Question

A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively:

• A. 1.2 times, 1.3 times
• B. 1.21 times, same
• C. both remain the same
• D. 1.1 times, 1.1 times

Answer 1

Answer: (B)

$<br/>R=\frac{\rho l}{A}<br/>$
Now, $1=1+\frac{1}{10}=\frac{111}{10}$
and therefore, $A=\frac{10A}{11}$
So ${}^{\prime }=\frac{p\times \left(\frac{111}{10}\right)}{\frac{10A}{11}}=\frac{\rho l}{A}\times \frac{(11{)}^2}{(10{)}^2}=1.21R$
Now resistance becomes $1.21$ times of initial and specific resistance is the intrinsic property so remains same.