A table tennis ball which has been covered with a conducting paint is suspended by a silk thread so that it hangs between two metals plates. One plate is earthed. When the other plate is connected to a high voltage generator, the ball

Q. A table tennis ball which has been covered with a conducting paint is suspended by a silk thread so that it hangs between two metals plates. One plate is earthed. When the other plate is connected to a high voltage generator, the ball
(a) Is attached to the high voltage
(b) Hangs without moving
(c) Swings backward and forward hitting the plates
(d) Is repelled to the earthed plate and stays there

Ans: (C)

A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. The distance travelled by it before coming to rest

Q. A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. The distance travelled by it before coming to rest
(a) (2mu2)/qE
(b) (mu2)/qE
(c) (mu2)/2qE
(d) (mu2)/2qE
Sol: (d)
Retardation a = qE/m
v2 = u2 − 2as
0 = u2 − 2qE/m s
s = mu2/2qE

A charged particle of mass m and charge q is released from rest in a uniform electric field E. The kinetic energy of the particle after time t is

Q. A charged particle of mass m and charge q is released from rest in a uniform electric field E. The kinetic energy of the particle after time t is
(a)(2E2 t2)/mq
(b) (Et2 m)/(2t2 )
(c) (E2 q2 t2)/2m
(d) (E q m)/2t
Sol: (c)
a = F/m = qE/m
v = u + at = 0 + qEt/m
K = 1/2 mv2 = 1/2 m (qEt/m)2
= (q2 E2 t2)/2m

Charge Q is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are

Q.Charge Q is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are
(a) Q/2 and Q/2
(b) Q/2 and 3Q/4
(c) Q/2 and 2Q/3
(d) e and (Q-e)
Sol: (a)
Let charges are q, Q − q
F = k q(Q −q)/r2 = k(qQ − q^2)/r2
For F to be maximum
dF/dq = 0 ⇒ Q − 2q = 0 ⇒ q = Q/2
Charges are Q/2 , Q/2

Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6g/cc and the angle of contact of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is

Q: Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6g/cc and the angle of contact of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is
(A) 1 : 0.15
(B) 1 : 3
(C) 1 : 6.5
(D) 1.5 : 1

Solution: h =2Scosθ/ρgr
or S =hρgr/2cosθ
or S ∝ hρ/cosθ
Sw/SHg = h1/h2 × ρ12 × cosθ1/cosθ2
Putting the values, we obtain 1 : 6.5
Hence, (C) is correct.