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)

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