Question

A mass of 1 g carrying charge q falls through a potential difference V. The kinetic energy acquired by it is E. When a mass of 2 g carrying the charge q falls through a potential difference V. What will be the kinetic energy acquired by it

• A. 0.25 E
• B. 0.50 E
• C. 0.75 E
• D. E

Answer 1

Answer: (B)

0.50 E

$E=q{V}_P\left(\begin{array}{c}q=che\\ V_p=potentialdifference\end{array}\right)$

In this momentum is constant

By conservation of linear momentum

$\therefore mv=constant$

$m\propto \frac{1}{v}$

$KE=\frac{1}{2}m{v}^2$

$\Rightarrow KE\propto m{\left(\frac{1}{m}\right)}^2$

$KE\propto \frac{1}{m}$

So,

$\begin{array}{c}\frac{E_e}{E}=\frac{1}{d}\\ Hence,\\ E_2=0.5E\\ So,optionBiscorrectanswer.\end{array}$