A wire of a certain material is stretched slowly by 10 percent. Its new resistance and specific resistance become respectively

Solution :
After stretching, specific resistance $$\left( \rho \right)$$  will remain same.
Original resistance of wire, $$R = \frac{{\rho l}}{A}$$
Ratio of resistance before and after streching,
$$\eqalign{ & \frac{{{R_2}}}{{{R_1}}} = \frac{{{l_2}}}{{{l_1}}} \times \frac{{{A_1}}}{{{A_2}}} = {\left[ {\frac{{{l_2}}}{{{l_1}}}} \right]^2} \cr & \frac{{{R_2}}}{{{R_1}}} = \frac{{{{\left( {l + 10\% l} \right)}^2}}}{{{l^2}}} \cr & \frac{{{R_2}}}{{{R_1}}} = \frac{{{{\left( {\frac{{11}}{{10}}l} \right)}^2}}}{{{l^2}}} \cr & {R_2} = 1.21\,{R_1} \cr} $$