Q. A wire $P$ has a resistance of $20\Omega$. Another wire $Q$ of same material but length twice that of $P$ has resistance of $8\Omega$. If $r$ is the radius of cross-section of $P$, the radius of cross-section of $Q$is
A) $r$
B) $\frac{r}{\sqrt{2}}$
C) $\sqrt{5r}$
D) $2r$
Explanation
(c) Resistance, R$= \frac{\rho l}{A}$
For wire P, $20 = \frac{\rho l}{\pi r^{2}}\ \ \ \ \ \ \ldots(i)$
Similarly for wire Q,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8 = \frac{\rho(2l)}{\pi\left( r^{'} \right)^{2}}\ \ \ \ \ \ldots.(ii)$
Dividing Eq.(i) by Eq. (ii), we have
$\ \ \ \ \frac{20}{8} = \frac{\rho l}{\pi r^{2}} \times \frac{\pi\left( r^{'} \right)^{2}}{\rho(2l)}$
$\Longrightarrow 5 = \left( \frac{r^{'}}{r} \right)^{2}$
$\Longrightarrow r^{'} = \sqrt{5}r$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.