JEE Main 30 June 2022 Shift 1 Solved Paper

Given,
f=10×106‌Hz
r=0.6m
Charge on proton (q)=e
We know,
Radius (r)=‌

mv

qB

=‌

√2mk

eB

Also, we know,
Cyclotron oscillation frequency should be equal to the pendulum evolution frequency.
∴f=‌

eB

2πm

⇒eB=2πmf
Putting this value of eB in equation (1) we get
r=‌

√2mk

2πmf

⇒r2=‌

2mk

4π2m2f2

⇒k=2π2mf2r2
=2×(‌

22

7

)2×1.67×10−27×(10×106)2×(0.6)2J
=1.2×10−12J
=‌

1.2×10−12

1.6×10−19

‌eV
=‌

12

16

×107‌eV
=0.75×107‌eV
=7.5×106‌eV
=7.5‌MeV