Question

A wire of resistance $9 \Omega$ is uniformly stretched such that.its radius becomes $1/3$ of the original radius. Ratio of the new to old resistance is

• A. 3
• B. 9
• C. 27
• D. 81

Answer 1

Answer: (D)

81

Answer 2

Here old resistance, $R_i = 9 \Omega$
Also, $R = \rho \frac{l}{A} = \rho \frac{l^2}{V} \, \, \, (\because \, \, V = Al)$
Now, new resistance , $R_f = ?$
$r' = \frac{1}{3}r or A' =\frac{1}{9}A$
$V' = V or A' l' = Al$
or $\frac{1}{9} Al' = Al or l' = 9 l$
$\therefore \, R_{f} = \frac{\rho l'^{2}}{V'} = \frac{\rho \left(9l\right)^{2}}{V} = 81 \frac{\rho l^{2}}{V} = 81 R_{i}$
$\therefore \, \frac{R_{f}}{R_{i}} = 81$