First slide

Ohm's law

Question

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

Moderate

A

1.2 times, 1.1 times
B

1.2 L times, same
C

both remain the same
D

1.1 times, 1.1 times

Solution

Specific resistance $(ρ)$ will remain same as it is the intrinsic property of material.
Original resistance of the wire, $R = \frac{ρl}{A}$
$\begin{array}{l} or R = \frac{l}{A} or R \propto \frac{l^{2}}{V} (as V = Al) \\ and R^{'} \propto \frac{{(l + 10 % l)}^{2}}{V} \\ Therefore, \frac{R^{'}}{R} = \frac{{(l + \frac{10}{100} l)}^{2}}{l^{2}} \\ or \frac{R^{'}}{R} = \frac{\frac{{(11 l)}^{2}}{10}}{l^{2}} = \frac{121}{100} \\ or R^{'} = 1 .21 R \end{array}$

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