Question

A wire of resistance $10\Omega$ is stretched such that its area becomes $\frac{1}{3}$rd and its volume remains constant. What will be the new resistance?

• A. $40\Omega$
• B. $60\Omega$
• C. $90\Omega$
• D. $30\Omega$

Answer 1

The correct option is C $90\Omega$

When the area of a wire is changed keeping its volume constant then:
$\frac{R_1}{R_2}=(\frac{A_2}{A_1}{)}^2\Rightarrow \frac{10}{R_2}=(\frac{\frac{A}{3}}{A}{)}^2=\frac{1}{9}$
$R_2=90\Omega$