% of original amount.
10. (b, c) : The initial compression in the spring (Dl) must be such that after burning of the thread, the upper cube rises to a height that produces a tension in the spring that is atleast equal to the weight of the lower cube. Actually, the spring will first go from its compressed state to its natural length and then get elongated beyond this natural length. Let l be the maximum elongation produced under these circumstances.
Then kl = mg ...(i)
Now, from energy conservation, 1
2kDl2 =mg (Dl + l) + 12kl2 ...(ii) (Because at maximum elongation of the spring, the speed of upper cube becomes zero)
From equations (i) and (ii),
Dl2 mg lD m g2 2 Dl mg mg
2
2 3 0 3
− − = = −
k k or, k , k
Therefore, acceptable solution of Dl equals 3mg k . Let v be the velocity of upper cube at the position (say, at C) when the lower block breaks off the floor, then from energy conservation
1 2mv2 = 12k(Dl2 – l2) – mg(l + Dl) Dl mg k = 7 or, v2 = 32 mg k 2
At the position C, the velocity of CM, vc = mv
m v + =0
2 2. Let the CM of the system (spring + two cubes) further rises up to Dyc2.
Now, from energy conservation, C v l l B 1 2(2m)v 2 c = (2m)g Dyc2 or, Dyc2 = vgc vg mgk 2 2 2 8 4 = =
But, uptil position C, the C.M. of the system has already elevated by,
Dyc1 = (Dl l m+2m) + =0 4mgk
Hence, the net displacement of the C.M. of the system, in upward direction
Dyc = Dyc1 + Dyc2 = 8mgk
11. (b) : Change in the resistance of the conductor on increasing its temperature,
R – R0 = R0aav(T – T0).
The disks will be effectively in series, so we will add the resistances to get the total. Looking only at one pair of disk, we have
Rc + Ri = R0c (ac(T – T0) + 1) + R0i (ai(T – T0) + 1) = R0c + R0i + (R0c ac + R0i ai) (T – T0).
This equation will only be constant if the coefficient for the term (T – T0) vanishes.
Then R0cac + R0iai = 0,
but R = rL/A, and the disks have the same cross sectional area, so Lcrcac + Liriai = 0 or L Lci i i c c = −r a r a = − × × × × × − × − − − − 9 68 10 6 5 10 3500 10 0 50 10 8 3 8 3 . . ( . ) = 0.036
12. (a, c) : To just lift off the rocket the launching pad, weight = thrust force
or mg = vr−dtdm or −dtdm=mgv
r
Substituting the values, we get − = + × dm dt (450 50 9 8)( . ) 2 103 = 2.45 kg s –1 Net acceleration, a = 20 m s–2 \ ma = Ft – mg or a = vmr −dtdm – g This gives − = + dm dt m g a vr ( )
Substituting the values, we get − = + + × dm dt (450 50 9 8 20)( . ) 2 103 = 7.45 kg s–1
The rate of fuel consumption is 10 kg s–1. So, the time for the consumption of entire fuel is
t = 450 10 = 45 s
Speed of the rocket after t = 45 s v = u – gt + vr ln m m0 Here, u = 0, vr = 2 × 103 m s–1, m0 = 500 kg and m = 50 kg
Substituting the values, we get
v = 0 – (9.8)(45) + (2 × 103) ln 500 50 or v = –441 + 4605.17 or v = 4164.17 m s–1= 4.164 km s–1 ≈ 4.2 km s–1
13. (a, b, c) : Given: n = 1 mole γ = 1.4 for diatomic gas.
T V V V V A A B C B = 300 K, = 16, = 2 A → B : adiabatic compression B → C : isobaric expansion C → D : adiabatic expansion D → A : isochoric process. To find TB B C D A P V V0 2V0 16V0 Process A → B is adiabatic \ TTB VV A A B − = γ 1 or TTB A = (16) 1.4 – 1 = (16)2/5 = (256)1/5 = 3.03 \ TB = TA × 3.03 or TB = 300 × 3.03 or TB = 909 K …(i) To find TD B → C is an isobaric process. \ V T V T B B C C = or TC = TB V VCB or TC = 909 × 2 or TC = 1818 K …(ii) C → D is an adiabatic process. \ TTD VV C C D = − γ 1 or TD = TC × VVC D − γ 1 or TD = 1818 × 162 1 4 1 − . = 1818 18 2 5 / or TD = 1818 × 0.435 or TD = 791.4 K …(iii)
% efficiency (h) = Net work doneHeat absorbed × 100% or h = Q1Q Q2
1
−
× 100%
where Q2 = Heat released in the cycle Q1 = Heat absorbed in the cycle For adiabatic process AB and CD, DQ = 0 or QAB = QCD = 0
Now QDA = nCV dT or QDA = (1)52R (TA – TD)
Q2 = 5 8.31×2 (300 – 791.4) = –10208.8 J Negative sign shows that heat is lost by the system
Again QBC = nCp DT Q1 = (1)72R (TC – TB) Q1 = 1 × 7 8 31×2. × (1818 – 909) Q1 = 7 8 31 909× .2× = 26438.3 J \ h = 1 2 1 − QQ 100% or h = 1 10208 8 26438 3 − . . 100% or h = 61.4% Hence TB = 909 K, TD = 791.4 K ≈ 791 K h = 61.4%
14. (a) : Initially, there will be induced charges of magnitude –q and +q on the inner and outer surface of the spherical layer respectively. Hence, the total electrical energy of the system is the sum of self energies of spherical shells, having radii a and b, and their mutual energies including the point charge q. Ui = 12 4 12 4 4 4 4 2 0 2 0 0 0 0 q b q a qq a qq b qq b pe + pe pe pe pe − + − + + − ( ) or, Ui = q b a 2 0 8pe 1 1−
Finally, charge q is at infinity hence, Uf = 0
Now, work done by the agent = increment in the energy = Uf – Ui = q a b 2 0 8pe 1 1−
15. (a, c) : Surface tension = Surface energy per unit area r = Radius of each small drop
R = Radius of big drop n × 4
3pr3 = 43pR3 or R = n r
1 3
Initial surface energy = Ei = n × 4pr2 × T = nE
Final surface energy = Ef = 4pR2 × T = 4pr2n2/3⋅T
Ef = n2/3E
Energy released = Ei – Ef = E(n – n2/3)
16. (d) : A 30 cm string fixed at both ends will resonate if the frequency is given by u = nv/2L, where n is an integer, L the length of the string, and v the wave-speed on the string. The string is observed to resonate at 880 Hz and then again at 1320 Hz, so the two corresponding values of n must differ by 1. We can then write two equations
(880 Hz) = nv2L and (1320 Hz) = (n2+1L)v or, (880 ) ( 2 1320 1 2 Hz and Hz) n v L n v L = + = .
Combining these two equations, we get (880 ) (1320 ) 1 Hz Hz n = n+ , (n + 1) (880 Hz) = n(1320 Hz), n = ( ) ( ) ( ) 880 1320 880 Hz Hz − Hz = 2.
For a string fixed at both ends (or a pipe closed or opened at both ends), the frequency difference between two adjacent harmonics is the same as the fundamental frequency, i.e., u1 = un + 1 – un
\ v = 2(0.3) (880)
2 = 264 m s–1 The tension, in the string
T = mv2 = 0.652 × 10–3 × (264)2 = 45.4 N
17. (b, c, d) : Let us fix the x-y co-ordinate system to the given plane, taking x-axis in the direction along which the force vector was oriented at the moment t = 0, then the fundamental equation of dynamics expressed via the projection on x and y-axis gives at latter time t, F cos wt = m dv dtx ...(i) and F sin wt = mdv dt y ...(ii) From (i), dv F m t dt x vx t =
∫
∫
0 0 cos w ...(iii) or vx = Fmwsin wtand similarly from (ii), vy = F
Hence, v v v F m t x y = + = 2 2 2 2 w w sin
It is seen from this that the velocity v turns into zero after the time interval Dt, which can be found from the relation,
O x
y F
= t
wDt2 = p. Consequently, the sought distance, is s = v dt F m t dt Fm t = =
∫
∫
2 2 8 0 2 2 0 w w w p w sin / D Average velocity, < > =∫
= =∫
∫
v v dt dt F m t dt F m w p w w p w p w 2 2 2 4 0 2 sin /18. (b) : The rope wraps around the dowel and there is a contribution to the frictional force Df from each small segment of the rope where it touches the dowel. There is also a normal force DN at each point where the contact occurs.
In the figure we can form a triangle with long side T and short side DN. In another we see a triangle with long side r and short side rDq. These triangles are similar, so rDq/r = DN/T.
T N r
r T
r
Now Df = mDN and T(q) + Df ≈ T(q + Dq). Combining, and taking the limit as Dq → 0, dT = df
1
m q
dT
T d
∫
=∫
Integrating both sides of this expression, 1 1 2 0 m q p dT T d T T
∫
=∫
; 1 12 m[ln ]T TT =p T2 = T1epµ.In this case T1 is the weight and T2 is the downward force. \ F = Wepµ 19. A → (P, R), B → (Q, S), C → (Q, S), D → (P, R). A. u = mZ e h n 2 4 0 2 3 3 4e ⇒ u ∝ Zn 2 3 B. L = nh2p ⇒ L ∝ n C. Magnetic moment: M = IA = 2per vpr2 = evr e m mvr e mL 2 =2 ( )=2 ⇒ M = e m nh 2 2p ⇒ M ∝ n D. i = ev r e mZe n h Ze nh 2 2 2 2 2 2 0 2 0 p p p e e = ⇒ i ∝ Zn 2 3 20. A → (S), B → (P), C → (Q), D → (R)
The optical path difference between the two waves arriving at P is d = (SS2 + S2P) – (SS1 + S1P) = (SS2 – SS1) + (S2P – S1P) = d sinq0 + d sinq = dyD0 dyD 1 2 + . d = 20 mm, y0 = 2 mm, D2 = 2 m, D1 = 1 m, y = 10 mm \ d = 20 21000× + × = 0.14 mm20 102000
For a bright fringe, d = nl ⇒ n = ld= × − 0 14 0 5 10 3 . . = 280 At the origin O, d' = dy D10 = 0.04 mm n' = ′= × − d l 0 04 0 5 10 3 . . = 80
Due to transparent paper, the change in optical path is (µ – 1)t = (1.45 – 1) (0.02) mm = 0.009 mm d" = 0.14 mm – 0.009 mm = 0.131 mm ⇒ n = 0 5 100 1313 . . × − = 262
Due to transparent paper, the path difference at O, d''' = d' –(µ – 1)t = (0.04 – 0.009) mm = 0.031 mm ⇒ n = 0 031
0 5 10 3
.
. × − = 62
1. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :
(a) 92 ± 2 s (b) 92 ± 5.0 s (c) 92 ± 1.8 s (d) 92 ± 3 s
2. A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x-y plane as shown in the figure.
Which of the following statements is false for the angular momentum L about the origin?
(a) L= −mv Rk 2
^ when the particle is moving from
A to B. (b) L mv R a k= − 2
^ when the particle is moving
from C to D. (c) L mv R a k= + 2
^ when the particle is moving
from B to C. (d) L mv Rk=
2
^ when the particle is moving from
D to A.
3. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals m. The particle is released, from rest, from the point P and it comes to rest at
a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR.
The values of the coefficient of friction m and the distance x (= QR), are, respectively close to :
(a) 0.2 and 6.5 m (b) 0.2 and 3.5 m (c) 0.29 and 3.5 m (d) 0.29 and 6.5 m
4. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 m s –2. (a) 2.45 × 10–3 kg (b) 6.45 × 10–3 kg (c) 9.89 × 10–3 kg (d) 12.89 × 10–3 kg
5. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to :
(a) turn left (b) turn right (c) go straight
(d) turn left and right alternately
6. A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere.)
(a) 2gR (b) gR
(c) gR/2 (d) gR 2 1
(
−)
7. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (a) of the metal of the pendulum shaft are respectively.
(a) 25°C; a = 1.85 × 10–5/°C (b) 60°C; a = 1.85 × 10–4/°C (c) 30°C; a = 1.85 × 10–3/°C (d) 55°C; a = 1.85 × 10–2/°C
8. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn = constant,
then n is given by (Here CP and CV are molar specific
heat at constant pressure and constant volume, respectively) : (a) n C CVP = (b) n C CC CP V = − − (c) n C C C CP V = − − (d) n C C C CVP = − −
9. ‘n’ moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of the gas during the process will be :
(a) 9 40 0 P V nR (b) 3 20 0 P V nR (c) 9 20 0 P V nR (d) 9P V0 0 nR
10. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2
3A from equilibrium position. The new amplitude of the motion is :
(a) A
3 41 (b) 3A (c) A 3 (d) 7
3 A
11. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is :
(take g = 10 m s–2)
(a) 2 2p s (b) 2 s (c) 2 2 s (d) 2 s
12. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density ρ = Ar,where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is : (a) Q a 2p 2 (b) Q b a 2p( 2 − 2) (c) a22Qb2 p( − ) (d) 2 2 Q a p
13. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field due to a point charge Q (having a charge equal to the sum of the charges on the 4 mF and 9 mF capacitors), at a point distant 30 m from it, would equal :
(a) 240 N/C (b) 360 N/C (c) 420 N/C (d) 480 N/C
14. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by :
(a) Linear increase for Cu, linear increase for Si. (b) Linear increase for Cu, exponential increase for Si. (c) Linear increase for Cu, exponential decrease
for Si.
(d) Linear decrease for Cu, linear decrease for Si.
15. Two identical wires A and B, each of length ‘l ’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of magnetic field at
the centres of the circle and square respectively, then the ratio B
BAB is : (a) p2 8 (b) p 2 16 2 (c) p2 16 (d) p 2 8 2
16. Hysteresis loops for two magnetic materials A and B are given below :
These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use :
(a) A for electric generators and transformers. (b) A for electromagnets and B for electric generators. (c) A for transformers and B for electric generators. (d) B for electromagnets and transformers.
17. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms),
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
18. Arrange the following electromagnetic radiations per quantum in the order of increasing energy : A : Blue light B : Yellow light
C : X-ray D : Radiowave (a) D, B, A, C (b) A, B, D, C (c) C, A, B, D (d) B, A, D, C
19. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears :
(a) 10 times taller (b) 10 times nearer (c) 20 times taller (d) 20 times nearer
20. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength l the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when : (a) a L b L =l2 and min=2l2 (b) a L b L = l and min=2l2 (c) a= lLandbmin= 4lL (d) a L=l2 andbmin= 4lL
21. Radiation of wavelength l, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3
4
l, the speed of the fastest emitted electron will be :
(a) > v 43 1 2/ (b) <v 4 3 1 2/ (c) = v 43 1 2/ (d) = v 34 1 2/
22. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be :
23. If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is :
(a) NOT (b) AND (c) OR (d) NAND
24. Choose the correct statement :
(a) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
(b) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
(c) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
(d) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.
25. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?
(a) 0.75 mm (b) 0.80 mm (c) 0.70 mm (d) 0.50 mm
26. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :
(a) f 2 (b) 3 4 f (c) 2f (d) f
27. A galvanometer having a coil resistance of 100 W gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is : (a) 0.01 W (b) 2 W
(c) 0.1 W (d) 3 W
28. In an experiment for determination of refractive index of glass of a prism by i – d, plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index?
(a) 1.5 (b) 1.6 (c) 1.7 (d) 1.8
29. Identify the semiconductor devices whose characteristics are given below, in the order (i), (ii), (iii), (iv) :
(a) Simple diode, Zener diode, Solar cell, Light dependent resistance
(b) Zener diode, Simple diode, Light dependent resistance, Solar cell
(c) Solar cell, Light dependent resistance, Zener diode, Simple diode
(d) Zener diode, Solar cell, Simple diode, Light dependent resistance.
30. For a common emitter configuration, if a and b have their usual meanings, the incorrect relationship between a and b is : (a) a b1 1 1= + (b) a= bb − 1 (c) a= bb + 1 (d) a b b = + 2 2 1
SolutionS
1. (a) : Here, t1 = 90 s, t2 = 91 s, t3 = 95 s, t4 = 92 s L.C. = 1 s
Mean of the measurements, t t N i i =
∑
t =90 91 95 92+ + + = 4 92 s Mean deviation = − = + + + =∑
t t N i i 2 1 3 0 4 1 5. s Since the least count of the instrument is 1 s, so reported mean time = (92 ± 2) s.2. (b, d) : Here v = speed of the particle a = side of square AE = R sin 45° = R 2 OE = R cos 45° = R 2 We know, L r p rp= × = sinqn L =( sin )r q p r p= ⊥
When the particle is moving along AB,
L AE p k=( )( )( )− = −mv Rk 2
When the particle is moving along BC,
L OF p k mv R a k= = + ( )( )( ) 2
When the particle is moving along CD,
L DE p k mv R a k= = + ( )( )( )
2
When the particle is moving along DA,
L OE p k=( )( )( )− = −mv Rk 2
Hence, options (b) and (d) are incorrect.
3. (c) : Here, PQ= h h °= = sin30 2 4 m QR = x = ?, m = ?
Energy of the particle is lost only due to friction between the track and the particle.
According to the question,
Energy lost by the particle over the part PQ = Energy lost by the particle over the part QR
or, f × PQ = f ′ × QR
or, m mg cos 30° × 4 = m mg x or, x = 4 cos 30° = 4 3
2 =2 3 m 3.5 m≈
Using work energy theorem for the motion of the particle, mgh – (f × PQ) – f ′(QR) = 0 – 0 or mgh – 2 f ′(QR) = 0 or, mgh – 2mmg x = 0 \ m = = × = ≈ h x 2 2 2 2 3 0 288 0 29. . 4. (d) : Here, m = 10 kg, h = 1 m, g = 9.8 m s–2 n = 1000 Energy of fat = 3.8 × 107 J kg–1 Efficiency, h = 20% = 15
Net work done by the man in lifting the mass = n × (Gain in potential energy of the mass) = n(mgh) = 1000 × 10 × 9.8 × 1 = 98000 J h = Net work done by the man
Energy in the fat 1 5 98000 3 8 107 = × × m . or, m = × × 98000 5 3 8 10. 7 \ m = 12.89 × 10–3 kg 5. (a)
6. (d) : Orbital velocity of the satellite, v GM R h
0= +
v GM
R
0≈ (... h << R)
Let ve be the minimum velocity required by the
satellite to escape from its orbit. \ 1 2mv2 GmM R h e = + ⇒ v GM R h GM R e= 2 + ≈ 2 (... h << R) so, required increment in the orbital velocity
= − =v v GM − R GM R e 0 2 = GM − = − R ( 2 1) gR( 2 1)
7. (a) : Time period of the pendulum clock at temperature q is given by T l g l g q=2p q =2p 0(1+aq) =2 0 1+ 1 2 p l aq g( )
T Tq≈ 01 1+ aq
2 ...(i)
Assume pendulum clock gives correct time at temperature q0
\ Tq0 T0 1 1aq0 2
= + ...(ii) At q= ° >40 C q0as clock loses time.
T40 T0 1 1 2 40
= + a× ...(iii) At q = 20°C < q0 as clock gains time.
T20 T0 1 1 2 20
= + a× ...(iv) From equations (ii) and (iii), we get
T T T 40 0 0 0 1 2 40 − = − q a q ( ) or 12 s = a(40 – q0) (12 h) ...(v) From equations (ii) and (iv), we get
T T T q0 20 a q 0 0 1 2 20 − = ( − )
or, 4 s = a(q0 – 20)(12 h) ...(vi) From equations (v) and (vi), we get
3(q0 – 20) = (40 –q0) 3q0 + q0 = 40 + 60
q0 100 4 25
= = °C
From equation (vi), 4 s = a(25 – 20)(12 × 3600 s) a =
× × = × − ° −
4
5 12 3600 1 85 10. 5 C 1
8. (b) : Here, PVn = constant or, PnVn–1 dV + Vn dP = 0
or, nPdV = – V dP
Also, from ideal gas equation PV = nRT PdV + VdP = nR dT or, PdV – nPdV = nRdT or, PdV nRdT n = − (1 ) ...(i) Also, dQ = dU + dW ⇒ nC dT = nCVdT + PdV nCdT nC dT nRdT n V = + − (1 ) or, C = C R n V +(1− ) or, (1− =n) C C−R V or, n R C C C C R C C C C C C V V V P V = − − = − + − = − − 1 ( )
9. (a) : Equation of line AB is given by
y y y y x x x x − = − − − 1 2 1 2 1( 1) P P P P V V V V − = − − − 0 0 0 0 0 0 2 2 ( 2 ) or, P P V V P = − 0 + 0 3 or, PV0 P V V P V = − 0 + 0 2 0 3 or, nRT P V V P V = − 0 + 0 2 0 3 or, T nR P V V P V = − + 1 0 3 0 2 0 ...(i)
For maximum value of T, dT dV = 0 or, −P + =
V00( )2V 3P0 0 \ V= 32V0 So, from equation (i)
T nR P V V P V max= 1 − 0 ×94 +92 0 0 2 0 0 = 94P VnR0 0.
10. (d) : Speed of a particle performing SHM is given by v=w A2−x2
At x= 2A
3 , initial speed of the particle, v=w A2− 2A =2 w A2 =wA 3 5 9 5 3 Now, its speed is trebled at the instant x= 2A
3 from equilibrium position, then new amplitude of the SHM is A′ (say). Hence, ′ = =v 3v A′ − 2A 3 2 2 w or, wA 5 w A 4A 9 2 2 = ′ − or, 5 4 9 2 2 2 A = ′ −A A or, ′ =A2 49A2 9 \ ′ =A 73A. 11. (c) : Speed of the wave pulse (wave) in the string,
v= T m Here, T m l y g = × × and m =ml \ v m l y g m l gy = × × = / Also, v dy=dt = gy or, dy y gdt t 0 20 0
∫
=∫
or, y1 2 g t t 0 20 0 1 2 / / =[ ]
⇒ 2 20 0( − =) 10× t \ t = 2 2 s12. (a) : Using Gauss’s theorem for radius r E d s⋅ = Q q+
∫
e10( ) ⇒ E×4 r2= 1 Q q+ 0 p e ( ) ...(i) q = charge enclosed between x = a and x = r. q A x x dx A xdx a r a r =∫
4p 2 =4p∫
=4 2 2 pA x a r =2pA r( 2−a2)Putting the value of q in equation (i), we get E×4 r2= 1 Q+2 A r −a 0 2 2 p e p ( ) E Q r A Aa r = + − 1 4 2 2 0 2 2 2 pe p p
E will be constant if it is independent of r \ Q r Aa r 2 2 2 2 = p or A Q a = 2p 2
13. (c) : 3 mF and 9 mF are in parallel combination so their equivalent capacitance = (3 + 9) = 12 mF
Now, 4 mF and 12 mF are in series so their equivalent capacitance =4 12× =
16 3 Fm
Charge on 3 mF = (3 mF) × (8 V) = 24 mC
\ charge on 4 mF and 12 mF are same (24 mC) as they are in series.
Charge on 9 9 9 3 24 18 Fm = Cm Cm + × =
Required charge Q = Charge on 4 mF + Charge on 9 mF Q = (24 + 18) mC = 42 mC
Required electric field, E Q r = 1 × 4pe0 2 E = ×9 10 ×42 10× − = − 30 420 9 6 2 1 ( ) N C
14. (c) : Resistivity of Cu increases linearly with increase in temperature because relaxation time decreases. Resistivity of semiconductor decreases exponentially with increase in temperature, as ρT =ρoe( /Eg k TB )
15. (d) : Wire A is bent into a circle of radius R, l=2 R⇒ =R l 2 p p B I R I l A= = × m m p 0 0 2 2 2 =m p0 I l
Wire B is bent into a square of side a, l=4a⇒ =a l 4 B I a B= ×4 4p( / )m0 2 (sin45° +sin45°) =2 × 2 = 2 16 2 0 0 m p m p I a I l \ B B I l I l A B = = m p m p p 0 0 2 16 2 8 2 / /
16. (d) : For both, the electromagnet and transformer, the magnetic field changes with time. Hence the energy losses must be less in both devices. Hysteresis loop represented in B has less area which means it dissipates less energy.
17. (d) : For a dc source I = 10 A, V = 80 V
Resistance of the arc lamp, R V I = =80= 10 8 W For an ac source, erms = 220 V u = 50 Hz w = 2p × 50 = 100 p rad s–1 Arc lamp will glow if I = 10 A, \ I R L = + e w rms 2 2 2 or, R L I 2+ 2 2= 2 w erms or, 8 100 220 10 2+ 2 2 2 = ( p L) or, L2 222 822 100 = − ( p) \ L = 30 14× = 100p 0 065. H.
