Physics for You June 2015

8/20/2019 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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Ring Road, New Delhi - 110029.

Managing Editor : Mahabir Singh

Editor : Anil Ahlawat (BE, MBA)

Contents

Physics Musing (Problem Set-23) 8

AIPMT 10

Solved Paper 2015

Exam Prep 23

Physics Musing (Solution Set-22) 29

Target PMTs 31

Practice Questions 2015 Kerala PMT 41

Solved Paper 2015

Brain Map 46

Ace Your Way CBSE XII 51

Series 1

Core Concept 62

WB JEE 65

Solved Paper 2015

Thought Provoking Problems 76

Concept Based FAQs 80

You Ask We Answer 84

Crossword 85

rialedit

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Editor : Anil Ahlawat

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All rights reserved. Reproduction in any form is prohibited.

Is the Advance of Science by Logic,

Intuition or Experimentation?

One had always assumed that the advance of science is logical andbased on assumptions that are verified by experimentation. This wastrue 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 ‘158

SINGLE OPTION CORRECT TYPE

1. A disc of radius R is rolling purely on a flat horizontalsurface, with constant angular velocity. Te 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 shellof radius R without slipping.It is released from rest frompoint A as shown in figure.Te angular velocity of centreof the ball in position B aboutthe centre of the shell is

(a) 25

g R r ( )−

(b) 107

g R r ( )−

(c)2

5

g

R r ( )− (d)

5

2

g

R r ( )−

3. An object is moving towards a converging lens on itsaxis. Te image is also found to be moving towards thelens. Ten, the object distance u must satisfy(a) 2 f < u < 4 f (b) f < u < 2 f (c) u > 4 f (d) u


4/80


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1. Tree blocks A, B and C , o masses 4 kg, 2 kg and

1 kg respectively, are in contact on a rictionless

surace, as shown. I a orce o 14 N is applied on

the 4 kg block, then the contact orce between

A and B is

B A C

(a) 8 N (b) 18 N(c) 2 N (d) 6 N

2. I radius o the 1327 Al nucleus is taken to be RAl,

then the radius o 53125e nucleus is nearly

(a)3

5RAl (b)

13

53

1 3

/

RAl

(c)53

13

1 3

/

RAl (d)5

3RAl

3. Which o the ollowing figures represent the

variation o 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 AIPM Guide 28 103 24 AIPM Guide 39 651


6 AIPM Guide 45 720 29 NCER Fingertips 38 119





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

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





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

Here, the reerences o ew are given :Exact Questions

Similar Questions

and more such questions ……


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PHYSICS FOR YOU | JUNE ‘15 11

4. Te two ends o a metal rod are maintained at

temperatures 100°C and 110°C. Te rate o heat

flow in the rod is ound to be 4.0 J/s. I the ends are

maintained at temperatures 200°C and 210°C, the

rate o 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 o monochromatic light o

wavelength ‘l’, diffraction is produced by a single

slit whose width ‘a’ is o the order o the wavelength

o the light. I ‘D’ is the distance o the screen rom

the slit, the width o the central maxima will be

(a)Da

l (b)

2Da

l

(c)

2D

a

l

(d)

D

a

l

6. Which logic gate is represented by the ollowing

combination o logic gates?

(a) AND (b) NOR (c) OR (d) NAND

7. A particle is executing SHM along a straight line.

Its velocities at distances x 1 and x 2 rom the mean

position are V 1 and V 2, respectively. Its time period

is

(a) 2 12

22

12

22

p V V

x x

+

+ (b) 2 1

222

12

22

p V V

x x

−

−

(c) 2 12

22

12

22

p x x

V V

+

+ (d) 2 2

212

12

22

p x x

V V

−

−

8. wo identical thin plano-convex glass lenses(reractive index 1.5) each having radius o

curvature o 20 cm are placed with their convex

suraces in contact at the centre. Te intervening

space is filled with oil o reractive index 1.7. Te

ocal length o the combination is(a) –50 cm (b) 50 cm(c) –20 cm (d) –25 cm

9. An electron moving in a circular orbit o radius r

makes n rotations per second. Te magnetic field

produced at the centre has magnitude

(a)m0n e

r (b)

m0

2

ne

r

(c)m

p

0

2

ne

r (d) Zero

10. A particle o 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

o the particle. Te acceleration o the particle as

a unction o x , is given by(a) –2b2 x –2n + 1 (b) –2nb2 e–4n + 1

(c) –2nb2 x –2n – 1 (d) –2nb2 x –4n – 1

11. Te electric field in a certain region is acting

radially outward and is given by E = Ar . A charge

contained in a sphere o radius ‘a’ centred at the

origin o the field, will be given by (a) 4pe0 Aa

3 (b) e0 Aa3

(c) 4pe0 Aa2 (d) Ae0a

2

12. A radiation o energy ‘E’ alls normally on a

perectly reflecting surace. Te momentum

transerred to the surace is (C = Velocity o light)

(a)2

2

E

C (b)

E

C 2

(c)E

C (d)

2E

C

13. A, B and C are voltmeters o resistance R, 1.5R and3R respectively as shown in the figure. When some

potential difference is applied between X and Y , the

voltmeter readings are V A, V B and V C respectively.

Ten

X Y

B

C

A

(a) V A = V B ≠ V C (b) V A ≠ V B ≠ V C

(c) V A = V B = V C (d) V A ≠ V B = V C

14. A rod o weight W is supported by two parallel

knie edges A and B and is in equilibrium in a

horizontal position. Te knives are at a distance d

rom each other. Te centre o mass o the rod is at

distance x rom A. Te 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 o the wire are

very long and parallel to X -axis while semicircular

portion o radius R is lying in Y -Z plane. Magnetic

field at point O isZ

RY

O

X

I

I I

(a)

B I

Ri k= − +

m

p p0

42

^ ^

(b)

B I

R i k= −

m

p p04 2^ ^

(c)

B I

Ri k= +

m

p p0

42

^ ^

(d)

B I

Ri k= − −

m

p p0

42

^ ^

16. A wind with speed 40 m/s blows parallel to the

roo o a house. Te area o the roo is 250 m2.

Assuming that the pressure inside the house is

atmospheric pressure, the orce exerted by the wind

on the roo and the direction o the orce 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 o wavelength 500 nm

is used. What will be the width o each slit or

obtaining ten maxima o double slit within the

central maxima o 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 v 0 at a radius R0. Te mass is

attached to a string which passes through a smooth

hole in the plane as shown.

v 0

m

R0

Te tension in the string is increased gradually

and finally m moves in a circle o radius R0

2. Te

final value o the kinetic energy is

(a) 2 02

mv (b)

1

2 02

mv

(c) mv 02

(d)1

402mv

19. Kepler’s third law states that square o periodo revolution (T ) o a planet around the sun, isproportional to third power o average distancer between sun and planet i.e. T 2 = Kr 3 here K isconstant.

I the masses o sun and planet are M and m respectively then as per Newton’s law o gravitationorce o attraction between them is

F GMmr

=2

, here G is gravitational constant.

Te relation between G and K is described as

(a) K = G (b) K G

=1

(c) GK = 4p2 (d) GMK = 4p2

20. Tree identical spherical shells, eacho mass m and radius r are placedas shown in figure. Consider an axis XX ′ which is touching to two shellsand passing through diameter o

third shell. Moment o inertia othe system consisting o these threespherical shells about XX ′ axis is

(a)16

52mr (b) 4mr 2

(c)11

5

2mr (d) 3mr 2

21. A ship A is moving Westwards with a speed o10 km h–1 and a ship B 100 km South o A, ismoving Northwards with a speed o 10 km h–1.Te time afer which the distance between them

becomes shortest, is(a) 5 2 h (b) 10 2 h

(c) 0 h (d) 5 h

22. Te ratio o the specific heatsC

C

p

v

= γ in terms o

degrees o reedom (n) is given by

(a) 12

+

n

(b) 12

+

n

(c) 11

+

n

(d) 13

+

n


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23. I in a p–n junction, a square input signal o 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 surace is illuminated with

monochromatic light o wavelength, l. Te

stopping potential or photo-electric current orthis light is 3V 0. I the same surace is illuminated

with light o wavelength 2l, the stopping potentialis V 0. Te threshold wavelength or this surace or

photo-electric effect is

(a)p4

(b)l6

(c) 6l (d) 4l

25. A parallel plate air capacitor o capacitance C is

connected to a cell o em V and then disconnected

rom it. A dielectric slab o dielectric constant K ,

which can just fill the air gap o the capacitor, is nowinserted in it. Which o the ollowing is incorrect?

(a) Te change in energy stored is1

2

112CV

K −

.

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

decreases K times.(d) Te energy stored in the capacitor decreases

K times.

26. A particle o mass m is driven by a machine that

delivers a constant power k watts. I the particle

starts rom rest the orce on the particle at time t is

(a) 2 1 2mk t − / (b)1

21 2mk t − /

(c)mk

t 2

1 2− / (d) mk t −1 2/

27. A Carnot engine, having an efficiency o η=1

10 as

heat engine, is used as a rerigerator. I the work

done on the system is 10 J, the amount o energy

absorbed rom the reservoir at lower temperature is

(a) 90 J (b) 1 J(c) 100 J (d) 99 J

28. One mole o an ideal diatomic gas undergoes atransition rom 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

Te change in internal energy o the gas duringthe transition is(a) 20 J (b) –12 kJ

(c) 20 kJ (d) –20 kJ29. A block o mass 10 kg, moving in x direction with

a constant speed o 10 m s–1, is subjected to aretarding orce F = 0.1x J/m during its travel romx = 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 o He+ (Helium), using non-relativistic approach, the speed o electron in thisorbit will be [given K = 9 × 109 constant, Z = 2 andh (Plack’s Constant) = 6.6 × 10–34 J s]

(a) 0.73 × 106 m/s (b) 3.0 × 108 m/s(c) 2.92 × 106 m/s (d) 1.46 × 106 m/s

31. A resistance ‘R’ draws power ‘P ’ when connectedto an AC source. I an inductance is now placed inseries with the resistance, such that the impedanceo 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 o mass m1 rests on a horizontal table. Alight string connected to it passes over a rictionlesspully at the edge o table and rom its other endanother block B o mass m2 is suspended. Tecoefficient o kinetic riction between the blockand the table is mk. When the block A is sliding onthe table, the tension in the string is

(a)m m g

m mk1 2

1 2

1( )

( )

++

m (b)

m m g

m mk1 2

1 2

1( )

( )

−+

m

(c)( )

( )

m m g

m mk2 1

1 2

++m

(d)( )

( )

m m g

m mk2 1

1 2

−+m


9/80


33. Te reracting angle o a prism is A, and reractive

index o the material o the prism is cot ( A/2). Te

angle o minimum deviation is(a) 90° – A (b) 180° + 2 A

(c) 180° – 3 A (d) 180° – 2 A34. On observing light rom three different stars P , Q

and R, it was ound that intensity o violet colour

is maximum in the spectrum o P , the intensity o

green colour is maximum in the spectrum o R

and the intensity o red colour is maximum in the

spectrum o Q. I T P , T Q and T R are the respective

absolute temperatures o P , Q and R, then it can be

concluded rom the above observations that(a) T P < T R < T Q (b) T P < T Q < T R(c) T P > T Q > T R (d) T P > T R > T Q

35. A conducting square rame o side ‘a’ and a longstraight wire carrying current I are located in the

same plane as shown in the figure. Te rame

moves to the right with a constant velocity ‘V ’. Te

em induced in the rame will be proportional to

(a)

1

2 2( )x a+ (b)

1

2 2( )( )x a x a− +

(c)12x

(d)1

2 2( )x a−

36. wo spherical bodies o mass M and 5 M and radii

R and 2R are released in ree space with initial

separation between their centres equal to 12R. I

they attract each other due to gravitational orce

only, then the distance covered by the smaller body

beore collision is(a) 7.5R (b) 1.5R

(c) 2.5R (d) 4.5R37. wo similar springs P and Q have spring constants

K P and K Q, such that K P > K Q. Tey are stretched

first by the same amount (case a), then by the

same orce (case b). Te work done by the springs

W P and W Q are related as, in case (a) and case (b)

respectively (a) W P > W Q ; W Q > W P (b) W P < W Q ; W Q < W P (c) W P = W Q ; W P > W Q(d) W P = W Q ; W P = W Q

38. wo particles o masses m1, m2 move with initial velocities u1and u2. On collision, one o theparticles get excited to higher level, afer absorbingenergy e. I final velocities o particles be v 1 and v 2 then we must have

(a)1

2

1

2

1

2

1

21 12

2 22

1 12

2 22m u m u m v m v + − = +e

(b)1

2

1

2

1

2

1

21 2 12

12 2

22 2

12

22

22m u m u m v m v + + = +e

(c) m u m u m v m v 12

1 22

2 12

1 22

2+ − = +e

(d)1

2

1

2

1

2

1

21 12

2 22

1 12

2 22m u m u m v m v + = + − e

39. Te approximate depth o an ocean is 2700 m.Te compressibility o water is 45.4 × 10–11 Pa–1 and density o water is 103 kg/m3. What ractional

compression o water will be obtained at the bottomo 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 bya gas to go rom a state A to a state C .

In process AB, 400 J o heat is added to the systemand in process BC , 100 J o heat is added to thesystem. Te heat absorbed by the system in theprocess AC will be(a) 460 J (b) 300 J(c) 380 J (d) 500 J

41. Te undamental requency o a closed organ pipeo length 20 cm is equal to the second overtone oan organ pipe open at both the ends. Te length o

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 y 1 = a sin(wt ) and y 2 = b cos(wt ) are superimposedthe 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 amplitudea

b


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43. I energy (E), velocity (V ) and time (T ) are chosen

as the undamental quantities, the dimensional

ormula o surace tension will be(a) [EV–2–2] (b) [E–2V–1–3]

(c) [EV–2

–1

] (d) [EV–1

–2

]44. Across a metallic conductor o non-uniorm cross

section a constant potential difference is applied.

Te quantity which remains constant along the

conductor is(a) drif velocity (b) electric field(c) current density (d) current

45. A potentiometer wire has length 4 m and resistance

8 W. Te resistance that must be connected in serieswith the wire and an accumulator o e.m.. 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, M A = 4 kg, M B = 2 kg, M C = 1 kg, F = 14 N

Net mass, M = M A + M B + M C = 4 + 2 + 1 = 7 kg

Let a be the acceleration o the system.

Using Newton’s second law o motion,

F = Ma

14 = 7a \ a = 2 m s–2

Let F ′ be the orce applied on block A by blockB i.e. the contact orce between A and B. Free body

diagram or block A

Again using Newton’s second law o motion,

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

2. (d) : Radius o the nucleus R = R0 A1/3

\ R

R

A

AAl

e

Al

e

=

1 3/

Here, AAl = 27, Ae = 125, Re = ?

R

RAl

e

=

=27

125

3

5

1 3/

⇒ R Re Al=5

3

3. (d) : de-Broglie wavelength, l =h

por l ∝

1

p p, l = constant

Tis represents a rectangular hyperbola.4. (b) : Rate o heat flow through a rod is given by

dQ

dt KA

dT

dx = −

Let length o the rod be L.

Case I :dT

dx

T

x L L= =

−=

DD

110 100 10

\ = −dQ

dt KA

L1 10 …(i)

Also J s, dQ

dt 1 14= − …(ii)

Case II : dT

dx

T

x L L= = − =D

D210 200 10

\ = −dQ

dt KA

L2 10 …(iii)

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

dQ

dt

dQ

dt 2 1 14= = −J s

5. (c) : Given situation isshown in the figure.

For central maxima,

sinq l=a

Also, q is very-very small so

sin tanq q≈ =

y

D

\ y D a

= l

, y D

a=

l

Width o central maxima = =22

y D

a

l.

6. (a) :

Te Boolean expression o this arrangement is

Y A B A B A B= + = ⋅ = ⋅Tus, the combination represents AND gate.

7. (d) : In SHM, velocities o a particle at distancesx 1 and x 2 rom mean position are given by

V a x 12 2 2

12= −w ( ) …(i)

V a x 2

2 2 222= −w ( )

…(ii)


11/80


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

V V x x 1

222 2

22

12− = −w ( )

w =

−

−

V V

x x

12

22

22

12

\ =

−

−T

x x

V V 2

22

12

12

22p

8. (a) : Given combination is equivalent to three

lenses. In which two are plano-convex with reractive

index 1.5 and one is concave lens o reractive

index 1.7.

Using lens maker ormula,

11

1 1

1 2 f R R= − −

( )m

For plano-convex lens

R1 = ∞, R2 = –20 cm,

\ = = −∞

−−

1 11 5 1

1 1

201 2 f f ( . ) = =

0 5

20

1

40

.

So, f 1 = f 2 = 40 cm

For concave lens,

m = 1.7, R1 = –20 cm, R2 = 20 cm\ = −

− −

11 7 1

1

20

1

203 f ( . )

= ×

−

= −0 7

2

20

7

100.

So, f 3

100

7= − cm

Equivalent ocal length ( feq) o the system is given by

1 1 1 1

1 3 2 f f f f eq= + +

= +

− +

1

40

1

100 7

1

40/

= − = − = −

1

20

7

100

2

100

1

50

\ f eq = – 50 cm

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

T =

I e e n e

ne= = = =( )

( )

2 2

2

2p ww

ppp/

Magnetic field at centre o current carrying circular

coil is given by

B I

r

ne

r = =

m m0 02 2

10. (d) : According to question, velocity o unit mass

varies as

v (x ) = bx –2n …(i)

dv

dx n x

n= − − −2 2 1b …(ii)

Acceleration o 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 rom a spherical surace o radius

a is

φ

enet =

qen

0

⇒ × =( ) ( ) [ ] Aa a q4 20

pe

Using equation (i)

\ q = 4pe0 Aa3

12. (d) : Energy o radiation, E h hC

= =ul

Also, its momentum p h E

C pi= = =l

p p E

C r i= − = −

So, momentum transerred to the surace

= − = − −

= p p

E

C

E

C

E

C i r 2

13. (c) : Te current flowing in the different brancheso circuit is indicated in the figure.

V A = IR


12/80


V I

R IRB = × =2

3

3

2

V I

R IRC = × =33

Tus, V A = V B = V C

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

N 1 = Normal reaction on A

N 2 = Normal reaction on B

W = Weight o the rod

In vertical equilibrium,

N 1 + N 2 = W …(i)

orque balance about centre o mass o the rod,

N 1x = N 2(d – x )

Putting value o N 2 rom equation (i)

N 1x = (W – N 1)(d – x )

⇒ N 1x = Wd – Wx – N 1d + N 1x ⇒ N 1d = 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

Rk1 3

0

4= = −

mp

^

Magnetic field at O due to semi-circular arc in

YZ -plane is given by

B I

Ri2

0

4= −

m ^

Net magnetic field at point O is given by

B B B B= + +1 2 3

= − − −

mp

m mp

0 0 0

4 4 4

I

Rk

I

Ri

I

Rk

^ ^ ^

= − +mp p0

42I

Ri k( )^ ^

16. (a) : By Bernoulli’s theorem,

P v P v 1 12

2 221

2

1

2+ = +r r

inside outside

Assuming that the roo width is very smallPressure difference,

P P v v 1 2 2

2121

2− = −r( )

Here, r = 1.2 kg m–3, v 2 = 40 m s–1, v 1 = 0,

A = 250 m2

P P 1 22 21

21 2 40 0− = × −. ( )

= × ×

1

21 2 1600. = 960 N m–2

Force acting on the roo F = (P 1 –P 2) × 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 l

= D

d

Width o central maxima in a single slit =2lD

aAs per question, width o central maxima o singleslit pattern = width o 10 maxima o double slitpattern

2

10l lDa

D

d =

a d

= = ×

= × =−

−2

10

2 10

100 2 10 0 2

33. .m mm

18. (a) : According to law o conservation o angularmomentummvr = mv′r ′

v R v R

0 00

2=

; v = 2v0 …(i)

\ K K

mv

mv

v v

0 0

2

2

02

1

21

2

= =

orK

K

v

v 0 0

222=

= ( ) (Using (i))

K = 4K 0 = 2mv 20

19. (d) : Gravitational orce o attraction between sunand planet provides centripetal orce or the orbito planet.

\ =GMm

r

mv

r 2

2

v

GM

r 2 =

…(i)

ime period o the planet is given by

T

r

v T

r

v = =

2 422 2

2

p p,

T r

GM

r

22 24

=

p[ ]Using equation (i)

T

r

GM 2

2 34=

p

…(ii)


13/80


According to question,T 2 = Kr 3 …(iii)

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

K

GM GMK

= \ =

244 2

p

p20. (b) : Net moment o inertia o the system,

I = I 1 + I 2 + I 3Te moment o inertia o a shell about itsdiameter,

I mr 1

22

3=

Te moment o inertia o a shell about its tangentis given by

I I I mr mr mr mr 2 3 1

2 2 2 22

3

5

3= = + = + =

\

I mr mr = × +25

3

2

32 2 = =

12

34

22mr mr

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

Velocity o ship A v A = 10 km h–1 towards west

Velocity o ship B v B = 10 km h

–1 towards north

OS = 100 km

OP = shortest distance

Relative velocity between A and B is

v v v AB A B= + =

−2 2 110 2 km h

cos ;45

1

2 100° = =

OP

OS

OP

OP = = =100

2100 2

250 2 km

Te time afer which distance between them equals

to OP is given by

t

OP

v t

AB

= = ⇒ =50 2

10 25 h

22. (a) : For n degrees o reedom, C n

Rv =2Also, C p – C v = R

C C R

nR R p v = + = +2

C

nR p = +

2

1

γ = =

+

= +C

C

nR

n R

n

n

p

v 2

1

2

2

( )/ \ = + γ 12

n

23. (b) : Diode is orward bias or positive voltage

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

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

equation

eV

hc hcs = −l l0

where V s = Stopping potential l = Incident wavelength l0 = Treshold wavelength

or V hc

es= −

1 1

0l l

For the first case

3

1 10

0

V hc

e= −

l l

...(i)

For the second case

V hc

e0

0

1

2

1= −

l l ...(ii)

Divide eqn. (i) by (ii), we get

3

1 1

1

2

1

0

0

=

−

−

l l

l l

3

1

2

1 1 1

0 0l l l l−

= −

3

2

3 1 1

0 0l l l l− = −

1

2

24

00l l

l l= =or

25. (b) :

q = CV ⇒ V = q/C Due to dielectric insertion, new capacitance

C 2 = CK


14/80


Initial energy stored in capacitor, U q

C 1

2

2=

Final energy stored in capacitor, U q

KC

2

2

2

=

Change in energy stored, DU = U 2 – U 1

DU

q

C K CV

K = −

= −

22

2

11

1

2

11

New potential difference between plates

′ =V

q

CK =

V

K

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

m is k watt.

or P = k

dW

dt k dW kdt = =;

Integrating both sides, dW k dt

W t

0 0 ∫ ∫ =

⇒ W = kt …(i)Using work energy theorem,

W mv m= −

1

2

1

202 2( )

kt mv =

1

22 [ ]Using equation (i)

v kt

m

=2

Acceleration o 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 = −

mkt

21 2/

27. (a) : For Carnot engine,

Eiciency, ;η = − = −11

1011

2

1

2

T

T

T

T

T

T

1

2 1

1

10

9

10= − = …(i)

For rerigerator,

\ =Q

Q

T

T 2

1

2

1

orQ W

Q

T

T 1

1

2

1

+=

Q

Q1

1

10 10

9

+= [Using equation (i)]

110 10

9

10 10

91

91 1+ = = − =

1Q Q

;

Q1 = 90 JSo, 90 J heat is absorbed at lower temperature.

28. (d) : We know, DU = nC vDT

=

− =n R

T T C R

B A v 5

2

5

2( ) [ ]or diatomic gas,

= −

=5

2

nR P V

nR

P V

nRPV nRT B B A A [ ]

= −5

2( )P V P V B B A A = × × − × ×

5

22 10 6 5 10 43 3( )

= − × = −5

28 10 203( ) kJ

29. (c) : Here, m = 10 kg, v i = 10 m s–1

Initial kinetic energy o the block is

K mv i i= = × × =−1

2

1

210 10 5002 1 2( ) ( )kg m s J

Work done by retarding orce

W F dx xdx x

r x

x

= = − = −

∫ ∫

1

2

0 1 0 1220

30 2

20

30

. .

= −

−

= −0 1

900 400

225. J

According to work-energy theorem,W = K f – K i

K f = W + K i = – 25 J + 500 J = 475 J

30. (d) : Energy o electron in He+ 3rd orbit

E Z

n3

2

213 6= − ×. eV = − ×13 6

4

9. eV

= − × × × ×− −13 6

4

91 6 10 9 7 1019 19. . .J J

As per Bohr’s model,Kinetic energy o electron in the 3rd orbit = – E3

\ × =−9 7 101

219 2. m v e

v = × ×

×= ×

−

−−2 9 7 10

9 1 101 46 10

19

316 1.

.. m s


15/80


31. (c) : Case I : P = V rmsI rms

= ×V V

Rrmsrms

P V R

V PR= ⇒ =rms rms2

2

...(i)

Case II : Power drawn in LR circuit

′ = = × ×P V I V V

Z

R

Z rms rms rmsrmscos φ

= V

R

Z rms2

2 = ×PR

R

Z 2

[Using eqn (i)]

′ =P P R

Z

2

2

32. (a) : Given situation is shown in the figure.Here, N = m1 g

f = mkN = mkm1 g …(i)

Let a be the acceleration o blocks.

Equation o motion or A and B

T – f = m1a …(ii)

m2 g – T = m2a …(iii)

Adding equation (ii) and (iii), we get

m2 g – f = (m1 + m2)a

a

m g f

m m=

−+

2

1 2

Put this value o 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 mk k1 2 1 2

1 2

1 2

1 2

1m m( )

(Using (i))

33. (d) :

As m

d

=

+

sin

sin

A

A2

2

cotsin

sin

cot A

A

A

A

22

2

2=

+

=

d

m

cos

sin

sin

sin

A

A

A

A2

2

2

2

=

+

d

sin sin ;p d p d2 2 2 2 2 2 2 2

−

= +

− = +

A A A A

\ d = p – 2 A = 180° – 2 A

34. (d) : According to Wein’s displacement law lmT = constant …(i)For star P , intensity o violet colour is maximum

For star Q, intensity o red colour is maximum.For star R, intensity o green colour is maximum.Also, lr > l g > lv Using equation (i), T r < T g < T v T Q < T R < T P

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

Em induced in the rame

e = B1(PQ)V – B2(RS)V

=

−m

p0

2 2

I

x aaV

( )/−

+m

p0

2 2

I

x aaV

( )/

=−

−+

mp0

2

2

2

2

2

I

x a x aaV

( ) ( )

= ×

− +

mp

0

22 2

2 2I a

x a x aaV

( )( )

\ ∝− +

e1

2 2( )( )x a x a

36. (a)

37. (a) : Here, K P > K QCase (a) : Elongation (x ) in each spring is same.

W K x W K x P P Q Q= =1

2

1

22 2,

\ W P > W Q


16/80


Case (b) : Force o elongation is same.

So and, x F

K x

F

K P Q1 2= =

W K x F

K P P P = =1

2

1

212

2

W K x F

K Q Q Q= =

1

2

1

222

2

\ W P < W Q

38. (a) : otal initial energy o two particles

= +

1

2

1

21 1

22 2

2m u m u

otal final energy o two particles

= + +

1

2

1

22 2

21 1

2m v m v e

Using energy conservation principle,

1

2

1

2

1

2

1

21 1

22 2

21 1

22 2

2m u m u m v m v + = + + e

\ + − = +1

2

1

2

1

2

1

21 1

22 2

21 1

22 2

2m u m u m v m v e

39. (a) : Depth o ocean d = 2700 m

Density o water, r = 103 kg m–3

Compressibility o water, K = 45.4 × 10–11 Pa–1

DV V

= ?

Excess pressure at the bottom, DP = r gd

= 103 × 10 × 2700 = 27 × 106 Pa

We know /

,( )

B P

V V =

DD

D D

DV

V

P

BK P K

B

= = =

.

1

= 45.4 × 10–11 × 27 × 106 = 1.2 × 10–2

40. (a) : As initial and final

points are same so

DU ABC = DU AC AB is isochoric process.

DW AB = 0DQ AB = DU AB = 400 JBC is isobaric process.

DQBC = DU BC + DW BC 100 = DU BC + 6 × 104 (4 × 10–3 – 2 × 10–3)100 = DU BC + 12 × 10DU BC = 100 – 120 = – 20 JAs, DU ABC = DU AC DU AB + DU BC = DQ AC – DW AC

400 20 2 10 2 10

1

22 10 4 10

4 3

3 4

− = − × × × +

× × × ×

−

−

DQ AC (

)

380 = DQ AC – (40 + 40), DQ AC = 380 + 80 = 460 J

41. (a) : For closed organ pipe, undamental requency

is given by

uc

v

l =

4

For open organ pipe, undamental requency is

given by

uo

v

l =

′22nd overtone o open organ pipe

u′ = 3uo ; ′ = ′u 32v l 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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17/80


42. (a) : Here, y 1 = asinwt

y b t b t 2 2= = +

cos sinw w

p

Hence, resultant motion is SHM with amplitude

a b2 2+ .

43. (a) : Let S = kE aV bT c

where k is a dimensionless constant.Writing the dimensions on both sides, we get[M1L0–2] = [ML2–2]a[L–1]b []c

= [M L ]2 2a a + b a b + c− −

Applying principle o homogeneity o dimensions,we get, a = 1 ... (i) 2a + b = 0 ... (ii) – 2a – b + c = – 2 ... (iii)Adding (ii) and (iii), we get c = – 2From (ii), b = – 2a = – 2\ S = kEV –2 T –2 or [S] = [EV–2–2]

44. (d) : Te area o cross section o conductor is nonuniorm so current density will be different but the

flow o electrons will be uniorm so current will beconstant.

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

= −1

10

1

V mLength o potentiometer wire, l = 4 m

So potential difference across potentiometer wire

= × =1

104 0 4. V …(i)

In the circuit, potential difference across 8 W

= I × 8 =+

×2

88

R …(ii)

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

0 4

2

88. =

+ ×R

4

10

16

88 40=

+ + =

RR,

\ R = 32 Wnn

Haryana police have nabbed four people, including two dentists and an MBBSstudent, from Rohtak for allegedly passing on answer keys to students usingvests with SIM card units and bluetooth-enabled earpieces during the All IndiaPre Medical Test (AIPMT).Apart from probing how 90 answer keys to the highly competitive all-Indiatest were leaked, police are also investigating at least nine candidates whoallegedly paid the gang around Rs 15-20 lakh for the “help” in their bid tobecome doctors.Police said the gang may have spread i ts reach to other states too, particularlyBihar and Rajasthan. They added that the accused claimed they had purchasedthe “engineered vests” from a shop in New Delhi.Of the four arrested, police have identified two as BDS doctors Sanchit andBhupender, 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, policesaid.Another MBBS doctor – his identity has been withheld – is also under thescanner for acting as a “mediator” between the accused and the students.Shrikant Jadhav, Inspector General of Police (Rohtak range), said the arrestsfollowed a tip-off. “We alerted the examination authorities and ordered athorough frisking of every student. Also, we received concrete informationabout the four accused who were staying in a hotel in Panipat. We tracked theirmobile phones to the Jhajjar bypass in Rohtak and caught them,” Jadhav toldThe Indian Express.Elaborating on the “special” vests, one of the investigating officers said, “Theyhad sim card units linked to earpieces via bluetooth. Specially configuredphones were also supplied to some students. Soon after the exam began, the

accused started sending answer keys to the nine students from whom they hadallegedly taken Rs 15-20 lakh each. For students with phones, answer keyswere sent through WhatsApp, and for those using the earpieces, they werepassed on through phone calls.”Police said the “engineered vests” and keys to the 90 questions were recoveredfrom the four accused who were produced in a Rohtak court – they were sentto police custody for four days for further interrogation.”So far, we have received a list of nine students who appeared in the AIPMTexams by allegedly paying money to the accused. Raids are being conducted tonab 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 9,000each,” 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 2005-06and got through the exam in 2007. But after eight years, he is still in the secondyear. It appears from interrogation that he might have got through the examusing unfair means,” the investigating officer said.As for Dangi, the alleged “kingpin”, police said he is a Meham resident whooperates from Alwar in Rajasthan. They believe that he allegedly procured andsupplied the answer keys to the four accused.During their interrogation, police said, the accused also claimed that the shopfrom where they purchased the vests had sold 700 such units in Bihar. “Therecould be a possibility that the magnitude of this racket is much larger in Biharand Rajasthan,” Jadhav said.

Courtesy : The Indian Express

AIPMT keys leaked, students get answers via bluetooth vests


18/80

PHYSICS FOR YOU JUNE 15 23

1. Which of the following units denotes the dimensions[ML Q– ], where Q denotes the electric charge?(a) weber (Wb) (b) Wb/m(c) henry (H) (d) H/m

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

maximum percentage error in the value ofL

T 2 is

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

3. Te dimensions of ab

in the equation P a t bx

= −2

,

where P is pressure, x is distance and t is time are(a) [M L– ] (b) [ML – ](c) [ML – ] (d) [ML– ]

4. In the following equation, x t and F representdisplacement, time and force respectively,

F a bt c d x A t = + +

+ ⋅+ +

1sin( )w f

Te dimensional formula for A·d is

(a) [–1] (b) [L–1] (c) [M–1] (d) [L–1]

5. What is the number of significant figures in0.310 × 10 ?

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

6. Te study of the earth’s surface is normallyperformed 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 dimensionallyincorrect?

(a) Capacitance C = [M–1L– A ](b) Magnetic field induction B = [ML – A–1](c) Coefficient of self-induction L = [ML – A– ](d) Specific resistance r = [M L – A– ]

8. A quantity is given by e0L V

t

D

D, where e is the

permittivity of free space, is length, D is potential

difference and D is time interval. Te dimensionalformula for is the same as that of (a) resistance (b) charge(c) voltage (d) current

9. e moment of inertia of a body rotating about a

given axis is 12.0 kg m in the SI system. What is the value of the moment of inertia in a system of unitsin which the unit of length is 5 cm and the unit ofmass is 10 g?(a) 2.4 10 (b) 6.0 10(c) 5.4 10 (d) 4.8 10

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

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

11. Match List I with List II and select the correctanswer :

List I ist II

. spring constant . –

B. pascal 2. M L T – ]

C. hertz 3. M L T – ]

. joule 4. M – – ]

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 PRACTICEUseful for All National and State Level Medical/Engg. Entrance Exams


19/80


12. Which of the following statements is incorrectregarding 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 ameasurement, smaller is the percentage error.

(d) Te power of 10 is counted while counting thenumber of significant figures.

13. Te length and breadth of a rectangular sheet are16.2 cm and 10.1 cm, respectively. Te area of thesheet in appropriate significant figures and error is(a) 164 ± 3 cm2 (b) 163.62 ± 2.6 cm2

(c) 163.6 ± 2.6 cm2 (d) 163.62 ± 3 cm2

14. Te density of a material in CGS system of units is

4 g cm

–3

. In a system of units in which unit oflength is 10 cm and unit of mass is 100 g, the valueof density of material will be(a) 0.04 (b) 0.4(c) 40 (d) 400

15. Distance Z travelled by a particle is defined byZ = a + bt + g t 2. Dimensions of g are(a) [L–1] (b) [L–1](c) [L–2] (d) [L2]

KINEMATICS

16. A cricketer can throw a ball to a maximum

horizontal distance of 100 m. With the same speedhow much high above the ground can the cricketerthrow 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 givenby x = (24t – 2.0t 3) m, where t is measured in s. Whatis the magnitude of the acceleration of the particleat 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 falls10 m short of the target. When the angle ofprojection is 45°, it falls 10 m ahead of the target.Both are projected from the same point with thesame speed in the same direction, the distance ofthe 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 anddirections indicated in the figure. Which of thefollowing statements is true?

(a)

b c f + =

(b)

d c f + =

(c)

d e f + =

(d)

b e f + =

b

c

f

ed

a

20. A particle is projected vertically upwards from apoint A on the ground. It takes time t 1 to reach apoint B, but it still continues to move up. If it takesfurther time t 2 to reach the ground from point B.Ten height of point B from the ground is

(a)1

21 2

2 g t t ( )+

(b) gt 1t 2

(c)1

81 2

2 g t t ( )+ (d) 1

21 2 gt t

21. A rectangular vessel when full of water, takes10 min to be emptied through an orifice in itsbottom. How much time will it take to be emptiedwhen half filled with water?(a) 9 min (b) 7 min(c) 5 min (d) 3 min

22. ime taken by theprojectile to reachfrom A to B is t , thenthe distance AB isequal to

(a) 2ut (b) 3 ut

(c)3

2ut (d)

ut

3

23. A ball A is thrown up vertically with a speed u andat the same instant another ball B is released from aheight h. At time t , the speed of A relative to B is(a) u (b) 2u

(c) u – gt (d) u gt 2 −

24. A car covers the first one-third of a distance x at aspeed of 10 km h–1, the second one-third at a speedof 20 km h–1 and the last one-third at a speed of

60 km h–1. Find the average speed of the car overthe entire distance x .(a) 10 km h–1 (b) 12 km h–1

(c) 18 km h–1 (d) 20 km h–1

25. Te co-ordinates of a moving particle are x = at 2, y = bt 2 where a and b are constants. Te velocity ofthe particle at any moment is

(a) 2 2 2t a b+ (b) 2t a b+

(c) 2 2 2t a b− (d) 2 2 2a b+

u

B

6 0 °

3 0 °

A


20/80


26. Te 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. Te escalator is moving

at 3 km/h relative to ground and has a total length

of 120 m. Te 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. Ten displacement

from P 1 to P 2 is

(a) 22

r cos q

(b) 22

r tan q

(c) 2r sinq (d) 22

r sin q

30. A particle moves in x - y plane. Te position vector

of particle at any time t is r t i t j= +{( ) ( ) }2 2 2 m. Te

rate of change of q at time t = 2 s (where q is theangle which its velocity vector makes with positive

x -axis) is

(a)2

17

1 rad s− (b)1

14

1 rad s−

(c)4

7

1 rad s− (d)6

5

1 rad s−

SOLUTIONS

1. (c) : [ML2Q–2] = [ML2–2A–2]

[Wb] = [ML2–2A–1]

Wb

m=[M A ]

2

2

− − 1

[henry] = [ML2 –2 A–2]

H

m[M A ]

22 2

=

− −

Obviously henry (H) has dimensionsML

Q.

2

2

P2

O r

r

P1

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

T L

g = 2π

Squaring both sides, we get

\ =T L

g

2 24π

or g L

T = 4 2

2π ...(i)

Te maximum percentage error in g is

D D D g g

L

L

T

T × = × +

×100 100 2 100

= 3% + 2 × 2% = 7%

From (i), we get

L

T

g 2 24= π

Te maximum percentage error inL

T 2 is

DD

L

T L

T

g

g

2

2

100 100 7

× = × = %

3. (b) : P a t

bx =

− 2

[a] = [2

], 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–1L04]

\

=

−a

b

[T

[M L T0

2

1 4

]

] = [ML0T–2]

4. (b) : F a bt

c d x

A t = + +

+ ⋅

+ +1

sin( )w f

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

\ [ A] = [F ] = [ML–2]As each term on RHS represents force

\+ ⋅

=1

c d x F

1

cF =

\ = = =−

− −[ ][ ] [ ]

[ ]cF

1 12

1 1 2

MLM L


21/80


As c is added to d ·x , therefore dimensions of c are

same that of d ·x .

\ [d ·x ] = [c]

or

M L

L M L [ ]

[ ]

[ ]

[ ]

[ ] [ ]d

c

x = = =

− −− −

1 1 21 2 2

Te dimensional formula for A·d is

[ A·d ] = [ML–2][M–1L–22] = [L–1]

5. (b)

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

latitudes and longitudes.

Carte also means map in French. Tis is derived

from Descrate the great mathematician.

7. (c) : [ ][ ]

[ ]

[ ]

[ ][ ]C

q

W = = =

−− −

2 2

2 21 2 4 2A

ML M L A

[ ][ ]

[ ][ ]

[ ]

[ ][ ][ ]B

F

I l = = =− − −

ML

A LML A0

2 2 1

[ ]

[ ]

[ ]

[ ]

[ ]

[

Ldi

dt

W

qt

i=

=

=

=

−

−

e ML A

A

ML

2 2

2 22 2A− ]

[ ][ ][ ]

[ ]

[ ][ ]

[ ][ ]r = = =

− −− −R A

L

ML A L

LML A

2 3 2 23 3 2

Choice (c) is dimensionally wrong.

8. (d) : As, C q

V =

DD

or e e

00 A

L

q

V C

A

L= =

DD

or e0 =( )

( )

DDq L

A V ...(i)

X L

V

t = e0

DD

( )Given

\ = X q L

A V L

V

t

( )

( )

DD

DD

(Using (i))

But [ A] = [L]2

\ = = X

q

t

DD

current

9. (d) : n n M

M

L

L

T

T

a b c

2 11

2

1

2

1

2

=

...(i)

Dimensional formula of moment of inertia

= [ML20]

\ a = 1, b = 2, c = 0Here, n1 = 12.0, M 1 = 1 kg, M 2 = 10 g

L1 = 1 m, L2 = 5 cm, T 1 = 1 s, T 2 = 1 s

n2

1 2 0

12 01

10

1

5

1

1=

.

kg

g

m

cm

s

s (Using (i))

= ×

×12

1000

10

100

5 1

1 2g

g

cm

cm

= 12 × 100 × 400 = 4.8 × 105

10. (d) : Here,D D Dmm

r

r

L

L= = =

0 003

0 3

0 005

0 5

0 06

6

.

.,

.

.,

.

As rπ

= m

r L( )2

\ D D D Drr

× = + +

×100

2100

m

m

r

r

L

L

= +

×+

×

0 003

0 3

2 0 005

0 5

0 06

6100

.

.

.

.

.

= 1 + 2 + 1 = 4%

11. (a) : Spring constant[ML

[L]M L = = =

−−F

x

21 0 2] [ ]

pascal = unit of pressure

= = =−

− −F

A

[ML

[LM L

2

2

1 1 2]

][ ]

hertz = unit of frequency =1

T = [M0L0–1]

joule = unit of work = force × distance =[ML–2] [L] = [M1L2–2].

12. (d) : Te power of 10 is irrelevant to thedetermination of significant figures.

13. (a) : Let length and breadth of a rectangular sheetare measured by using a metre scale as 16.2 cm and10.1 cm respectively. Each measurement has threesignificant 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

Terefore, as per rule, area will have only threesignificant figures and error will have only onesignificant figure. Rounding off, we get

A = 164 ± 3 cm2

14. (c) : As n1u1 = n2u2

4100

1032 3

g

cm

g

cm= n

( ) ⇒ n2 = 40

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

Distance (LHS) = [ g ] × []2 \ = −[ ] [ ] g L 2


22/80


16. (a) : Here, Rmax = 100 m

R u

g max =

2

(where u is the velocity of the projection

of ball)

or u2 = 100 g ...(i)

Using v 2 – u2 = 2as

(0)2 – 100 g = 2 (– g ) (H ) [Using (i)]

or H = 50 m

17. (a) : Given : x = 24t – 2.0t 3 m

Velocity, m sv dx

dt

d

dt t t = = − −( . )24 2 0 3 1

= 24 – 6t 2 m s–1

Acceleration, m sa d x

dt

d

dt

t = = − −2

22 224 6( )

= –12t m s–2

For v = 0, we get

24 – 6t 2 = 0 or t = 2 s

Hence, at t = 2 s, the acceleration will be

a = –12(2) m s–2 = –24 m s–2

Its magnitude is 24 m s–2.

18. (b) : Let d be distance of the target from the point

of projection.

\ × °

= −u

g

d 2 2 75

10sin( )

or u

g d

2

210= − ...(i)

andu

g d

2 2 4510

sin( )× °= +

or u

g d

2

10= + ...(ii)

Divide (i) by (ii), we get

d

d

−

+

=10

10

1

2

or d = 30 m

19. (c) : As per the laws of

vector addition,

d e f + =

d

f

e

Tis is as shown in adjacent figure.

20. (d) : ime taken for the particle to reach the highest

point ist t 1 2

2

+.

As v = u – gt

At highest point, v = 0

Terefore, initial velocity of the particle is

u g t t

= +

1 2

2 ...(i)

Terefore, height of point B from the ground is

h ut gt g t t

t gt = − = +

−1 1

2 1 21 1

21

2 2

1

2 (Using (i))

or h g t t t

gt = +

−12

1 212

2 2

1

2 or h gt t =

1

21 2

21. (b) : If A0 is the area of orifice at the bottom below

the free surface and A that of vessel, time t taken to

empty the tank,

t

A

A

H

g = 0

2 \ t

t

H

H

H

H

1

2

1

2

1

1 2= = /

\ t t

21

2

10

27= = ≈

minmin

22. (d) : Refer the figure below. Horizontal component

of velocity at A.

B

C A

u

60°30°

u u u AC u t ut H H = ° = \ = × =cos60 2 2

AB AC ut ut

= ° = × =sec 302

2

3 3

23. (a) : At time t , Bu

B= 0

A

h

u u A

=

Velocity of A, v A = u – gt (upwards)

Velocity of B, v B = gt (downwards)

= – gt (upwards)

Relative velocity of A with respect

to B is

v AB = v A – v B = (u – gt ) – (– gt ) = u

24. (c) : For first one-third of distance

Distance covered = x

3 km

speed = 10 km h–1.

Te time taken for the journey,

t x x

13

10 30= =

/h h

For the next one-third of distance :

Distance covered = x

3 km.


23/80


Speed = 20 km h–1

Te time taken for travel is

t x x

23

20 60= =

/h h

For the last one-third of distance :Distance covered =

x

3 km.

Speed is 60 km h–1

Te time taken for travel is

t x x

33

60 180= =

/h h

\ Average Speed =total distance

total time=

+ +

x

x x x

30 60 180

=180

10

x

x = 18 km h–1

25. (a) : v dx dt

at v dy dt

bt x y = = = =2 2;

\ = + = +v v v a t b t x y 2 2 2 2 2 24 4

= +2 2 2t a b

26. (b) : Maximum height,

H u

g gH

u= =

2 2 2 2

2 2

sin sinq qor ...(i)

Velocity at highest point, v H = u cosqLet v x , v y be the horizontal and vertical velocity of

projectile at heightH

2

. Ten

v x = ucosq

and v u g H

u gH y 2 2 2 2 22

2= − × = −sin sinq q

v u u u

y 2 2 2

2 2 2 2

2 2= − =sin

sin sinq

q q (Using (i))

\ = +( )Net velocity at height H v v x y 22 2

1 2/

As per question,

2

5

2

52 2

1 22 2 2v v v v v v x y H x y H +( ) = +( ) =

/or

or 25 2

2 2 2 2 2 2u u ucos sin cosq q q+

=

or sin2q = 3cos2q or sin cosq q= 3

or tan tanq = = °3 60 or q = 60°27. (b) : Velocity of the passenger with respect to

ground

v PG = v PE + v EG = 6 + 3 = 9 km h–1

ime takenm

m s

s st x

v PG= =

×= =

−

120

95

18

240

548

1

28. (a) :

According to the given problem

1

2

3

4

1

22 2m u mu( cos )q = ×

or cos23

4q = or cos cosq = = °

3

230

or q = 30°

29. (d) :

P

P

O r

r

2

1

x

According to cosine formula,

cosq =

+ −r r x

r

2 2 2

22

or 2 2 2 2 2r r r x cosq = + −or x 2 = 2r 2 – 2r 2cosq = 2r 2 [1 – cosq]

=

2 22

2 2r sin q

Displacement from P 1 to P 2 is

x = 22

r sin q

30. (a) : Given,

r t i t j= +{( ) ( ) }^ ^

2 2 2

Comparing it with standard equation of position

vector, r x i y j= +^ ^

, we get x = 2t and y = 2t 2

⇒ v dx

dt x = = 2 and v

dy

dt t y = = 4

\ tanq = = =v

v

t t

y

x

4

22

Differentiating with respect to time we get,

(sec )2 2q

qd dt

=

or ( tan )1 22+ =q

qd dt

or ( )1 4 22+ =t d

dt

q

ord

dt t

q=

+

2

1 4 2

at s, = rad st d

dt =

+= −2

2

1 4 2

2

1721q

( )


24/80

SOLUTION SET-22

1. (b) : Te woman experiences three orces; mg , herweight acting vertically downwards; N 1, reaction

due to her weight; N 2, horizontal reaction which

provides the centripetal acceleration.

From Newton’s second law,

∑ = =F N mv

r x 22

SF y = N 1 – mg = 0 v = (2pr )u (where u is requency) = (2p × 9) (6/60) = 1.8p m s–1

Tereore,

N 2

250 1 8

9178= =

( )( . )pN

N 1 = mg = 490 N

Te magnitude o her weight is the magnitude othe resultant orce exerted on her by the chair.

N N N = +1

222

= +490 1782 2 = 521 N

2. (c) : Let T be the time o flight and u be the initial velocity o the stone then

A B

2R

u

222

Ru

g =

sin q ...(i)

and T u

g =

2 sinq ...(ii)

Eliminating u, we get

tanq q= = ×

× = \ = °

gT

R

2 2

4

10 10

4 250 3

1

330

3. (b) :

GB

O

R

W

A

Asa a/

sin sin( )

2

λ p q=

−

⇒ = \ = −sin sin sin ( sin )q λ q λ2 21

4. (b) : At NP, temperature = 273 K andpressure = 105 N m–2

v RT

M

RT

M

P = =

g

ρ,

⇒ = ⇒ = = ×

=v P v

P

g

ρ g ρ2 2

5

330 1 3

101 4

( ) ..

and g = 1 + (2/ f ) ⇒ f = 2/( g – 1) ⇒ f = 5

5. (d) : Along the vertical direction,Net impulse due to normal = change in momentum

Ndt m v mv ∫ = +( ) cos/2 q ...(i)(where N is the normal by the floor on the ball)Along the horizontal direction,Friction orce, f = mN Let horizontal velocity o the ball afer collision = v ′Net impulse due to riction = change in momentum

in horizontal direction

− = ′ − ∫ m qNdt mv mv sin

⇒ –m[m(v /2) + mv cosq] = mv ′ – mv sinq (Using eqn. (i))

\ v ′ = v sin q – m(v /2) + v cos q}

6. (c) :

T sinq =mg

2

T ′ = T cosq =mg

2sincos

q q

\ T ′ =mg

2tan q

7. (c) : ension T 3 required to move third block= mmg .

ension T 2 to move 2nd block = mmg + 2T 3

= 3mmg .Force F required to move the first block

= 2T 2 + mmg = 7mmg .



25/80

8. (d) : Change in position o CM in verticaldirection

∆h l l = °

+

cos30

2

CM

l

l cos30= l 3/2

60º

A

B C

F

l

60º

Work done by gravity = = +

mg h mg

l l ∆

3

4

=

+

mgl

3 4

4

9. (b) :

1 kg

10 N s

10 cos30º N s

30º

30º1 kg

J T

J T

Here 10cos30° – J T = J T

[Both the masses will move with same velocity

along the string]

⇒ = ° = J T 5 30 52 3cos N s

10. (b) : Equation o path,

y x

=2

20dy

dt

x dx

dt =

10, v

x v y x = 10

At x = 10 v y = v x

otal v v v x y = +2 2 ,

6 2 2= v x

v x 2 36

2

18= = , v x = 3 2

v y = 3 2

Now since the magnitude o velocity is 6 m s–1

constant, tangential acceleration will be zero. It has

only normal acceleration.

Now

y x

=2

20

dy

dx

x x = =

2

20 10 \ dy

dx x

==10

1

d y

dx

2

2

1

10=

Radius o curvature at the given point is

R

dy

dx

d y

dx

=+

= +

=1

1 1

1

10

10 2

2 3 2

2

2

3 23 2

/

//( ) ( )

= × ×10 2 2 2 = 20 2

Normal acceleration

av

RRT = = =2 36

20 2

9 2

10units

SOLUTION OF MAY 2015 CROSSWORD

WINNERS May 2015 Guneet Kaur Harsh Gupta Rizwan Khan

Solution Senders (April 2015) Aditya Srivastava

Chandra Shekhar Panigrahi Rajneesh Kumar

Solution Senders of Physics Musing

SET-22

1. Swayangdipta Bera

2. Debajyoti Dash (West Bengal)

3. Sayantan Bhanja (West Bengal)

SET-21

1. Komal Khatri (Pune)

2. Girish Ranjan (Bihar)



26/80


1. A wave is represented by y (x , t ) = asin(kx –wt + f).

Te phase o the wave is

(a) f (b) kx – wt

(c) wt + f (d) kx – wt + f

2. Name o units o some physical quantities are given

in List I and their dimensional ormulae are given

in List II. Match List I with List II and select the

correct answer.

List I List II

P. Pa s 1. [M0L2–2K–1]

Q. N m K–1 2. [ML–3K–1]

R. J kg–1K–1 3. [ML–1–1]

S. W m–1 K–1 4. [ML2–2K–1]

P Q R S

(a) 4 3 1 2

(b) 3 2 4 1

(c) 3 1 4 2

(d) 3 4 1 2

3. Te apparent weight o a person in a lif moving

downwards is hal his apparent weight in the same

lif moving upwards with the same acceleration.

Te acceleration o the lif is

(a) g (b) g

4 (c)

g

2 (d)

g

34. Te ratio o the angular velocities o the earth about

its own axis and the hour hand o a watch is

(a) 1 : 2 (b) 2 : 1 (c) 1 : 12 (d) 12 : 1

5. Consider the ollowing statements :

Nuclear orce is

1. charge independent

2. long range

3. central

Which o the above statements is/are correct?

(a) 1 only (b) 1 and 2

(c) 2 and 3 (d) 1 and 3

6. Te magnetic susceptibility o a material o a rod is

299. Te permeability o the material o the rod is

(m0 = 4p × 10–7 H m–1)

(a) 3771 × 10–5 H m–1 (b) 3771 × 10–6 H m–1

(c) 3771 × 10–7 H m–1 (d) 3771 × 10–8 H m–1

7. A particle motion on a space curve is governed by

x = 2sint , y = 3cost and z = 5 sin t . Te speed o

the particle at any instant is

(a) 3 2 sint (b) 3 2cos t

(c) 3 2sin t (d) independent o time

8. A V transmitting antenna is 128 m tall. I the

receiving antenna is at the ground level, the

maximum distance between them or satisactorycommunication in LOS mode is

(Radius o the earth = 6.4 × 106 m)

(a) 64 10 km (b)128

10km

(c) 128 10 km (d)64

10km

9. A battery o em 3 V and internal resistance 0.2 W

is being charged with a current o 5 A. What is the

potential difference between the terminals o the

battery?

(a) 2 V (b) 3 V (c) 3.5 V (d) 4 V

10. When NaCl is added to water, the surface tension of

water

(a) increases (b) decreases

(c) remains constant (d) nothing can be said

11. A person walks on a straight road rom his house

to a market 2.5 km away with a speed o 5 km h–1.

Finding the market closed, he instantly turns back

and reaches his house with a speed o 7.5 km h–1.

Te average speed o the person is


27/80


(a)14

3

1m s− (b)5

31m s−

(c)5

6

1m s− (d)1

31m s−

12. Te hal lie o a radioactive element is 10 h. Te

raction o initial activity o the element that will

remain afer 40 h is

(a)1

2 (b)

1

16 (c)

1

8 (d)

1

4

13. Light travels rom air to water, rom water to glassand then again rom glass to air. I x representsreractive index o water with respect to air, y represents reractive index o glass with respect towater and z represents reractive index o air withrespect to glass, then which one o the ollowing is

correct?(a) xy = z (b) yz = x

(c) zx = y (d) xyz = 1

14. Which o the ollowing parameters is the same ormolecules o all gases at a given temperature?(a) Mass (b) Speed(c) Momentum (d) Kinetic energy

15. In a meter bridge experiment, the length AB o thewire is 1 m. Te resistors X and Y have values 5 W and 2 W respectively. When a shunt resistance S isconnected to X , the balancing point is ound to be

0.625 m rom A. Ten, the resistance o the shuntis

(a) 5 W (b) 10 W (c) 7.5 W (d) 12.5 W

16. Te equation o trajectory o a projectile is

y x x = −

105

9

2

m. Te range o the projectile is

(a) 36 m (b) 24 m (c) 18 m (d) 9 m

17. A ray o light travels rom an optically denser

medium towards a rarer medium. Te critical angle

or the two media is C . Te maximum possible

angle o deviation o the ray is

(a)p2

− C (b) p – 2C

(c) 2C (d)p2

+ C

18. Te successive resonance requencies in an open

organ pipe are 1944 Hz and 2600 Hz. I the speed o

sound in air is 328 m s–1, then the length o the pipe is

(a) 0.40 m (b) 0.04 m

(c) 0.50 m (d) 0.25 m19. A machine gun fires 240 bullets per minute with a

velocity o 600 m s–1. I the mass o each bullet is

10 g, the power o the gun is

(a) 7.2 kW (b) 72 kW

(c) 3.6 kW (d) 36 kW

20. Te sensitivity o a galvanometer that measures

current is decreased by 1

40times by using shunt

resistance o 10 W. Ten, the value o the resistanceo the galvanometer is

(a) 400 W (b) 410 W (c) 30 W (d) 390 W

21. Which one o the ollowing is correct about

wave-particle duality?

(a) Wave-particle duality holds or matter particles

but not or light.

(b) Wave-particle duality holds or light but not or

matter particles.

(c) Wave-particle duality holds or electrons but

not or protons.

(d) Wave-particle duality holds or light as well as

or matter particles.

22. Work done to increase the temperature o

one mole o an ideal gas by 30°C, i it is

expanding under the condition V ∝ T 2/3 is(R = 8.31 J mol–1K–1)

(a) 116.2 J (b) 136.2 J

(c) 166.2 J (d) 186.2 J

23. Electric charges A and B attract each other. Electric

charges B and C also attract each other. But electric

charges C and D repel each other. I A and D are

held close together then which one o the ollowingis correct?

(a) Tey cannot affect each other.

(b) Tey attract each other.

(c) Tey repel each other.

(d) Cannot be predicted due to insufficient data.

24. In a transistor i a varies between20

21

100

101and ,

then the value o b lies between(a) 1–10 (b) 0.95–0.99

(c) 20–100 (d) 200–300


28/80


25. Te minimum orce required to move a body up

an inclined plane is three times the minimum orce

required to prevent it rom sliding down the plane.

I the coefficient o riction between the body and

the inclined plane is 12 3

, the angle o the inclined

plane is

(a) 60° (b) 45° (c) 30° (d) 15°

26. Te threshold requency o the metal o the cathode

in a photoelectric cell is 1 × 1015 Hz. When a certain

beam o light is incident on the cathode, it is ound

that a stopping potential 4.144 V is required to

reduce the current to zero. Te requency o the

incident radiation is (h = 6.63 × 10–34 J s)

(a) 2.5 × 1015

Hz (b) 2 × 1015

Hz(c) 4.144 × 1015 Hz (d) 3 × 1016 Hz

27. A 50 mF capacitor is connected to an ac source

V = 220 sin50t where V is in volt and t is in second.

Te rms current is

(a) 0.55 A (b)0 55

2

.A

(c)2

0 55.A (d) 2 A

28. O the ollowing, NAND gate is

(a)

(b)

(c)

(d)

29. On a temperature scale Y, water reezes at –160°Y

and boils at –50°Y. On this Y scale, a temperature

o 340 K is(a) –106.3°Y (b) –96.3°Y

(c) –86.3°Y (d) –76.3°Y

30. A satellite is revolving very close to a planet o

density r. Te period o revolution o the satellite is

(a)

3pr

G (b)

3

2

p

rG

(c)

3p

rG

(d)

3p

r

G

31. I the ratio o maximum and minimum intensities

o an intererence pattern is 36 : 1, then the ratio o

amplitudes o the two interering waves will be

(a) 3 : 7 (b) 7 : 4

(c) 4 : 7 (d) 7 : 532. A wire o length 6.28 m is bent into a circular coil

o 2 turns. I a current o 0.5 A exists in the coil, the

magnetic moment o the coil is

(a)p

4A m2 (b)

1

4 A m2

(c) p A m2 (d) 4p A m2

33. Te electric field or an electromagnetic wave in

ree space is

E kz t i= − × −30 6 108 1cos( ) .^

Vm Te

magnitude o wave vector is

(a) 2 rad m

–1

(b) 3 rad m

–1

(c) 4 rad m–1 (d) 6 rad m–1

34. wo parallel plane sheets 1 and 2 carry uniorm

charge densities s1 and s2(s1 > s2) as shown in the

figure. Te magnitude o the resultant electric field

in the region marked I is

1 2

(a)s

e

1

02

(b)s

e

2

02

(c)s s

e

1 2

02

+ (d)

s s

e

1 2

02

−

35. In Millikan’s oil drop experiment, a charged oil

drop o mass 3.2 × 10–14 kg is held stationary

between two parallel plates 6 mm apart, by applying

a potential difference o 1200 V between them. How

many electrons does the oil drop carry ?( g = 10 m s–2)

(a) 7 (b) 8 (c) 9 (d) 10

36. wo long wires each parallel to the z -axis and each

carrying current I , are at (0, 0) and (a, b). Te orce

per unit length o each wire is

(a)m

p

02

2 22

I

a b( )+ (b)

m

p

02

2 22

I a b

a b

( )

( )

+

+

(c)m

p

02

2 2 3 22

I

a b( ) /+ (d)

m

p

02

2 2 1 22

I

a b( ) /+


29/80


37. wo blocks o masses 1 kg and

2 kg are connected by a metal

wire going over a smooth pulley

as shown in the adjacent figure.

Te breaking stress o the metal is

40

3106 2

p×

−N m What should be

the minimum radius o wire used

i it should not break?

( g = 10 m s–2)

(a) 0.5 mm (b) 1 mm

(c) 1.5 mm (d) 2 mm

38. A rectangular coil is rotating in a uniorm magnetic

field B. Te em induced in the coil is maximum

when the plane o the coil(a) is parallel to B

(b) makes an angle 30° with B

(c) makes an angle 45° with B

(d) is perpendicular to B

39. A steady dc current is flowing through a cylindrical

conductor. Which of the following statements is/are

correct?

1. Te electric field at the axis o the conductor is

zero.

2. Te magnetic field at the axis o the conductor

is zero. Select the correct answer using the code given

below :

(a) 1 only (b) 2 only

(c) both 1 and 2 (d) neither 1 nor 2

40. Te ratio between kinetic and potential energies o

a body executing simple harmonic motion, when

it is at a distance o1

N o its amplitude rom the

mean position is

(a) N 2 + 1 (b)1

2

N (c) N 2 (d) N 2 – 1

41. PQR is a right angled triangular plate o uniorm

thickness as shown in the figure. I I 1, I 2 and I 3

are moments o inertia about PQ, QR and PR axes

respectively, then

(a) I 3 < I 2 < I 1 (b) I 1 = I 2 = I 3(c) I 2 > I 1 > I 3 (d) I 3 > I 1 > I 2

42. A tank o height 5 m is ull o water. Tere is a hole

o cross-sectional area 1 cm2 in its bottom. Te

volume o water that will come out rom this holeper second is ( g = 10 m s–2)

(a) 10–3 m3 s–1 (b) 10–4 m3s–1

(c) 10 m3s–1 (d) 10–2 m3s–1

43. Te temperature o a perect black body is 727°C

and its area is 0.1 m2. I Stean’s constant is

5.67 × 10–8 W m–2 K–4, then heat radiated by it in

0.3 minutes is

(a) 1701 J (b) 17010 J

(c) 102

Documents

Physics for You June 2015