JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, maximum value of accelerating potential,
V = 12 kV = 12000 V,
Speed achieved by a proton cyclotron, v=

c

6

where, c is the speed of light =3×108ms−1
Mass of proton, mp=1.67×10−27‌kg
Charge of proton, qp=1.6×10−19C
According to the question, assume a proton makes n revolutions in the cyclotron to achieve the speed v.
The energy absorbed by proton due to the accelerating potential provided by the radio oscillator
=n×2×qp×V
This energy absorbed by proton will be equal to the kinetic energy gain by the proton i.e.,
2nqpV=

mpv2

2

Substituting the given values in above expression, we get
n×2×1.6×10−19×12000
=

1

2

×1.67×10−27×(

3×108

6

)2
n×3.84×10−15=2.0875×10−12
n = 543.62
As, n is number of revolutions, so it should be an integer.
∴ n = 543