Question

A cylindrical wire $P$ has resistance $10\Omega$. A second wire $Q$ has length and diameter half that of $P$. If the material of both the wires is same, then resistance of wire $Q$ is

• A. $10\Omega$
• B. $20\Omega$
• C. $5\Omega$
• D. $\frac{5}{2}\Omega$

Answer 1

Answer: (B)

Given, length of wire, $P=l$
Length of wire, $Q=\frac{l}{2}$
Diameter of wire, $P=d$
Diameter of wire, $Q=\frac{d}{2}$
Resistance $R=\rho \cdot \frac{l}{A}=\rho \cdot \frac{l}{\pi d^2/4}$
$\frac{\text{ Resistance of }Q}{\text{ Resistance of }P}=\left(\frac{\rho \cdot \frac{l}{2}}{\frac{\pi }{4}{\left(\frac{d}{2}\right)}^2}\right)/\frac{\rho l}{\frac{\pi }{4}{d}^2}=\frac{\frac{1}{2}}{\frac{1}{4}}=\frac{4}{2}=\frac{2}{1}$
$\Rightarrow$ Resistance of $Q$
$=$ Resistance of $P\times 2$
$=10\times 2=20\Omega$