A ray of light passes through an equilateral prism such that angle of incidence is equal to the angle

25. Four metallic plates each with a surface area of

one side A are placed at a distance d from each other as shown in figure. The capacitance of the system is

0 » ^ ( 0 ^ („, ^ d d

26. An object is d i s p l a c e d f r o m p o s i t i o n vector

A A A A

rx = (2i + 3j)m to r2 = (4i + 6 ; ) m under a force

r\ A A

F = (3x i + 2y j) N. The work done by this force is

(a) 63 J (b) 73 J (c) 83 J (d) 93 J 27. If the velocity of the particle is increased three

t i m e s , t h e n t h e p e r c e n t a g e d e c r e a s e in its de Broglie wavelength will be

(a) 33.3% (b) 66.6%

(c) 99.9% (d) 133.2%

28. The electrical analog of a spring constant k is (a) L (b) - (c) C (d) 1

L -

, ^C

29. If power dissipated in the 9 £2 resistor in the circuit shown is 36 W, the potential difference across the 2 Q resistor is

90

-WA-6 £1

-VWV-a

— I H '

WA-v

(a) 4 V (b) 8 V (c) 40 V (d) 2 V 30. A steel ball is dropped on a hard surface from a

height of 1 m and rebounds to a height of 64 cm.

The maximum height attained by the ball after nth bounce is (in m) (}-\j > 9 mi t

(a) (0.64)2™ IOf (c) (0.5)2

ai fchmMf,-(d) (0.8)"

31. Four simple harmonic vibrations xt = 8sincof, x3 = 4sin(cot + rt) a n d x2 = 6 s i n cot h—

2 n • I 3n

x4 = 2 sin | at + — are superimposed on each other. The resulting a m p l i t u d e and its phase difference with xt are respectively

(a) 20, tan-'

j i j

(b) 4 ^ 2 , -(c) 20, tan"1 (2) (d) 4V2, -4

32. 200 g of a solid ball at 20°C is dropped in an equal amount of water at 80°C. The resulting temperature is 60°C. This means that specific heat of solid is

(a) one-fourth of water (b) one-half of water (c) twice of water (d) four times of water 33. The length of a potentiometer wire is I. A cell of

emf e is balanced at a length 1/5 from the positive end of the wire. If length of the wire is increased by 1/2, at what distance will the same cell given a balance point.

(a) —I (b) (c) —I (d) —1

15 15 w 10 10

34. A uniform thin bar of mass 6m and length 12L is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is

(a) 20mL2 (b) 6 m l2 (c) — ml1 (d) 30mL2

5

35. A ray of light passes through an equilateral prism such that angle of incidence is equal to the angle of emergence and the latter is equal to [ 3

of prism. The angle of deviation is (a) 45° (b) 39° (c) 20° (d) 30

0 I angle

36. A light rod of length 2 m / / / / / / / / / / / / / / / / / / / / /

Brass suspended from the ceiling

horizontally by means of two vertical wires of equal length. A weight W is h u n g from a light rod as shown in figure.

The rod h u n g by means of steel wire of cross-sectional area A1 = 0.1 cm2 and brass wire of cross-sectional area Az = 0.2 cm2. To have equal stress in both wires, — 1 _

(a) i

; 3+ (b) (c) (d)

PHYSICS FOR YOU I MAY '13 3 9

37. In the circuit shown below, the key K is closed at t = 0. The current through the battery is

•K 500 kg slowly at constant speed on a level road.

If a force at 500 N is applied, the acceleration of the car will be

(a) zero (b) 0.2 m s"2

(c) 0.5 m s "2 (d) 1.0 m s 2

39. A particle of m a s s 1 x 10~26 kg and c h a r g e 1.6 x 10~19 C travelling with a velocity 1.28 x 106 m s"1

along the positive x-axis enters a region in which a uniform electric field £ and a uniform magnetic field B are present. If E = -102.4 x 103 k N C"' and

-7 A -9

B = 8 x 10 j Wb m the direction of motion of the particle is

(a) along the positive x-axis (b) along the negative x-axis (c) at 45° to the positive x-axis (d) at 135° to the positive x-axis

40. The p o w e r obtained in a reactor u s i n g U2 3 5

disintegration is 1000 kW. The mass decay of U235 per hour is

(a) 10 ^g (b) 20 ng (c) 40 ng (d) 1 ng 41. A charge Q is enclosed by a gaussian spherical

surface of radius R. If the radius is doubled, then the outward electric flux will

(a) increase four times (b) be reduced to half (c) remain the same (d) be doubled

42. A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to the escape velocity for that planet.

The maximum height attained by the body is R R (b) T (c) , . R

(a) —

' 2 3 v ' 5 (d) ?

43. A metal rod of Young's modulus Y and coefficient of thermal expansion a is held at its two ends such that its length remains invariant. If its temperature is raised by f°C, the linear stress developed in it is (a) - f (b) ^ (c) Yat (d)

Y a t (Yat) 44. If T denotes the temperature of the gas, the volume

thermal expansion coefficient of an ideal gas at constant pressure is

(a) T (b) T2 (c) | (d) ~

45. Water rises in a capillary tube to a height of 2 cm. In another capillary tube whose radius is one third of it, how much the water will rise ? (a) 2 cm (b) 4 cm (c) 6 cm (d) 8 cm 46. Energy required to break one bond in DNA is

(a) 10-10 J (b) 10~18J (c) 10~7J (d) 10~20 J 47. A ray of light strikes a silvered surface inclined to

another one at an angle of 90°. Then the reflected ray will turn through

(a) 0°

48. An open and a closed pipe have same length.

The ratio of frequencies of their nth overtone is n + 1 49. An optical fibre communication system works on a

wavelength of 1.3 jim. The number of subscribers it can feed if a channel requires 20 kHz are (a) 2.3 x 1010 (b) 1.15 x 1010

(c) 1 x 105 (d) 2.3 x 1014

50. A gas is expanded from volume V0 to 2V„ under three different processes as shown in the figure.

Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let A(ij, Aii2 and AU3

be the change in internal energy of the gas in these three processes. Then

P4 1

4 0 PHYSICS FOR YOU | MAY '13 ?

(a) Afij > AU2 > All3 (b) ALT! < AU2 < AU3

(c) AU2 < ALZj < AU3 (d) AU2 < AU3 < AU, SOLUTIONS

(c) : Kinetic energy of rotation,

K = H O ^ I ^ ) 2 - 1 * 2

2

1 L

2 I 2 I 2 (mr2) (v L = 7(0)

K • K'

1 L .

- j (As is constant during rotation)

• = 4

* ( r / 2 )2

or K' = 4K

Thus kinetic energy of mass increases by a factor of 4.

(a): MAAA/V . ARC

In t h e g i v e n n e t w o r k CDBE is a b a l a n c e d Wheatstone bridge. Thus, the resistance connected across DE becomes ineffective. The equivalent circuit is as s h o w n in figure.

MMMf A R

•-AMAAr

•A R

MAAAAr A R

2 R

-WWV-2 R -<WWV- -WAAr-. 2R

-JWV\Ar

B

Hence, the equivalent resistance between A and

B is

q 3 3

(d): Here, N = 6, CP = 8 cal m o l1 K1 R = 8.31 J mol-1 K"1 = 2 cal mol"1 K"1

As CP - CV = R

6.

7.

Cy Cp - R

= 8 cal mol"1 K"1 - 2 cal m o H K-1

= 6 cal mol"1 K1

Change in internal energy of the gas is AU = nCvAT = 6 x 6 x 15 = 540 cal 4. (b): [Angular m o m e n t u m ]

[Magnetic moment]

2m

2m ' M "

It AT

5.

= [MA"1 T"

(b): Transition f r o m 4E to E (4 E - E ) = he

or A,, = he 3E

Transition from — E to £

3 A. 2

he 3hc

4 E

:..(i)

— 4E

— £

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

V _ 4

?l2 ~ 9

(a): According to polygon law resultant of four forces, each of F1 acting at 72° is along the fifth side of polygon taken in opposite order. As F2 is acting along this side of polygon, therefore, net force on the particle = F2 - Fv

F -F .•. Acceleration = — —

-m

(a): F u n d a m e n t a l f r e q u e n c y p r o d u c e d b y a stretched string is given by

\> = = k\Jr (As I and|i are kept constant) dv

v ldT 2 T

dT „dv „ 6 or — = 2 — = 2 x

T V 600

(d) (b):

• 0.02

Earth's surface

Acceleration due to gravity at point A is

PHYSICS FOR YOU I MAY '13 4 1

£4 =

f h ] 2

f RF ) 1+ - l + { ><E) V ^t2RF /

_ 4

2

V

^...(i)

where g is the acceleration d u e to gravity on the earth's surface.

Acceleration due to gravity at point B is 1 R

2 ...(ii) E

J k . 2Rr

= £ 2

Divide (ii) by (i), we get KB=9

-8A 8

10. (c) : The power gain of the amplifier Power gain = Current gain x Voltage gain

= 50 x 5 = 250

11. (d): Let the electric field due to these two point charges be zero at point P at a distance x f r o m

+8 q - 2 q

A

x = 0 B

x = L

M

x-P -H

L (81) ^{I 1 ( -2? ) Q}

47t£0(L+x)2 4 t o0 x 2

1 (8(7) 1 2 q or ——— = —

4rce0 (L+x)2 47te0 x2

2 _ 1 L+x x

2x = L + x or x = L :. AP = AB + BP = L + L = 2L

12. (d): Zeroth law of thermodynamics leads to the concept of temperature.

13. (c) : Area of cross-section of the tube is A = 7t[(5 cm)2 - (4.5 cm)2]

= 4.7571 cm2 = 4.75rt x 10"4 m2

Resistance of the tube is

R pi __ 1.7x10 flmx5m 4.75 x 3.14 x 10- 4 m2

= 0.56 x 10"4 £2 = 5.6 x 10"5 Q.

1 4 . (a): Let AB = y, BC = CD = h and tAB * t

As per question • ; n

y-\gt2

y + h = ±g(t+2)2

and y + 2h = ±g(t + 3)2

Solving these three equations, we get

t = 0.5 s J- D

15. (c) : Here, R = 300 m p. = 0.3 g - 10 m s~2

The m a x i m u m speed of the car is

®max= \t^Rg = a/0-3 X 300 m x 10 m s~2

= 30 m s_ 1= 30 x — k m h "1 = 108 km h"1

5

16. (c) : According to ideal gas equation PV = — RT m

M

...(i) ...(ii) As per question

20 x V = — R x 300 M

p,x V = (jnl2)Rx350

M Divide (ii) by (i), we get

P' 175 175 „„

— = or P' = x 20 = 11.7 atm 20 300 300

17. (a): Point (0, 0, -a) lies on z-axis. T h e r e f o r e , magnetic field d u e to current along z-axis is zero a n d d u e to rest two wires is - - i n m u t u a l l y

2ti a

perpendicular directions along positive indirection and negative x-direction.

Un / ^{A A}

271 a

18. (c) : Vertical component of initial velocity, My = usin30°

uy 8 0 - i

or u = — - — = = 160 m s sin 30° ( 1 / 2 L Q - '

(••• Uy =! 80 m s (Given)) Horizontal component of initial velocity,

u = wcos30° =p 160 x — = 80V3m s"J 3 1

Time of flight, 2

^ 2u sin 0 2 x 160 x sin 30°

T = = = 1 6 s X 10 i - — r - : 4s> ni?'

4

4 2 PHYSICS FOR YOU | MAY '13 ?

Let v be the velocity of the projectile at t = —.

Its horizontal and vertical components are given by

vx = ux = 80 V3 m s"1

vy = My - gt = 80 - 10 x 4 = 40 m s"1

Its magnitude is given by

v = y lvx+ vy =>/( 80V3 f + (40)2

= 40 7 l 2 + l = 40 Vl3 = 144 m s"1

19. (c) : The electrostatic potential at the centre of the first ring (i.e., at O) with charge Q, is due to charge Q, itself as well as due to charge Q2 on the second ring which is given by

Qi

v , = -r r

4 t c0 a 47te0 a-42

Similarly, the electrostatic potential at the centre of the second ring (i.e., at O') is given by

V2 = ——+ —

47re0 a-J2 47te0 a

Required work done , W = q( V^ - V2)

W = q \ — & — — ^

[47ie0 a 4t i e0 a\l2 47ce0 a\l2 4ne0 a

471 e0a

4ne0a

42Qx+Q1-Q,-42Q1

4~2

4718, ¹—1t[42(Q1 -Q2) - l ( Q j - Q2) ] ,«v2

qU2

4nena\f2

(Q1-Q2)-20. (b): Here, 2/ = 0.1 m, m = 10-4 A m B = 30 Wb m 2, 6 = 30°, x = ? x = MBsinO = m(2/)Bsin6

= 10~4 (0.1) x 30sin30° = 1.5 x l ( H N m . f ( n )

i n cot + -I 6 j 21. (b): x = asm

V o ,

Velocity = — = — asinf raJ->-i-™4=a(ocos[ <at +

y dt dt { &

f)

Maximum velocity = am As per question

ac0 [ K

— = fl(OCOS cot + —

2 6 or cos (01 + 7C

or cot + — = 60° = —E rad 6 6

2n n n 2K K T or cot = = or — f = or f =

-6 -6 -6 T -6 12 22. (b): — = n 3 h

In :. n =3

The kinetic energy of the electron in nth orbit is n 2 n

K, = e V = ^ eV = 1.51 eV 13.6 w 13.6 eV = - —

3Z 9

23. (d): If a represents angle of repose, then tana = 0.8

a = tan- 1 (0.8) = 39°

The given angle of inclination is less than the angle of repose. So, the 1 kg block has no tendency to move.

Note that mg sinO is exactly balanced by the force of friction. So, T = 0

24. (d): = jf for 0 <t<~

16Vn

V2 =

< v >0-T = 4 T / 4

dt

r -, T / 4

16 V2 t3 j 2 3

L J 11 WT0'4

V

S

25. (b):

The equivalent circuit of the given network is as shown in the figure.

PHYSICS FOR YOU I MAY '13 4 3

1 1

4' '3

The capacitance of each capacitor is d

Here, the two capacitors are connected in parallel.

Hence, the e q u i v a l e n t c a p a c i t a n c e b e t w e e n A and B is C _AB 2 e0A

26. (c):

h (4-6)

W = j? dr= J (3x2i + 2y])-(dxt + dyj + dzk)

h (2,3) (4,6)

= J (3x2dx + 2ydy) = [x3 + y2]^) (2,3)

= [(4)3 + (6)2 - (2)3 - (3)2] = 83 J 27. (b): X = — and A,' = — = - X

mv m3v 3

% decrease in de Broglie wavelength X - X'

• x 100

- H

^{x 100}

= 11 - ^ | x 100 = 66.6 %

28. (d): The a n a l o g y b e t w e e n m e c h a n i c a l a n d electrical quantities is as shown in the table.

Mechanical system Electrical system

Mass m Inductance L

Force constant k 1

Reciprocal capacitance — Displacement x Charge q

Velocity v = —

y dt

^ , da Current i =

—-dt Mechanical energy

1 1 1 2

E = —kx+—mv 2 2

Electromagnetic energy U = —— +—LI2

2 C 2 29. (c) : As, P = I9 P ZR or I = J*

V R

Current through 9 Q resistor in figure given

below is

( h + h )

A h

9Q -AAAAr

6Q -AAAAr

V

2Q

-J^VWV-(h + h)

h=

36 W

\ 9Q = 2 A

As resistors 9 £2 and 6 £2 are connected in parallel, therefore, potential difference

VA-Vb=9I1=6I2

or I2 = - J j = ~ x 2 A : 3 A 6 1 6

Current drawn from the battery

= I1 + 72 = 2 A + 3 A = 5 A

Potential difference across 2 £2 resistor

= (5 A) (2 £2) = 10 V

30. (b): When a ball is dropped from a height h and it rebounds to a height hlf then

Here, h = 1 m, hx

10.64 m e = l m

= 64 cm = 0.64 m 0.8

The maximum height attained by the ball after n[h bounce is

h„ = e2"h = (0.8)2" (1 m) = (0.8)2" m

31. (d): The resulting amplitude and corresponding phase difference can be calculated by vector method as follows:

A, = 6 /St = 8

IAX = 8 - 4 = 4 and IAy = 6 - 2 = 4 Therefore, resulting amplitude is and phase difference with x1 is <f> = —. K

4 4 PHYSICS FOR YOU | MAY '13 ?

t

32. (b): Heat lost by water is

Qi = 200 x swaler x (80 - 60) = 200 x Swaler x 20 Heat gain by solid ball is

Q2 = 200 x Ssol;d x (60 - 20) = 200 x sS0lid x 40 According to principle of calorimetry

Qi = Qi

200 x Swater x 20 = 200 x Ssolid x 40

= 1 ''solid — 2 ^water 33. (c) : In first case,

Potential gradient, K =

where e0 is the emf of the battery in potentiometer circuit.

As per question

...(i)

ry l ^ EO e = K - = -y- x - = -u

-5 1 -5 -5 In second case,

I 31 Length of potentiometer wire = I + — = —

Potential gradient, K' = £° = - ^ 0

8 (31 / 2) 3 I If V is the new balancing length, then

e = K'l' = ^ x l ' 31

Equating (i) and (ii), we get

...(ii)

^n 2 £n „ „ 3 ,

•JL = —— x / or I = — /

5 3 1 10 34. (a): Length of each side of hexagon = 2 L

Mass of each side = m

Let O be centre of mass of hexagon.

Therefore, perpendicular distance of O f r o m each side

r = Ltan60° = l V 3 O

The desired moment of inertia of hexagon about 0 is

1 — 6 [lone side]

\2

= 6

= 6

= 6

m(2L)' 2

+ mr 12

mL -mi <LV3)2

+ 3 mL1

3 = 20 mL

35. (d): i = e = -A 4

where i is the angle of incidence, e is the angle of emergence and A is the angle of prism.

i = - x 60° = 45°

4

When i = e, prism is in m i n i m u m deviation position,

.-. 8m = 2i - A = 2 x 45° - 60° = 30°

36. (d): Given,

Stress in steel wire = Stress in brass wire Ti

\

^A2

T2 a² 0 1 0.2

37. (a): When key K is closed at f = 0, the current through inductance L will increase with time, resulting induced emf across L, which will oppose the current through the arm having inductance L. Therefore, the current from battery will flow through R2. This current is / = - at t = 0.

R2

When t = °°, the current will reach to steady state in the circuit. The presence of L will become ineffective. Now effective resistance of circuit

R1 + R2

Current through battery,

V(R1+R2) r =

-R1R2/(RL + R2) Net force

R,R2

38. (c): Acceleration = Mass _ (500-200) N

500 kg

= - m s~2 = 0.6 m s~2

5

PHYSICS FOR YOU I MAY '13 4 5

39. (a): Here, m = 1 x 10~26 kg q = 1.6* 10"19 C

1.28 x 106 i m s_ 1

£ = -102.4 x l 03f c N C_ 1

B = 8 x l 0 "2/ W b m "2

Force on a charged particle in a uniform electric and magnetic fields is

F = cjE + q(v x B)

= q(E + vxB)

= (1.6 x 10~19)[(-102.4 x 103 fc)

+ (1.28 x 106 i x 8 x 10~2 /)]

= (1.6 x 10"19)[(-102.4 x 103 k + 102.4 x 103 k)]

= 0

Acceleration of the particle, a = - • 0 Hence, the particle will move along x-axis.

40. (c) : A c c o r d i n g to E i n s t e i n ' s m a s s e n e r g y relation

•• mc^c or m = -E

Mass decay per second Am _

~AF~

10'

J_ A E _ I cz Af ~ c

„-l

1000 x 1 0J W (3 x 108 m s""1)2

9 x 1 0 ^,16 kgs~

Mass decay per hour Am

At • x 60 x 60 = ^10°

9 x 1 0 ^,16

k g s -1 (3600 s)

= 4 x 10"8 kg - 40 x 10"6 g = 40 fig

41. (c) : According to Gauss's law, the total outward electric flux linked with gaussian surface

<))£ = — x charge enclosed by surface.

eo

If the radius of the gaussian surface is doubled the total outward electric flux will remain the same as charge enclosed by the guassian surface is unchanged.

42. (d): Escape velocity from planet's surface is 12 GM

where M and R be the mass and radius of the planet respectively,

The velocity of projection of the body from the planet's surface is

1 1 2 GM u = —v„ = — , /

3 3V R

According to the law of conservation of mechanical energy

Total energy on planet's surface = Total energy at maximum height h GMm j_Q | ^ GMm)

—mu + 1 2 2

—1 m

2

\ 2 GM 3 v R I - i — L .

9 R + h

R + h ) GMm GMm

R R + h or h R

43. (c) : Due to change in temperature t°C, increase in length,

Al = lat or A/

I = at Y stress stress

strain AI / / :. Stress = Y x — = Yat Al

44. (c) : According to an ideal gas equation PV = nRT

:. PdV = nRdT (P = constant) dV = |— | dT

or dV = \ Y\dT ...(i)

The volume coefficient of expansion is given by

= dV

J~VdT

dV = yVdT ...(ii) Equating (i) and (ii) we get

Y = 45. (c) : h = I T 2Tcos9

rpg , 2Tcos6 hr = = constant

V i - h2r2 or h2 =

Substituting the given values, we get h2 = (2.0)(3)

= 6 cm

^ 2 - 1

4 6 PHYSICS FOR YOU | MAY '13 _?

46. (d): The energy required to break one bond in DNA is 10"20J.

47. (d): For mirror Mv Zi = 0°

Z r = 0° i.e. the reflected ray would retrace its path turning through 180°. Mirror M2 has no effect.

48. (b): Let I be the length of the pipes and v the speed of sound.

Then frequency of open organ pipe of «th overtone is

v'n=(n + 1) v 21

and frequency of closed organ pipe of nth overtone is

The desired ratio is

•»„ = 2(n + 1) x>' In +1

49. (b): Optical source frequency,

v_ c _ 3 x 10 m s ⁴ X 1.3 x 10"6m

= 2.3 x 1014 Hz

Number of channels or subscribers - 2 3 X l° " = 1 . 1 5 x 1 0 "

20 x 103

50. (a): P r o c e s s 1 is i s o b a r i c (P = c o n s t a n t ) expansion.

Hence temperature of gas will increase.

A U1 = positive

Process 2 is an isothermal process.

A U2 = 0

Process 3 is an a d i a b a t i c e x p a n s i o n . Hence temperature of gas will fall.

A(J3 = negative .-. AUj > AU2 > AU3.

PHYSICS FOR YOU j MAY '13 4 7

^Oiigyl^CL^ ^ I Electrostatic s ^ w n i* i i > i in i i w—.••••> . • f

=

z^Y .

Superposition of H Electric Charges! . £ = ^(between two parallel •F i = F i2 + F i3 + Jr i4+ - + J: 1N S conductors)

• F = JF

12 + f

2

+2

r

F os0

[Electri c Potentia l | —

Here, F12, f13, F14 forces exerted • Potential difference Electric Potential Energy on charge q1 by the individual W \argesq2,q3...qNrespectively.V = y , Electric potential energy of .1 • Electric potential due to charge q at system of two point charges, distance r from it. \ „ n 1 q U = njA Coulomb's Law f~V= 4m~f4ne °ryi • Frvacium^ 1 M, • Electric potential atYpoint due to a •Electric potential energy of a • i-(vacuum) - ^ r r system of N point charges, 4JIEQ f r n • cji 1 q q1 1 pcos8 (j = —i_ £ . F(medium) = — -j-1r2 47te0 all pairs rjk , .0 r • Potential energy of an eleclric dipole in a uniform electric field, U=-pE (cos 62 — cos 9}) • If initially the dipole is perpendi- Quantisation of Charges - . cular to the field E, 0, = 90° and r-—— n . (Electrostatics) e2=e,then

; • q = ne 4—Electri c Charge s _

• Mass transferred during SM^^HM^^^^ . ^^"HBmmam^*^^ =-pb cost) = -p-h charging = mex n • If initially the dipole is parallel to the field Ev Bj = 0° and 02 = 6, then U=-pE( cosG-1)

rj Electri c Fiel d k

r/1m \ J= pE( 1-cose) _ c £ = - <7o * ~ jg r— Electric Fields of Point Charges Electric Flux and Gauss's Theory 1 1 .

£ =— t

• Electric flux through a plane surface areaS held in a uniform electric field47t£ o r , c_rc a • By the principle of superposition, electric field — coso due to a number of ^ointchar7 es • According to Gauss's theorem, the total electric flux through a closed surface S on _ _. _ . enclosing charge q,E = + E2 + E3 +...

dS ~ Eo Continuous Charge Distributions ——. • Flux density • Volume charge density da ,P =dV — Application of Gauss's Law 1 • Surface charge density dq • Electric field of a long straight wire of uniform linear charge density XCT ~ ^s £ = ——— • Linear charge density 271E 0r da where r is the perpendicular distance of the observation point from the wire. X = -jj- ^ • Electric field^of an infinite plane sheet of uniform surface charge density CT, . Force exerted on a charge % due to a continuous ^ = charge distribution • Electric field of two positively charged parallel plates with charge densities ctj and CT2 F = i—r such that c1>CT2>0, 4TO0Jr2 „ , 1 . , , ., „ . , , • Electric field due to continuous charge E = ±-—(0,+CT,) (Outside the plates) ,& 2e0K 1 2J t distribution £ = (Inside the plates)i = • Electric field of two equally and oppositely charged parallel plates, E=0 a (Foroutsidepoints) Dipole Moment, £ = ^ (For inside P°ints ) Dipole Field and Torque on a Dipole • Electric field of a thin spherical shell of charge density a and radius R, _ Ia • Dipole moment, p=^ x 2a E = For r > R (Outside points) • Dipole field at an axial point at distance r from 47te o r the centre of the dipole E = 0 Forr<R (Insidepoints) j2^ E = —— %r For r = R (At the surface)£axiai = 4£ , 2_ 2,2 4TO o R2 Whenr»a °° Here q=4n R2 a. 1 2p • Electric field of a solid sphere of uniform charge density p and radius Raxiai ~4neor3 * 1 q E = "~2 For r > R (Outside points) .Dipole field at an equatorial point at distance r 0r r from the centre of the dipole is E = For r < R (Inside points) 1 p 4f0 R ^equatorial = 4 22 3/2 £ = — •4 For r = R (At the surface) whenr»«} 4to0 R2 r= J F 4 3 equatorial ^ Here q = ~nR p . Torque, x = p£ sin 6. ,„„„

T H E LAWS OF C O N S E R V A T I O N 1.

2.

3.

4.

5.

Conservation of energy and conservation of mass are the corner stones of classical physics.

Demonstrate this in the formula.

v2-u2 = las This formula obeys (a) conservation of mass (b) conservation of energy (c) conservation of kinetic energy (d) conservation of potential energy A and B are two identical capacitors.

If A has a charge Q and now it is connected to B which has n o initial

charge. Compare the initial energy ^ g with the final energies of A and B.

Choose the correct statement.

(a) The initial energy is equal to the final energies of the capacitors.

(b) The initial energy is greater than the total final energy.

How much energy is lost on sharing the charges?

Two capacitors are connected as shown in the problem (2) to share charges. The laws which are followed here are

(a) law of conservation of charge (b) law of conservation of mass (c) law of conservation of energy (d) none of these

The essential differences between emission of X-rays and y-rays are

(a) Production of X-rays such as Kw Kt> rays are (b)

spontaneous.

X-rays are mostly by the excitation or ionisation of the inner level electrons.

(c) y-rays are spontaneously produced.

(d) The element is changed to another element in the production of X-rays.

6. In p+-decay of Na,

(a) electron is emitted (b) positron is emitted (c) a negative neutrino is generated

(d) a positive neutrino is generated (e) it is spontaneous emission.

Neutrinos are neutral particles of negligible mass.

The neutrino is emitted to conserve (a) mass

(b) charge (c) energy (d) none of these

The life-time of neutrinos are very small. Their mass is negligible compared to even the electrons.

H o w is it that one has discovered neutrinos produced by solar emissions w h e n the distance is very large?

We have seen that in higher physics, there are many conservation laws. H o w is it in classical physics we find that there is conservation of mass and energy?

S O L U T I O N S

(a,b) : (a) is true for classical physics. Mass, length and time do not change due to changes in energy in kinematics.

(b) v2 -u2 = las ...(i)

Multiplying by 1 m in equation (i) 1 2 1 2 1 „

=> —mv —mil =—mlas

1 1 1

=> Final K.E. - initial K.E. = Force * distance.

The difference in energy is the work done.

Conservation of energy is given in this formula.

(b): Let the initial charge be Q. As there is conservation of charges, the final charges are q on each capacitor = j

r • • , 1 Q2

Initial energy =

-1 q2 1 q2

The total final energy = — — + — —

=> Total energy =

The total final energy = (Q/2f

C

191

4 C

5 0 PHYSICS FOR YOU | MAY' 13

}

3. Initial energy = — —

Total final energy = — = — I '•' q = ^

6 > C 4C I 2 J

1 Q2 O2 1 Q2

Energy lost = — — - — = — — 4. (a,c): (a) Charge is conserved.

Contrary to general feeling, energy is also conserved. The initial energy before charges are shared is more than the sum of the final charges.

But the difference in energy is used to p u m p the charges from one charged capacitor to the uncharged one. (c) is true. This is just like water in a tank flowing to a tub on the ground. The higher potential energy of water makes it flow to the ground.

5. (b,c): Normally X-rays are produced by the ionisation or excitation to higher empty levels, of electrons from the inner shell (b).

Y-rays are spontaneously produced (c).

X-rays production does not change the element.

It is the production of y-rays that the element also changes.

For example, ® Co emits an electron ((3 ) producing 2gNi in excited level. This reaches the ground level by the emission of y-rays.

6. ( b , c , e ) : The P+ emission by 2 2 N a ,

2 2N a —» 2gNe + e+ +1>

Therefore (b) and (c) are correct.

This is a spontaneous reaction.

Therefore (e) is correct.

7. (c): The neutrino has negligible mass and no charge. It conserves energy in the reaction (c). In the emission \Y by Na, u is emitted and in emission of e~ by P, antineutrino is emitted. The difference between l> and v is in the direction of spin.

8. The detection of neutrinos is possible because

In document Physics for You - June 2013 (Gnv64) (Page 29-41)