Question

A body of mass $1gm$ and carrying a charge ${10}^{-8}C$ passes from the point P to Q which are at electric potentials $600V$ and $0V$ respectively. The velocity of the body at Q is $20cm/sec$. Its velocity in m/sec at 'P' is:

• A. $\sqrt{0.028}$
• B. $\sqrt{0.056}$
• C. $\sqrt{0.56}$
• D. $\sqrt{5.6}$

Answer 1

The correct option is A $\sqrt{0.028}$

Given P at $600V$ & Q at $0V$

$v_q=20cm/s$

we know change in potential energy $u=q\Delta v$

Using law of conservation of energy

$U_1+K.E_1=U_2+K.E_2$

$\Rightarrow 600\times {10}^{-8}+\frac{1}{2}m{v}^2=0\times {10}^{-8}-\frac{1}{2}m(20\times {10}^{-2}{)}^2$

$\Rightarrow 6\times {10}^{-6}+\frac{1}{2}\times 1\times {10}^{-3}\times v_p^2=\frac{1}{2}1\times {10}^{-3}\times 4^2\times {10}^{-2}$

$\Rightarrow \frac{{10}^{-3}\times u{p}^2}{2}=2\times {10}^{-5}-6\times {10}^{-6}$

$\Rightarrow v_p^2=(2\times {10}^3)\times 1.4\times {10}^{-5}$

$\Rightarrow v_p^2=20.8\times {10}^{-2}$

$\Rightarrow v_p=\sqrt{0.028}$