Physics for You June 2015

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Physics for you|june ‘15 7

Vol. XXIII No. 6 June 2015

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Editor : Anil Ahlawat

is the Advance of science by Logic, intuition or Experimentation?

O

ne had always assumed that the advance of science is logical and based on assumptions that are verified by experimentation. This was true during the days of classical physics. However, modern physics in the twentieth century has grown by leaps and bounds flouting the ancient logical methods. If one were to analyse the march of physics, einstein, Bohr, Louis de Broglie, Dirac, Schrodinger, Max Born and many other did not wait for experimentation for the formulation of theories.

According to einstein’s photoelectricity, photon is both a particle and a wave. Diffraction and interference of light as well as that of electrons were unified by Max Born. Max Born developed quantum mechanics based on the probability waves suggested by einstein. Dirac’s contribution of his idea of ‘Dirac vacuum’ consisting of all particles and antiparticles is near the idea of ‘Soonya’ of Indian Philosophy. However our philosophy says that it is “Poorna” or infinity.

Infinity + infinity = infinity, according to modern mathematics. But infinity – infinity is undefined. According to our concept, if one takes Poorna from Poorna, Poorna remains.

einstein’s concept of c, the velocity of light and his enunciation that the mass of a body increases with velocity are beyond classical physics and transcends normal logic. einstein, Bohr, Louis de Broglie and Max Born set right the controversy of whether light is a particle or wave, by their correction that particles and light are simultaneously matter and wave. Intuition has proved to be more powerful than logic, assumptions and experimentation.

Anil Ahlawat Editor

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physics for you|june ‘15 8

single option correct type

1. A disc of radius R is rolling purely on a flat horizontal surface, with constant angular velocity. The angle between the velocity and acceleration vectors at point P is (a) zero

(b) 45° (c) 135° (d) tan–1_(1/2)

2. A solid ball of radius r rolls inside a hemispherical shell of radius R without slipping.

It is released from rest from point A as shown in figure. The angular velocity of centre of the ball in position B about the centre of the shell is (a) 2 5 g R r ( − ) (b) 710 g R r ( − ) (c) 2 5 g R r ( − ) (d) 2 5 g R r ( − )

3. An object is moving towards a converging lens on its axis. The image is also found to be moving towards the lens. Then, the object distance u must satisfy

(a) 2f < u < 4f (b) f < u < 2f (c) u > 4f (d) u < f

4. A uniform solid brass sphere of radius a0 and mass m is set spinning with angular speed w0 about a diameter at temperature T0. If its temperature be increased to T without disturbing the sphere, its new angular speed, assuming that its new radius is a, will be

(a) w = w0 (b) w= T_T w 0 0 (c) w_{= }a _ w a0 2 0 (d) w=_T T_T− 0_w 0 0

5. A metal ball A (density 3.2 g cm–3_{) is dropped in water,} while another metal ball B (density 6.0 g cm–3_{) is dropped} in a liquid of density 1.6 g cm–3_{. If both the balls have the}

same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is

(a) 2.0 (b) 0.5 (c) 4.0 (d) 0.25 6. Two glass plates are touching at one end and separated

by a thin wire at the other end. When a monochromatic parallel beam of wavelength 4200 Å incident normally on the glass plates is reflected, an interference pattern of 30 fringes is observed. If the wavelength of light used is taken 7000 Å instead of 4200 Å, the number of fringes observed will be

(a) 50 (b) 40 (c) 30 (d) 18

7. Consider a YDSE that has different slits width, as a result, amplitudes of waves from slits are A and 2A, respectively. If I0 be the maximum intensity of the interference pattern, then intensity of the pattern at a point where phase difference between waves is f, is

(a) I0cos2_f _(b) I0 2

3sin f2

(c) I₉0[5 4+ cos ]f (d) I₉0[5 8+ cos ]f subjective type

8. Two identical sonometer wires have a fundamental frequency of 500 Hz when kept under the same tension. What fractional increase in the tension of one wire would cause an occurrence of 5 beats per second, when both wires vibrate together?

9. Find the temperature at which the fundamental frequency of an organ pipe is independent of small variation in temperature in terms of the coefficient of linear expansion (a) of the material of the tube.

10. A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope.

A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?

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.

By Akhil Tewari, Author Foundation of Physics for jee Main & Advanced, Senior Professor Physics, RAO IIT ACADeMY, Mumbai.

PHYSICS

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1. Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between

A and B is

B

A _C

(a) 8 N (b) 18 N

(c) 2 N (d) 6 N

2. If radius of the 27₁₃Al nucleus is taken to be RAl,

then the radius of 125₅₃Te nucleus is nearly

(a) 3₅R_Al (b) 13₅₃ 1 3    / R_Al (c) 53 13 1 3    / R_Al (d) 5₃R_Al

3. Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength? (a) p  (b) p  (c) p  (d) p 

Exam Q. No. MTG Book Q. No. P. No. Exam Q. No. MTG Book Q. No. P. No.

1 AIPMT Guide 28 103 24 AIPMT Guide 39 651

6 AIPMT Guide 45 720 29 NCERT Fingertips 38 119

Exam Q. No. MTG Book Q. No. P. No. Exam Q. No. MTG Book Q. No. P. No.

2 NCERT Fingertips 17 291 20 Physics For You May'15 39 34

15 Physics For You Jan'15 21 15 45 AIPMT Guide 18 446

Here, the references of few are given :

Exact Questions

Similar Questions

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Physics for you|june ‘15 11 4. The two ends of a metal rod are maintained at

temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be

(a) 8.0 J/s (b) 4.0 J/s

(c) 44.0 J/s (d) 16.8 J/s

5. For a parallel beam of monochromatic light of wavelength ‘l’, diffraction is produced by a single slit whose width ‘a’ is of the order of the wavelength of the light. If ‘D’ is the distance of the screen from the slit, the width of the central maxima will be

(a) Da_l (b) 2Da_l

(c) 2D

al (d) Da

l

6. Which logic gate is represented by the following combination of logic gates?

(a) AND (b) NOR

(c) OR (d) NAND

7. A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean

position are V1 and V2, respectively. Its time period

is (a) 2 12 22 12 22 p V V x x + + (b) 2 12 22 12 22 p V V x x − − (c) 2 12 22 12 22 p x x V V + + (d) 2 2 2 12 12 22 p x x V V − −

8. Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

(a) –50 cm (b) 50 cm

(c) –20 cm (d) –25 cm

9. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude

(a) m0n e2 r (b) m₀ 2 ne r (c) m p 0 2 ne r (d) Zero

10. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to

v(x) = bx–2n_,

where b and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by

(a) –2b2_x–2n + 1 _{(b) –2nb}2_e–4n + 1

(c) –2nb2_x–2n – 1 _{(d) –2nb}2_x–4n – 1

11. The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius ‘a’ centred at the origin of the field, will be given by

(a) 4pe0Aa3 (b) e0Aa3

(c) 4pe0Aa2 (d) Ae0a2

12. A radiation of energy ‘E’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light) (a) 2E₂ C (b) E C2 (c) E C (d) 2E C

13. A, B and C are voltmeters of resistance R, 1.5R and

3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then X Y B C A (a) VA = VB ≠ VC (b) VA ≠ VB ≠ VC (c) VA = VB = VC (d) VA ≠ VB = VC

14. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is (a) W d x(_x− ) (b) W d x(_d− )

(c) Wx

d (d)

Wd x

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15. A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is Z R _Y O X I I _I   (a) B I R i k = −  +    m p0 p 4 2 ^ ^ (b) B I R i k =  −    m p0 p 4 2 ^ ^ (c) B I R i k =  +    m p0 p 4 2 ^ ^ (d) B I R i k = −  −    m p0 p 4 2 ^ ^

16. A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2_.

Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (rair = 1.2 kg/m3)

(a) 2.4 × 105 _{N, upwards}

(b) 2.4 × 105 _{N, downwards}

(c) 4.8 × 105 _{N, downwards}

(d) 4.8 × 105 _{N, upwards}

17. In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

(a) 0.5 mm (b) 0.02 mm

(c) 0.2 mm (d) 0.1 mm

18. A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is

attached to a string which passes through a smooth hole in the plane as shown.

v₀ m R₀

The tension in the string is increased gradually and finally m moves in a circle of radius R0

2 . The final value of the kinetic energy is

(a) 2mv 02 (b) 1₂mv02

(c) mv02 (d) 1₄mv02

19. Kepler’s third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance

r between sun and planet i.e. T2 _{= Kr}3_{here K is}

constant.

If the masses of sun and planet are M and m respectively then as per Newton’s law of gravitation force of attraction between them is

F GMm r

= ₂ , here G is gravitational constant. The relation between G and K is described as

(a) K = G (b) K

G

= 1

(c) GK = 4p2 _{(d) GMK = 4p}2

20. Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis

XX′ which is touching to two shells

and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about XX′ axis is

(a) 16₅ mr 2 (b) 4mr2

(c) 11

5mr 2 (d) 3mr2

21. A ship A is moving Westwards with a speed of 10 km h–1_{and a ship B 100 km South of A, is}

moving Northwards with a speed of 10 km h–1_.

The time after which the distance between them becomes shortest, is

(a) 5 2 h (b) 10 2 h

(c) 0 h (d) 5 h

22. The ratio of the specific heatsC

C

p

v = γ in terms of degrees of freedom (n) is given by

(a) 1 2_ + _ n (b) 1 2+   n (c) 1 1_ + _ n (d) 1 3+   n

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Physics for you|june ‘15 13 23. If in a p–n junction, a square input signal of 10 V is

applied, as shown,

then the output across RL will be (a)

–5 V (b)

5 V

(c) (d)

24. A certain metallic surface is illuminated with monochromatic light of wavelength, l. The stopping potential for photo-electric current for this light is 3V0. If the same surface is illuminated

with light of wavelength 2l, the stopping potential is V0. The threshold wavelength for this surface for

photo-electric effect is (a) p 4 (b) l 6 (c) 6l (d) 4l

25. A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? (a) The change in energy stored is 1₂CV2 1 1

K−   .

(b) The charge on the capacitor is not conserved. (c) The potential difference between the plates

decreases K times.

(d) The energy stored in the capacitor decreases

K times.

26. A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is (a) 2mk t− /1 2 (b) 1₂ mk t− /1 2

(c) mk t₂ − /1 2 (d) mk t−1 2/

27. A Carnot engine, having an efficiency of η = 1 10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

(a) 90 J (b) 1 J

(c) 100 J (d) 99 J

28. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, A B 5 2 4 6 P(in kPa) V(in m )3

The change in internal energy of the gas during the transition is

(a) 20 J (b) –12 kJ

(c) 20 kJ (d) –20 kJ

29. A block of mass 10 kg, moving in x direction with a constant speed of 10 m s–1_{, is subjected to a}

retarding force F = 0.1x J/m during its travel from

x = 20 m to 30 m. Its final KE will be

(a) 275 J (b) 250 J

(c) 475 J (d) 450 J

30. Consider 3rd_{orbit of He}+_{(Helium), using}

non-relativistic approach, the speed of electron in this orbit will be [given K = 9 × 109_{constant, Z = 2 and} h (Plack’s Constant) = 6.6 × 10–34_{J s]}

(a) 0.73 × 106_{m/s (b) 3.0 × 10}8_m/s

(c) 2.92 × 106_{m/s (d) 1.46 × 10}6_m/s

31. A resistance ‘R’ draws power ‘P’ when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ‘Z’, the power drawn will be (a) P R Z    (b) P (c) P R Z    2 (d) P R Z

32. A block A of mass m1 rests on a horizontal table. A

light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass m2 is suspended. The

coefficient of kinetic friction between the block and the table is mk. When the block A is sliding on the table, the tension in the string is

(a) m m g m mk 1 2 1 2 1 ( ) ( ) + + m _(b)m m g m mk 1 2 1 2 1 ( ) ( ) − + m (c) (m₍2_{m m}km g1₎) 1 2 + + m (d) ( ) ( ) m m g m mk 2 1 1 2 − + m

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33. The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is

(a) 90° – A (b) 180° + 2A

(c) 180° – 3A (d) 180° – 2A

34. On observing light from three different stars P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If TP, TQ and TR are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that (a) TP < TR < TQ (b) TP < TQ < TR (c) TP > TQ > TR (d) TP > TR > TQ

35. A conducting square frame of side ‘a’ and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity ‘V’. The emf induced in the frame will be proportional to

(a) 1 2 2 ( x a+ ) (b) 1 2 2 ( x a x a− )( + ) (c) 1₂ x (d) 1 2 2 ( x a− )

36. Two spherical bodies of mass M and 5M and radii

R and 2R are released in free space with initial

separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is

(a) 7.5R (b) 1.5R

(c) 2.5R (d) 4.5R

37. Two similar springs P and Q have spring constants

KP and KQ, such that KP > KQ. They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs

WP and WQ are related as, in case (a) and case (b) respectively

(a) WP > WQ ; WQ > WP (b) WP < WQ ; WQ < WP (c) WP = WQ ; WP > WQ (d) WP = WQ ; WP = WQ

38. Two particles of masses m1, m2 move with initial

velocities u1and u2. On collision, one of the

particles get excited to higher level, after absorbing energy e. If final velocities of particles be v1 and v2

then we must have (a) 1 2 1 2 1 2 1 2 1 12 2 22 1 12 2 22 m u + m u − =e m v + m v (b) 1 2 1 2 1 2 1 2 1 2 1 2 12 2 22 2 12 22 22 m u + m u + =e m v + m v (c) m u m u₁2 ₁+ ₂2 ₂− =e m v m v₁2 ₁+ ₂2 ₂ (d) 1 2 1 2 1 2 1 2 1 12 2 22 1 12 2 22 m u + m u = m v + m v − e

39. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10–11_Pa–1

and density of water is 103_kg/m3_{. What fractional}

compression of water will be obtained at the bottom of the ocean?

(a) 1.2 × 10–2 _{(b) 1.4 × 10}–2

(c) 0.8 × 10–2 _{(d) 1.0 × 10}–2

40. Figure below shows two paths that may be taken by a gas to go from a state A to a state C.

In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be

(a) 460 J (b) 300 J

(c) 380 J (d) 500 J

41. The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

(a) 120 cm (b) 140 cm

(c) 80 cm (d) 100 cm

42. When two displacements represented by

y₁ = a sin(wt) and y2 = b cos(wt) are superimposed

the motion is

(a) simple harmonic with amplitude a b2₊ 2 (b) simple harmonic with amplitude (a b+ )

2 (c) not a simple harmonic

(d) simple harmonic with amplitude a

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Physics for you|june ‘15 15 43. If energy (E), velocity (V) and time (T) are chosen

as the fundamental quantities, the dimensional formula of surface tension will be

(a) [EV–2_T–2_] _{(b) [E}–2_V–1_T–3_]

(c) [EV–2_T–1_] _{(d) [EV}–1_T–2_]

44. Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is

(a) drift velocity (b) electric field (c) current density (d) current

45. A potentiometer wire has length 4 m and resistance 8 W. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is (a) 44 W (b) 48 W (c) 32 W (d) 40 W solutions 1. (d) : Here, MA = 4 kg, MB = 2 kg, MC= 1 kg, F = 14 N Net mass, M = MA + MB + MC = 4 + 2 + 1 = 7 kg Let a be the acceleration of the system.

Using Newton’s second law of motion,

F = Ma

14 = 7a \ a = 2 m s–2

Let F′ be the force applied on block A by block

B i.e. the contact force between A and B. Free body

diagram for block A

Again using Newton’s second law of motion, F – F ′ = 4a

14 – F′ = 4 × 2 14 – 8 = F′ \ F′ = 6 N

2. (d) : Radius of the nucleus R = R0A1/3

\ R_RAl A_A Te Al Te =     1 3/ Here, AAl = 27, ATe = 125, RTe = ? R RAl_Te =   = 27 125 3 5 1 3/ ⇒ R_Te= 5R_Al 3 3. (d) : de-Broglie wavelength, l = h p or l ∝ 1 p, l =constantp

This represents a rectangular hyperbola.

4. (b) : Rate of heat flow through a rod is given by

dQ

dt = −KA dTdx

Let length of the rod be L. Case I : dT dx T x L L =D = − = D 110 100 10 \ dQ = − dt1 KA10 L …(i) Also, dQ_dt1=4J s−1 …(ii) Case II : dT dx T x L L =D = − = D 210 200 10 \ dQ = − dt2 KA10 L …(iii)

So, from equations (i), (ii) and (iii)

dQ dt dQ dt 2₌ 1₌₄_{J s}−1 5. (c) : Given situation is shown in the figure. For central maxima,

sinq=l

a

Also, q is very-very small so sinq≈tanq= y D \ y D a= l, y D a = l

Width of central maxima =2y=2 D

a

l _.

6. (a) :

The Boolean expression of this arrangement is

Y A B A B A B= + = ⋅ = ⋅

Thus, the combination represents AND gate.

7. (d) : In SHM, velocities of a particle at distances

x1 and x2 from mean position are given by

V₁2=w (2 2a −x₁2₎ _…(i)

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From equations (i) and (ii), we get

V₁2−V₂2=w (2 x₂2−x₁2) w = − − V V x x 12 22 22 12 \ = − − T x x V V 2 22 12 12 22 p

8. (a) : Given combination is equivalent to three lenses. In which two are plano-convex with refractive index 1.5 and one is concave lens of refractive index 1.7.

Using lens maker formula,

1 _{1 1} 1 1 2 f = − R −R   (m )

For plano-convex lens R1 = ∞, R2 = –20 cm, \ = = − ∞−−    1 1 _{1 5 1 1} 1 20 1 2 f f ( . ) =0 520 = 1 40 . So, f1 = f2 = 40 cm

For concave lens,

m = 1.7, R₁ = –20 cm, R₂ = 20 cm \ = − − −    1 _{1 7 1} 1 20 1 20 3 f ( . ) =0 7. × −₂₀2 = −₁₀₀7 So, f3= −100₇ cm

Equivalent focal length (f eq) of the system is given by

1 1 1 1 1 3 2 f_eq = f + f + f = +− + 1 40 1 100 7 1 40 / = 1 − = − = − 20 7 100 2 100 1 50 \ feq = – 50 cm

9. (b) : Current in the orbit, I e

T = I= e = e= n e ne= ( ) ( ) 2 2 2 2 p w w p p p /

Magnetic field at centre of current carrying circular coil is given by B I r ne r =m0 =m0 2 2

10. (d) : According to question, velocity of unit mass varies as v(x) = bx–2n _…(i) dv dx= − n x n − − 2 b 2 1 …(ii)

Acceleration of the particle is given by

a dv dt dv dx dx dt dv dx v = = × = ×

Using equation (i) and (ii), we get a = (–2nbx–2n – 1_{) × (bx}–2n₎

= –2nb2_x–4n – 1 11. (a) : According to question,

electric field varies as E = Ar Here r is the radial distance.

At r = a, E = Aa …(i)

Net flux emitted from a spherical surface of radius

a is φ e net= qen 0 ⇒ ( ) (Aa × 4 a2)= q [ ] 0 p

e Using equation (i)

\ q = 4pe0Aa3

12. (d) : Energy of radiation, E h= u=hC l

Also, its momentum p h E

C pi = = = l p p E C r= − = −i

So, momentum transferred to the surface = −p p = − −E _ _{ =} C E C E C i r 2

13. (c) : The current flowing in the different branches of circuit is indicated in the figure.

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Physics for you|june ‘15 17 V_B=2I × R IR= 3 3 2 V_C= ×I R IR= 3 3 Thus, VA = VB = VC

14. (b) : Given situation is shown in figure.

N₁ = Normal reaction on A

N2 = Normal reaction on B W = Weight of the rod

In vertical equilibrium,

N1 + N2 = W …(i)

Torque balance about centre of mass of the rod, N1x = N2(d – x)

Putting value of N2 from equation (i)

N1x = (W – N1)(d – x) ⇒ N1x = Wd – Wx – N1d + N1x ⇒ N1d = W(d – x) \ N =W d x− d 1 ( )

15. (a) : Given situation is shown in the figure.

Parallel wires 1 and 3 are semi-infinite, so magnetic field at O due to them

 

B B I

Rk

1= 3= −₄m_p0 ^

Magnetic field at O due to semi-circular arc in

YZ-plane is given by



B I

R i

2= −m₄0 ^

Net magnetic field at point O is given by_{ } _ _

B B B= ₁+ ₂+B₃ = −₄m_p0_RI k^−m₄0_RIi^−₄m_p0_RI k^ = − + m p0 p 4 2 I R( i k) ^ ^

16. (a) : By Bernoulli’s theorem,

P₁ 1 v₁2 P₂ v₂2 2 1 2 + r = + r inside outside      

Assuming that the roof width is very small Pressure difference, P P₁ ₂ 1 v₂2 v₁2 2 − = r( − ) Here, r = 1.2 kg m–3_{, v}₂_{= 40 m s}–1_{, v}₁_{= 0,} A = 250 m2 P P₁ ₂ 1 2 2 2 1 2 40 0 − = × . ( − ) = ×12 1 2 1600. × = 960 N m–2

Force acting on the roof F = (P1 – P2) × A = 960 × 250

= 2.4 × 105_{N upwards} 17. (c) : For double slit experiment,

d = 1 mm = 1 × 10–3_{m, D = 1 m, l = 500 × 10}–9_m

Fringe width b= Dl

d

Width of central maxima in a single slit = 2lD_a As per question, width of central maxima of single slit pattern = width of 10 maxima of double slit pattern 2lD ₁₀ l a D d =   a=2d= × − = × − = 10 2 10 10 0 2 10 0 2 3 3 . m . mm

18. (a) : According to law of conservation of angular momentum mvr = mv′r′ v R v R_{0 0} 0 2 =  ; v = 2v0 …(i) \ K K mv mv v v 0 0 2 2 0 2 1 2 1 2 = _{= } _ or _KK _vv 0 0 2 2 2 =    = ( ) (Using (i)) K = 4K0 = 2mv20

19. (d) : Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet. \ GMm= r mv r 2 2 v GM r 2₌ _…(i)

Time period of the planet is given by

T r v T r v =2p , 2=4p2 2₂ T _GMr r 2₌ 4 2 2   

p _[_{Using equation (i)}_]

T r

GM

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According to question,

T2_{= Kr}3 _…(iii)

Comparing equations (ii) and (iii), we get

K

GM GMK

=4p2 \ =4p2

20. (b) : Net moment of inertia of the system, I = I1 + I2 + I3

The moment of inertia of a shell about its diameter,

I₁ 2mr2

3 =

The moment of inertia of a shell about its tangent is given by I₂ I₃ I mr₁ 2 2mr2 mr2 mr2 3 5 3 = = + = + = \ I= ×2 5mr + mr 3 2 3 2 2 ₌12 ₌ 3 4 2 2 mr _mr

21. (d) : Given situation is shown in the figure.

Velocity of ship A vA = 10 km h–1 towards west Velocity of ship B vB = 10 km h–1 towards north OS = 100 km OP = shortest distance

Relative velocity between A and B is

v_AB= v_A2 +v_B2 =10 2 km h−1 cos45 ; 1 2 100 ° =OP = OS OP OP =100= = 2 100 2 2 50 2 km

The time after which distance between them equals to OP is given by

t OP

v_AB t

= =50 2 ⇒ =

10 2 5 h

22. (a) : _{For n degrees of freedom, C n R}_v₌

2 Also, Cp – Cv = R C_p=C_v+ =R n R R+ 2 C_p=_n+ _R 2 1 γ = = +    ₌ + C C n _R n R n n p v 2 1 2 2 ( )/ \ γ 1 2= +n

23. (b) : Diode is forward bias for positive voltage

i.e. V > 0, so output across RL is given by

24. (d) : According to Einstein’s photoelectric equation

eV_s =hc hc−

l l₀

where Vs = Stopping potential l = Incident wavelength l0 = Threshold wavelength or V hc e s= _1− 1 _ 0 l l For the first case

3 ₀ 1 1 0 V hc e =  −    l l ...(i)

For the second case

V hc e 0 0 1 2 1 =  −    l l ...(ii)

Divide eqn. (i) by (ii), we get 3 1 1 1 2 10 0 = −     −     l l l l 3 1 2 1 1 1 0 0 l l− l l    = −     3 2 3 1 1 0 0 l l− = −l l 21 2 ₄ 0 0 l l= or l = l 25. (b) : q = CV ⇒ V = q/C

Due to dielectric insertion, new capacitance

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Initial energy stored in capacitor, U₁=₂q_C2 Final energy stored in capacitor, U q

KC

2 2 2 = Change in energy stored,

DU = U2 – U1

DU q_{C K} CV

K

= 2 _ − _{ =} 2_ − _

2 1 1 12 1 1

New potential difference between plates ′ =

V q

CK = VK

26. (c) : Constant power acting on the particle of mass

m is k watt.

or P = k

dW

dt =k dW kdt; =

Integrating both sides, dWW tk dt

0 0

∫

=

∫

⇒ W = kt …(i)

Using work energy theorem,

W=1mv − m 2 1 2 0 2 _{( )}2 kt= 1mv

2 2 [Using equation (i)]

v kt

m

= 2

Acceleration of the particle, a dv=_dt

a k m t k mt =1 = 2 2 1 2

Force on the particle, F ma mk

t

= =

2 = mk t2 −1 2/

27. (a) : For Carnot engine, Efficiency, η = −1 ; 1 = − 10 1 1 2 1 2 T T T T T T1₂ 1 110 9 10 = − = …(i) For refrigerator, \ Q = Q T T 2 1 2 1 or Q W Q T T 1 1 2 1 + = Q Q 1 1 10 10 9

+ ₌ _{[Using equation (i)]}

1 10 10 9 10 10 9 1 9 1 1 + = = − = 1 Q ; Q Q1 = 90 J

So, 90 J heat is absorbed at lower temperature.

28. (d) : We know, DU = nCv DT

= n R T5₂ ( B−TA) [for diatomic gas,Cv=5₂R]

=5 _ − _ = 2 nR P V nR P V nR PV nRT B B A A _[_ _] =5 − 2(P VB B P VA A) = × × − × × 5 2(2 103 6 5 103 4) =5 − × = − 2( 8 103) 20kJ 29. (c) : Here, m = 10 kg, vi = 10 m s–1 Initial kinetic energy of the block is

K_i=1mv_i = × × − =

2

1

2 10 10 500

2 ₍ _kg_{) (} _{m s} 1 2₎ _J

Work done by retarding force

W F dx_r xdx x x x = = − = −    

∫

1 2 0 1 0 1 2 20 30 2 20 30 . . = −0 1 900 400. _ −₂ _= −25J According to work-energy theorem,

W = Kf – Ki

Kf = W + Ki = – 25 J + 500 J = 475 J

30. (d) : Energy of electron in He+₃rd_orbit

E Z n 3 2 2 13 6 = − . × eV = −13 6 4× 9 . eV = −13 6 4× × × − × − 9 1 6 10 19 9 7 10 19 . . J . J

As per Bohr’s model,

Kinetic energy of electron in the 3rd_{orbit = – E} 3 \ 9 7 10× − =1 2 19 2 . m v_e v = × × × = × − − − 2 9 7 10 9 1 10 1 46 10 19 31 6 1 . . . m s

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Physics for you|june ‘15 20 31. (c) : Case I : P = VrmsIrms =V ×V R rms rms P V R V PR = rms ⇒ = rms 2 2 ...(i) Case II : Power drawn in LR circuit

′ =P V I =V ×V × Z R Z rms rmscosφ rms rms = V R Z rms2 ₂ =PR R×_Z₂ [Using eqn (i)]

′ =

P P R Z

2 2

32. (a) : Given situation is shown in the figure. Here, N = m₁g

f = mkN = mkm1g …(i)

Let a be the acceleration of blocks. Equation of motion for A and B

T – f = m1a …(ii)

m2g – T = m2a …(iii)

Adding equation (ii) and (iii), we get m2g – f = (m1 + m2)a a m g f m m = − + 2 1 2

Put this value of a in equation (iii)

T m g m m g f m m = − − + 2 2 2 1 2 ( ) = + + = + + m m g m m g m m m m g m m k k 1 2 1 2 1 2 1 2 1 2 1 m ( m ) (Using (i)) 33. (d) : As m d = +       sin sin A A2 2 cot sin sin cot A A A A 2 2 2 2 = +       =     d m  cos sin sin sin A A A A 2 2 2 2       = +       d sin p sin d p; d 2 2− 2 2 2 2 2 2   A = A+  − = +A A \ d = p – 2A = 180° – 2A

34. (d) : According to Wein’s displacement law

lmT = constant …(i)

For star P , intensity of violet colour is maximum For star Q, intensity of red colour is maximum. For star R, intensity of green colour is maximum. Also, lr > lg > lv

Using equation (i), Tr < Tg < Tv

TQ < TR < TP

35. (b) : Here, PQ = RS = PR =QS = a

Emf induced in the frame e = B1(PQ)V – B2(RS)V = − m p 0 2 2 I x a aV ( / ) − + m p 0 2 2 I x a aV ( / ) = − − +    _ m p 0 2 2 2 2 2 I x a x a aV ( ) ( ) = × − +    _ m p 0 2 2 2 2 2 I a x a x a aV ( )( ) \ ∝ − + e 1 2 2 ( x a x a)( ) 36. (a) 37. (a) : Here, KP > KQ

Case (a) : Elongation (x) in each spring is same.

W_P =1K x W_P _Q= K x_Q 2 1 2 2_, 2 \ WP > WQ

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Case (b) : Force of elongation is same.

So, x F and K x F K P Q 1= 2= W K x F K P P P =1 = 2 1 2 12 2 W K x F K Q Q Q =1 = 2 1 2 22 2 \ WP < WQ

38. (a) : Total initial energy of two particles

=1 + 2 1 2 1 12 2 22 m u m u

Total final energy of two particles

=1 + + 2 1 2 2 22 1 12 m v m v e

Using energy conservation principle, 1 2 1 2 1 2 1 2 1 12 2 22 1 12 2 22 m u + m u = m v + m v + e \ 1 + − = + 2 1 2 1 2 1 2 1 12 2 22 1 12 2 22 m u m u e m v m v

39. (a) : Depth of ocean d = 2700 m Density of water, r = 103_{kg m}–3

Compressibility of water, K = 45.4 × 10–11_Pa–1

DV

V = ?

Excess pressure at the bottom, DP = rgd = 103_{× 10 × 2700 = 27 × 10}6_Pa We know / , ( ) B P V V = D D DV D _D V P B K P K B    = = .  =1 = 45.4 × 10–11_{× 27 × 10}6_{= 1.2 × 10}–2 40. (a) : As initial and final

points are same so DUABC = DUAC AB is isochoric process. DWAB = 0 DQAB = DUAB = 400 J BC is isobaric process. DQBC = DUBC + DWBC 100 = DUBC + 6 × 104 (4 × 10–3 – 2 × 10–3) 100 = DUBC + 12 × 10 DUBC = 100 – 120 = – 20 J As, DUABC = DUAC

DUAB + DUBC = DQAC – DWAC 400 20 2 10 2 10 1 2 2 10 4 10 4 3 3 4 − = − × × × + × × × × − − DQ_AC ( ) 380 = DQAC – (40 + 40), DQAC = 380 + 80 = 460 J

41. (a) : For closed organ pipe, fundamental frequency is given by

u_c v

l

= 4

For open organ pipe, fundamental frequency is given by u_o v l = ′ 2

2nd_{overtone of open organ pipe}

u′ = 3uo ; ′ =u 3₂_lv_′

According to question, uc = u′

v l v l 4 3 2 = ′ l′ = 6l Here, l = 20 cm, l′ = ? \ l′ = 6 × 20 = 120 cm

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42. (a) : Here, y1 = asinwt

y₂ b t b t

2 = cosw = sin_w +p _

Hence, resultant motion is SHM with amplitude

a2+b2.

43. (a) : Let S = kEa_Vb_Tc

where k is a dimensionless constant.

Writing the dimensions on both sides, we get [M1_L0_T–2_{] = [ML}2_T–2_]a_[LT–1_]b _[T]c

= [M La a + b2 T− −2 a b + c]

Applying principle of homogeneity of dimensions,

we get, a = 1 ... (i)

2a + b = 0 ... (ii)

– 2a – b + c = – 2 ... (iii)

Adding (ii) and (iii), we get c = – 2

From (ii), b = – 2a = – 2

\ S = kEV–2_T–2 _{or [S] = [EV}–2_T–2_]

44. (d) : The area of cross section of conductor is non uniform so current density will be different but the

flow of electrons will be uniform so current will be constant.

45. (c) : Required potential gradient = 1 mV cm–1

= 1 −

10V m 1

Length of potentiometer wire, l = 4 m

So potential difference across potentiometer wire

= 1 × =

10 4 0 4. V …(i)

In the circuit, potential difference across 8 W

= I × 8 = + ×82R 8 …(ii)

Using equation (i) and (ii), we get

0 4 2 8 8 . = ₊ × R 4 10 16 8 8 40 = +R, + =R \ R = 32 W _nn

Haryana police have nabbed four people, including two dentists and an MBBS student, from Rohtak for allegedly passing on answer keys to students using vests with SIM card units and bluetooth-enabled earpieces during the All India Pre Medical Test (AIPMT).

Apart from probing how 90 answer keys to the highly competitive all-India test were leaked, police are also investigating at least nine candidates who allegedly paid the gang around Rs 15-20 lakh for the “help” in their bid to become doctors.

Police said the gang may have spread its reach to other states too, particularly Bihar and Rajasthan. They added that the accused claimed they had purchased the “engineered vests” from a shop in new Delhi.

Of the four arrested, police have identified two as BDS doctors Sanchit and Bhupender, one as second-year MBBS student Ravi and the fourth as Rajesh. The alleged “kingpin” of this racket, Roop Singh Dangi, is on the run, police said.

Another MBBS doctor – his identity has been withheld – is also under the scanner for acting as a “mediator” between the accused and the students. Shrikant jadhav, Inspector General of Police (Rohtak range), said the arrests followed a tip-off. “We alerted the examination authorities and ordered a thorough frisking of every student. Also, we received concrete information about the four accused who were staying in a hotel in Panipat. We tracked their mobile phones to the jhajjar bypass in Rohtak and caught them,” jadhav told The Indian express.

elaborating on the “special” vests, one of the investigating officers said, “They had sim card units linked to earpieces via bluetooth. Specially configured phones were also supplied to some students. Soon after the exam began, the

accused started sending answer keys to the nine students from whom they had allegedly taken Rs 15-20 lakh each. For students with phones, answer keys were sent through WhatsApp, and for those using the earpieces, they were passed on through phone calls.”

Police said the “engineered vests” and keys to the 90 questions were recovered from the four accused who were produced in a Rohtak court – they were sent to police custody for four days for further interrogation.

”So far, we have received a list of nine students who appeared in the AIPMT exams by allegedly paying money to the accused. Raids are being conducted to nab all those involved,” jadhav said.

”We have also sent teams to nab those who sold these vests to the accused. each vest was purchased by the accused at a cost of approximately Rs 9,000 each,” he added.

Police also suspect that at least one of the accused – Ravi, the MBBS student – may have resorted to similar means to pass his own medical entrance test. ”He undertook PMT coaching from an institute in Kota, Rajasthan in 2005-06 and got through the exam in 2007. But after eight years, he is still in the second year. It appears from interrogation that he might have got through the exam using unfair means,” the investigating officer said.

As for Dangi, the alleged “kingpin”, police said he is a Meham resident who operates from Alwar in Rajasthan. They believe that he allegedly procured and supplied the answer keys to the four accused.

During their interrogation, police said, the accused also claimed that the shop from where they purchased the vests had sold 700 such units in Bihar. “There could be a possibility that the magnitude of this racket is much larger in Bihar and Rajasthan,” jadhav said.

Courtesy : The Indian Express

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units and MeasureMent

1. Which of the following units denotes the dimensions [ML2_Q–2_{], where Q denotes the electric charge?}

(a) weber (Wb) (b) Wb/m2

(c) henry (H) (d) H/m2

2. In an experiment of simple pendulum, the errors in the measurement of length of the pendulum (L) and time period (T) are 3% and 2% respectively. The T) are 3% and 2% respectively. The T

maximum percentage error in the value of L

T2 is

(a) 5% (b) 7% (c) 8% (d) 1%

3. The dimensions of a_b in the equation P a t bx

= − 2,

where P is pressure, P is pressure, P x is distance and x is distance and x t is time t is time t are (a) [M2_LT–3_] _{(b) [ML}0_T–2_]

(c) [ML3_T–1_] _{(d) [MLT}–3_]

4. In the following equation, x,x,x t and t and t F represent F represent F

displacement, time and force respectively,

F a bt

c d x A t

= + +

+ ⋅ + +

1 _sin(_w _f₎

The dimensional formula for A·d isd isd

(a) [T–1_{] (b) [L}–1_{] (c) [M}–1_] _{(d) [TL}–1_] 5. What is the number of significant figures in

0.310 × 103_?

(a) 2 (b) 3 (c) 4 (d) 5

6. The study of the earth’s surface is normally performed with

(a) rectangular cartesian co-ordinates (b) gaussian system

(c) cartesian co-ordinates, but spherical (d) none of these

7. Which one of the following is dimensionally incorrect?

(a) Capacitance C = [MC = [MC –1_L–2_T4_A2_]

(b) Magnetic field induction B = [ML0_T–2_A–1_]

(c) Coefficient of self-induction L = [ML2_T–2_A–1_]

(d) Specific resistance r = [M L3_T–3_A–2_] 8. A quantity X is given by XX is given by e₀L V

t

D

D , where e0 is the

permittivity of free space, L is length, DVVV is potential is potential

difference and Dttt is time interval. The dimensional is time interval. The dimensional

formula for X is the same as that ofXX is the same as that of

(a) resistance (b) charge

(c) voltage (d) current

9. The moment of inertia of a body rotating about a given axis is 12.0 kg m2_{in the SI system. What is the}

value of the moment of inertia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g?

(a) 2.4 × 103 _{(b) 6.0 × 10}3

(c) 5.4 × 105 _{(d) 4.8 × 10}5

10. A wire has a mass (0.3 ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is

(a) 1% (b) 2% (c) 3% (d) 4%

11. Match List I with List II and select the correct answer :

List I List II

A. spring constant spring constant spring constant 1. [M1_L2_T–2_]

B. pascal 2. [M0L0T–1] C. hertz 3. [M1_L0_T–2_] D. joule 4. [M1L–1T–2] A B C D (a) 3 4 2 1 (b) 4 3 1 2 (c) 4 3 2 1 (d) 3 4 1 2

chapterwise McQ’s for practice

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12. Which of the following statements is incorrect regarding significant figures?

(a) All the non-zero digits are significant.

(b) All the zeros between two non-zero digits are significant.

(c) Greater the number of significant figures in a measurement, smaller is the percentage error. (d) The power of 10 is counted while counting the

number of significant figures.

13. The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is (a) 164 ± 3 cm2 _{(b) 163.62 ± 2.6 cm}2

(c) 163.6 ± 2.6 cm2_{(d) 163.62 ± 3 cm}2

14. The density of a material in CGS system of units is 4 g cm–3_{. In a system of units in which unit of}

length is 10 cm and unit of mass is 100 g, the value of density of material will be

(a) 0.04 (b) 0.4

(c) 40 (d) 400

15. Distance Z travelled by a particle is defined by

Z = a + bt + gt2_{. Dimensions of g are}

(a) [LT–1_] _{(b) [L}–1_T]

(c) [LT–2_] _{(d) [LT}2_]

KineMatics

16. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same speed how much high above the ground can the cricketer throw the same ball?

(a) 50 m (b) 100 m

(c) 150 m (d) 200 m

17. A particle moving along the x axis has position given by x = (24t – 2.0t3_{) m, where t is measured in s. What}

is the magnitude of the acceleration of the particle at the instant when its velocity is zero?

(a) 24 m s–2 _{(b) zero}

(c) 12 m s–2 _{(d) 48 m s}–2

18. When the angle of projection is 75°, a ball falls 10 m short of the target. When the angle of projection is 45°, it falls 10 m ahead of the target. Both are projected from the same point with the same speed in the same direction, the distance of the target from the point of projection is

(a) 15 m (b) 30 m

(c) 45 m (d) 10 m

19. Six vectors, a through f have the magnitudes and

directions indicated in the figure. Which of the following statements is true?

(a) _{b c f}  _{+ =} (b) d c f  + = (c) d e f  + = (d) b e f  + = b  c  f  e  d  a 

20. A particle is projected vertically upwards from a point A on the ground. It takes time t1 to reach a

point B, but it still continues to move up. If it takes further time t2 to reach the ground from point B.

Then height of point B from the ground is (a) 1₂g t t(1+ 2)2 (b) gt1t2

(c) 1

8g t t(1+ 2)2 (d) 1 2gt t1 2

21. A rectangular vessel when full of water, takes 10 min to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water?

(a) 9 min (b) 7 min

(c) 5 min (d) 3 min

22. Time taken by the projectile to reach from A to B is t, then the distance AB is equal to (a) 2ut (b) 3 ut (c) ₂3ut (d) ut 3

23. A ball A is thrown up vertically with a speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is

(a) u (b) 2u

(c) u – gt (d) u2−gt

24. A car covers the first one-third of a distance x at a speed of 10 km h–1_{, the second one-third at a speed}

of 20 km h–1_{and the last one-third at a speed of}

60 km h–1_{. Find the average speed of the car over}

the entire distance x.

(a) 10 km h–1 _{(b) 12 km h}–1

(c) 18 km h–1 _{(d) 20 km h}–1

25. The co-ordinates of a moving particle are x = at2_, y = bt2 where a and b are constants. The velocity of

the particle at any moment is (a) 2t a b2+ 2 (b) 2t a b+ (c) 2t a b2− 2 (d) 2 a b2+ 2 u B 60°_30° A

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Physics for you|june ‘15 25 26. The speed of a projectile when it is at its greatest

height is 2 5/ times its speed at half the maximum height. What is its angle of projection?

(a) 30° (b) 60°

(c) 45° (d) 0°

27. A passenger is walking on an escalator at a speed of 6 km/h relative to escalator. The escalator is moving at 3 km/h relative to ground and has a total length of 120 m. The time taken by him to reach the end of the escalator is

(a) 16 s (b) 48 s

(c) 32 s (d) 80 s

28. A body is projected such that its kinetic energy at the top is (3/4)th_{of its initial kinetic energy. What is}

the angle of projection with the horizontal?

(a) 30° (b) 60°

(c) 45° (d) 120°

29. A particle is moving on circular path as shown in the figure. Then displacement from P1 to P2 is (a) 2 2 r cosq (b) 2r tan₂q (c) 2r sinq (d) 2 2 r sinq

30. A particle moves in x-y plane. The position vector of particle at any time t is r={( ) ( ) }2t i+ 2t j2  m. The rate of change of q at time t = 2 s (where q is the angle which its velocity vector makes with positive

x-axis) is (a) 2 17 rad s−1 (b) 114 rad s−1 (c) 4₇ rad s−1 (d) 6₅ rad s−1 solutions 1. (c) : [ML2_Q–2_{] = [ML}2_T–2_A–2_] [Wb] = [ML2_T–2_A–1_] Wb m2 =[M A ] 2    _ T− −1 [henry] = [ML2_T–2_A–2_] H m2 [MT A ] 2 2    _ = − −

Obviously henry (H) has dimensions ML

Q . 2 2         P2 O _r r  _P 1

2. (b) : Time period of simple pendulum is

T L

g

= 2π

Squaring both sides, we get

\ T = L g 2 _4π2 or g L T = 4π2 ₂ ...(i)

The maximum percentage error in g is

Dg D D g L L T T × = × +    × 100 100 2 100 = 3% + 2 × 2% = 7% From (i), we get

L T

g

2 =₄_π2

The maximum percentage error in L

T2 is D D L T L T g g 2 2 100 100 7     × = × = % 3. (b) : P a t bx = − 2

[a] = [T2_{], as t}2_{is subtracted from a.}

From, P a t bx t bx = − =2 2 [ ]b =       = − − t Px 2 2 1 2 [T ML T L ] [ ][ ] = [M–1L0T4] \    = − a b [T [M L T0 2 1 4 ] ] = [ML 0_T–2_] 4. (b) : F a bt c d x A t = + + + ⋅ + + 1 _sin(_w _f₎

As sin(wt + f) is dimensionless, therefore A has dimensions of force.

\ [A] = [F] = [MLT–2_]

As each term on RHS represents force \ + ⋅ = 1 c d x F 1 c=F \ [ ]= = ₋ = − − [ ] [ ] [ ] c F 1 1 2 1 1 2 MLT M L T

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As c is added to d·x, therefore dimensions of c are same that of d·x. \ [d·x] = [c] or M L T L M L T [ ] [ ] [ ] [ ] [ ] [ ] d c x = = − −1 1 2 = − −1 2 2 The dimensional formula for A·d is

[A·d] = [MLT–2_][M–1_L–2_T2_{] = [L}–1_] 5. (b)

6. (c) : Spherical cartesian co-ordinates are used with latitudes and longitudes.

Carte also means map in French. This is derived from Descrate the great mathematician.

7. (c) : [ ] [ ] [ ] [ ] [ ] [ ] C q W = 2 = AT₂ 2₋₂ = − −1 2 4 2 ML T M L T A [ ] [ ] [ ][ ] [ ] [ ][ ] [ ] B F I l = = MLT− = − − A L ML T A0 2 2 1 [ ] [ ] [ ] [ ] [ ] [ L _di dt W q t i =    =     ₌   _ = − − e ML T AT T A ML T 2 2 2 22_A−2_] [ ] [ ][ ] [ ] [ ][ ] [ ] [ ] r = R A = − − = − − L ML T A L L ML T A 2 3 2 2 3 3 2

Choice (c) is dimensionally wrong.

8. (d) : As, C=_DD_Vq or e₀ A e0 L q V C A L = D _ = _ D  or e₀= ( ) ( ) D D q L A V ...(i) X L V t = e₀ D D (Given) \ X= q L A V L Vt ( ) ( ) D D D D (Using (i)) But [A] = [L]2 \ X= q= t D D current 9. (d) : n n M M L L T T a b c 2 1 1 2 1 2 1 2 =                   ...(i)

Dimensional formula of moment of inertia = [ML2_T0_] \ a = 1, b = 2, c = 0 Here, n1 = 12.0, M1 = 1 kg, M2 = 10 g L₁ = 1 m, L2 = 5 cm, T1 = 1 s, T2 = 1 s n₂ 1 2 0 12 0 1 10 1 5 1 1 =                   . kg g m cm s s (Using (i)) = ×             × 12 1000 10 100 5 1 1 2 g g cm cm = 12 × 100 × 400 = 4.8 × 105 10. (d) : Here, Dm D D m r r L L =0 003 = = 0 3 0 005 0 5 0 06 6 . . , . . , . As r π = m r L ( 2) \ _D_rr D D D   ×100=_ _mm+2_rr+ _LL_×100 = + × +   × 0 003 0 3 2 0 005 0 5 0 06 6 100 . . . . . = 1 + 2 + 1 = 4%

11. (a) : Spring constant [MLT

[L] M L T = =F − = − x 2 1 0 2 ] [ ]

pascal = unit of pressure

= =F − = − − A [MLT [L M L T 2 2_] ] [ 1 1 2]

hertz = unit of frequency = _{T = [M}1 0_L0_T–1_]

joule = unit of work = force × distance =[MLT–2_{] [L]}

= [M1_L2_T–2_].

12. (d) : The power of 10 is irrelevant to the determination of significant figures.

13. (a) : Let length and breadth of a rectangular sheet are measured by using a metre scale as 16.2 cm and 10.1 cm respectively. Each measurement has three significant figures.

\ Length l can be written as

l = 16.2 ± 0.1 cm = 16.2 cm ± 0.6%

Similarly, the breadth b can be written as

b = 10.1 ± 0.1 cm = 10.1 cm ± 1%

Area of the sheet, A = l × b = 163.62 cm2_{± 1.6%}

=163.62 ± 2.6 cm2

Therefore, as per rule, area will have only three significant figures and error will have only one significant figure. Rounding off, we get

A = 164 ± 3 cm2 14. (c) : As n1u1 = n2u2 4 100 10 3 2 3 g cm g cm = n ( ) ⇒ n2 = 40

15. (c) : By homogeneity of dimensions of LHS and RHS,