KCET 2018 Physics Solved Paper

Given, charge q is accelerated through a potential difference V. Therefore,
‌

1

2

mv2=qV
⇒mv2=2qV
⇒(mv)2=2qmV
⇒mv=√2mqV
The radius of circular path followed by the charge is
r=‌

mv

qB

=‌

√2mqV

qB

⇒r∝√V
Thus, the potential difference required to move the charge in circle of radius 2r is 4 V