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Indian National Physics Olympiad - 2012

Solutions

Please note that alternate/equivalent solutions may exist. Brief solutions are given below.

1. (a) i+1 =

7

13i +

6

13

v

r

(b) =v

r

Argument : Initially i increases until it reaches a value v = r, i.e. the speed

of the falling ball. Thereafter the ball merely touches the sphere and does notimpart it any momentum.

(c) i =v

r

1

7

13

ii=0,1,2,3,......

or i =v

r

1 7

13i1

i=1,2,3,......

(d) =v

r

2. (a) vy =R2

(2x + R)2vx

(b) See figure below:

vyvx

x0

0.5

1

0

0.25

R/2 R

(c) Speed = 2.22 km hr1

3. (a) =mag

R

( 1)

(b) For ma = 29.0 kgkmol1 ; = approx 10 Kkm1

(c) =

1

(d) approximately 30.0 km

(e) ps = ps0 exp

LmvR

1

Ts0 1

T

where Ts0 and ps0 are the initial points for the integration. A convenient choicewould be the triple point of water.

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(f) At zc atmospheric pressure should be equal to saturation pressure. Condition is

p0

T0 zc

T0

/1= ps0 exp

Lmv

R

1

Ts0

1

T0 zc

4. (a) Magnetic field = 0N I

lk < r

0 > r

Value of magnetic field =

1.26 102 T < r0 > r

where is the radial distance.

(b) L =0N

2r2

lValue of L = 1.97 102 H

(c) E = 3.95 J

(d) i =e

R(1 eRt/L) + i0e

Rt/L if i0 = 0

(e) e = iR + Ldi

dt i

Lv

l + vtwhere L = 0N

2r2/(l + vt)

(f) Electric field =

0Ni0

2lsin(t) < r

0Ni0r2

2lsin(t) > r

(g) The plot of E with radial distance:

r

0.04 V/m

Electricfield

radial distance

Lines of forces: Note, the lines of force are dense upto = r and increasingly sparsethereafter.

r

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5. (a) Since 0 < Eb, hence no ionisation by a single photon is possible.

(b) Speed of electron =

eF0m sin(t)

(c) Average kinetic energy =e2F204m2

(d) F0 = 1.5 103

V m1

(e) Potential energy = e2

40r+ eF0z

(f) See figure below:

V(z)

z0.5 0.5

0 1

1

21

0.22

0.22

19

(g) F0 =E20

e3

(h) approx 174 Vm1 which is physically possible.

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