Question

Two charged particles A and B are allowed to fall from rest through the same potential difference. The mass of A is twice that of B, and the charge on particle B is 2 times that of A. What is the ratio of the speed of A and B?
1. 1 ∶ 2
2. 1 ∶ 4
3. 1 ∶ 1
4. 1 ∶ 3

Answer 1

Correct Answer - Option 1 : 1 ∶ 2

Concept:

The relationship between the charge and velocity of the charged particle is given by

\(qV = \frac{1}{2}m{v^2}\)................(i)

Where q = charge of the particle

V = Potential difference 

v = speed of charge

m = mass of charge

Calculation:

According to the question

Mass of charge particle B, m_B is m

then, the mass of charged particle A, m_A is 2m

Charge of charge particle A, q_A is q

then, the charge of charged particle B, q_B is 2q

Using equation (i)

For charge particle A, \({q_A}{V_A} = \frac{1}{2}{m_A}v_A^2\)

\(q{V_A} = \frac{1}{2} \times 2m \times v_A^2\)................(ii)

Using equation (i)

For charge particle B, \({q_B}{V_B} = \frac{1}{2}{m_B}v_B^2\)

\(2q{V_B} = \frac{1}{2} \times m \times v_B^2\)................(iii)

It is given that V_A = V_B

On solving equation (ii) and (iii)

\(\frac{{{V_A}}}{{{V_B}}} = \frac{{4v_A^2}}{{v_B^2}}\)

\(\frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{1}{4}} = \frac{1}{2}\)