Question

A wire of resistance 5 ohms is stretched such that longitudinal strain is $200\%$.The new resistance in ohms is

• A. 10
• B. 25
• C. 30
• D. 45

Answer 1

Answer: (D)

Here, $R=5\Omega$,
or $R=\frac{\rho l}{A}=\frac{\rho l^2}{V}(\because V=Al)$
On stretching, longitudinal strain,
$\frac{\Delta l}{l}=200\%=2$
$\Delta l=2l$ ....(i)
New length of wire $I^{\prime }=I+\Delta l=3l$ (Using (i))
or, $R^{\prime }=\frac{\rho l^{\prime 2}}{V}=\frac{\rho (3l{)}^2}{V}=9\times 5=45\Omega$