l
r (b) l l πε
1 2
0 0
4 ln −1
l r (c) l l
πε
1 2
0 0
2 ln +1
l
r (d) l l πε
1 2
0 0
2 ln −1
l r
OLYMPIAD
PROBLEMS
6. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of
the outer surface of hollow shell be V. What will be the new potential difference between the same two surfaces if the shell given a charge –3Q?
(a) V (b) 2V
(c) 3V (d) 4V
7. Non-relativistic protons move rectilinearly in the region of space where there are uniform mutually perpendicular electric and magnetic fields. The trajectory of the protons lie in the plane X-Z as shown in figure and forms an angle f with X-axis.
Find the pitch of the helical trajectory along which the protons will move after the electric field is switched off.
(a) 2π mE2
qB (b) 2π 2
mE f qB sin (c) 2πmE2 f
qB cos (d) 2πmE2 f qB tan
8. A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 5 cm. A vertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20 W, the wire PQ starts sliding on the rails. Find the coefficient of friction.
(a) 0.12 (b) 0.50
(c) 0.36 (d) 0.78
solutions 1. (d):
M gw = 108g
Mg=ρsV g0 = 109g ρw 0V g
rwV g M g Mg0 = w + rwV g0 =108g+109g
rwV0= ×11 108 ...(i) Condition when temperature rised from 10°C to 15°C.
r′wV g0′ = ′ ′ +rsV g mg0
r′ ′ =wV g0 109g mg+ ...(ii)
Now, r r
γ γ
′ =w + w ′ = +
w
T V V s T
( ), ( )
1 D 0 0 1 D
γw = ×2 10−4 °C−1,γs =3αs = ×3 1 2 10. × −5°C−1 DT = 5°C,
\ r γ
γ
w s
w
V T
T m
0 1 9
1 ( ) 10
( )
+
+ D = +
D ...(iii)
Putting rwV0 from eqn. (i) in (iii), we get
11 10 1
18 109
× × +
+ ( )= +
( )
γ γw s
T
T D m
D
⇒ 11 10 1 5 3 6 10
1 5 2 10 10
8 5
4 9
× × + × ×
+ × × −− = +
[ . ]
[ ] m
⇒ 1 × × + ×
+ × − − = +
1 10 1 1 8 10
1 10 108 ( . 4 4) 109
( ) m
⇒ 11 × 108 + 1.8 × 11 × 104 = 109 + 106 + m + m × 10–3 or 108 – 80.2 × 104 = m (1 + .001)
or 104 (10000 – 80.2) = m (1.001)
\ m = 99098901.1
So at this position, mass will be 99098901.1 g. So finally reduction in the mass will be
= 108 – 99098901.1 = 901098.9 g = 9.01 × 105 g ++++++++
++++++
Shell Sphere
a Q b
2. (c) : Time interval between low tide and high tide is half the time period of oscillation. Interval between 11:20 AM and 5:40 PM is 380 minute.
Time period = 760 minute. Amplitude of oscillation is 3 m. Centre of oscillation is at a height 7 m above the bar.
Water is at half tide point after an interval of T/4 = 190 minute, i.e., at 2:30 PM.
x = a sin wt
where, a = 3 m, w= 2π = π 760 380.
The ship can cross the bar when the depth is 9.4 m, i.e., x = 2.4 m
\ 2.4 = 3 sin wt or sin wt = 0.8
⇒ t = 380π sin ( . )−1 0 8 = 107.1 minute.
Hence the earliest time is 2:30 PM + 1 h 48 min, i.e., at 4:18 PM.
3. (d) : Taking height of ionospheric layer H, length of direct path on the Earth’s surface D, the path difference between two routes is
Ionosphere
Transmitter Earth's
surface Receiver
H H2+D2/ 4
D
Dx= H +D D
2 − 4
2 2 1 2/
...(i)
The fluctuation in intensity takes place due to interference between the two signals arriving at the receiver. Each time path difference Dx changes by l (wavelength of the radiation), the received signal strength will vary through one cycle, thus the frequency of the observed fluctuation is
f d x dt c
d x
=1 = dt l
( )D u ( )D ...(ii) where u is the frequency of the radiation and c is speed of light.
Differentiating the expression for path difference Dx with respect to time gives
d x
dt H H D dH
dt ( )D = + /
−
2 2 4
2 1 2
...(iii) From eqns. (ii) and (iii),
The fluctuation frequency, f Hv
c H D
= +
2 −
4
2 2 1 2
u / ...(iv)
where v dH
= dt is the vertical velocity of the layer.
On solving the eqn. (iv) for v, we have v fc
H H D
= +
2 4
2 2 1 2
u
/
= ×
× × × × + ×
( / )(8 60 3 10) ( ) ( ) / 2 2 10 10 2 10 5 10
4
8
5 7 5 2 5 2 1 2
= 3.2 m s–1
4. (a) : At the point P, point of intersection of streams, the range of the two streams are same. As the ranges of liquid issuing from a depth h from the surface and from the base is same, hence
CD = AB = h
or y = H + h + h = H + 2h Time taken by water coming out of C to reach P,
t = 2h g Velocity of water coming out of C,
v = 2g H h( + ) Hence required value of x = vt
= 2h × 2 + =2 + g g H h( ) h H h( ) 5. (c) : Electric field near a long wire is given by
E= l r πε 2 0
The second wire lies in the non-uniform field of first wire. Each element of second wire experiences
Wire 1
l λ1
r x
0
Wire 2 dx
“Have the courage to follow your heart and intuition. They somehow know what you truly want to become.”
Steve Jobs
different magnitude of field. Therefore we consider a differential element dx, charge dQ = l2dx, at a distance x from the long wire. The force acting on this element dF is
dF = EdQ
The force acting on each element depends on x, the 2 separation between wires 1 and 2.
Integrating the expression for dF in the limits x = r0 to x = r0 + l, we obtain shell respectively, potential at their surfaces will be,
Vsphere = 1
And so according to given problem
V = V V Q potential at its surface and also inside will change by
V0 = 1
i.e., if any charge is given to external shell, the potential difference between sphere and shell will not change. This is because by presence of charge on outer shell, potential everywhere inside and on the surface of shell will change by same amount and hence potential difference between sphere and shell will remain unchanged.
7. (d) : When electric field is switched off, the path followed by the particle will be helical.
Pitch = v||T ...(i) where v|| =vcos (90° – f) = vsin f Hence, from eq. (i)
Pitch = ( sin )v m f 2qBπ
...(ii)
When both fields are present qE = qv Bcosf
or v E
=B
cosf ...(iii)
Substituting the value of v from eq. (iii) in eq. (ii), we get on the wire overcomes friction force.
mN = F ⇒ mmg = IlB
1. A coastguard ship locates a pirate ship at a distance 560 m. It fires a cannon ball with an initial speed 82 m s–1. At what angle from horizontal the ball must be fired so that it hits the pirate ship?
(Take g = 10 m s–2)
(a) 54° (b) 125° (c) 28° (d) 18°
2. 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) 1
2g t t(1+ 2)2 (b) gt1t2 (c) 1
8g t t(1+ 2)2 (d) 1 2gt t1 2
3. A hollow cylinder has a charge q coulomb within it.
If f is electric flux in units of volt meter associated with the curved surface B, the flux linked with the plane surface A in units of volt meter will be
A B
C
(a) q
2e0 (b) f
3 (c) q
e f
0−
(d) 1 2
q
e f
0 −
4. A thin metal disc of radius 0.25 m and mass 2 kg starts from rest and rolls down an inclined plane. If its rotational kinetic energy is 4 J at the foot of the inclined plane, then its linear velocity at the same point is (a) 1.2 m s–1 (b) 2 2 m s−1 (c) 20 m s–1 (d) 2 m s – 1
5. Two radioactive elements X and Y have half lives of 50 minutes and 100 minutes respectively. Initially
both of them have same numbers of atoms. After 200 minutes what is the value of following fraction ? Number of atoms of X unchanged/Number of atoms of Y unchanged.
(a) 4 (b) 2 (c) 1
2 (d) 1
6. For the circuit shown in figure, the current through 4 the inductor is 0.9 A. While the current through the capacitor is 0.4 A. The current drawn from the generator is
(a) I = 1.13 A
L C
(b) I = 0.9 A (c) I = 0.5 A (d) I = 0.6 A
7. A boy of mass 50 kg is standing on a weighing machine placed on the floor of a lift. The machine reads his weight in N. What is the reading of the machine if the lift is moving upwards with a uniform speed of 10 m s–1? (Take g = 10 m s–2)
(a) 510 N (b) 480 N (c) 490 N (d) 500 N 8. Two cells A and B are connected in the secondary
circuit of a potentiometer one at a time and the balancing length are respectively 400 cm and 440 cm. The emf of the cell A is 1.08 V. The emf of the second cell B is
(a) 1.08 V (b) 1.188 V (c) 11.88 V (d) 12.8 V
9. A simple pendulum is executing SHM with a period of 6 s between two extreme positions B and C about a point O. If the length of the arc BC is 10 cm, how long will the pendulum take the move from position C to a position D towards O exactly midway between C and O.
(a) 0.5 s (b) 1 s (c) 1.5 s (d) 3 s
10. The critical angle of a transparent crystal is 45°.
Then its polarising angle is Examon
5th June