UPSEE Solved Model Paper 2

Refer to the figure. Let v be the velocity of the particle. Its kinetic energy is

1

2

mv2 = qV or v = (

2qV

m

)

1

2

(1)

The particle follows a circular path from A to B of radius r which is given by

mv2

r

= qvB or r =

mv

qB

(2)
Using (1) and (2), we have
r =

m

qB

(

2qV

m

)

1

2

=

1

B

(

2mV

q

)

1

2

In triangle BCD, sin θ =

BD

BC

=

d

r

. Therefore,
sin θ = Bd(

q

2mV

)

1

2

, which is choice (1).