Question

A particle 'A' of charge q and particle B of charge 4q have identical mass m each. When allowed to fall from rest through same potential difference the ratio of their speeds v_A : v_B is-

• A. 2 : 1
• B. 1 : 2
• C. 4 : 1
• D. 1 : 4

Answer 1

Answer: (B)

1 : 2

Answer 2

DK = q(v_i – v_f)

$\frac { 1 } { 2 }$ $\mathrm { mv } _ { \mathrm { A } } ^ { 2 }$ = qV …(1)

$\frac { 1 } { 2 }$ $\operatorname { mv } _ { \mathrm { B } } ^ { 2 }$ = 4qV …(2)

\ $\left( \frac { \mathrm { v } _ { \mathrm { A } } } { \mathrm { v } _ { \mathrm { B } } } \right) ^ { 2 } = \frac { 1 } { 4 }$ Þ \ $\frac { v _ { \mathrm { A } } } { \mathrm { v } _ { \mathrm { B } } }$ = 1 : 2