MHT CET 2021 Physics Solved Paper

(d) Given, mass of both particle =m
Charge of particle, A=+q
Charge of particle, B=+4q
Potential difference =V
Kinetic energy is given by
K=‌

1

2

mv2‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
This energy is equal to electrostatic potential energy is
K=V×Q‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
From Eqs. (i) and (ii), we have
‌

1

2

mv2=V×Q
For particle A,
qV=‌

1

2

mvA2‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)
For particle B,
4qV=‌

1

2

mvB2‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv)
Dividing Eq. (iii) by Eq. (iv), we get
‌

1

4

‌‌=‌

vA2

vB2

⇒‌‌‌

vA

vB

‌‌=‌

1

2

Hence, the ratio of their speed ‌

vA

vB

=‌

1

2