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 $P \times 2$
$=10 \times 2=20\, \Omega$