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 = 2 l $ ....(i)
New length of wire $I'= I + \Delta l = 3 l$ (Using (i))
or, $R' = \frac{\rho l'^2}{V} = \frac{\rho(3l)^2}{V} = 9 \times 5 = 45 \, \Omega$