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A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively:
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Verified Answer
The correct answer is:
$1.21$ times, same
\(\mathrm{R}=\frac{\rho \mathrm{l}}{\mathrm{A}}\)
Now, \(1=1+\frac{1}{10}=\frac{111}{10}\)
and therefore, \(\mathrm{A}=\frac{10 \mathrm{~A}}{11}\)
So \(\mathrm{R}^{\prime}=\frac{\mathrm{p} \times\left(\frac{111}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho \mathrm{l}}{\mathrm{A}} \times \frac{(11)^{2}}{(10)^{2}}=1.21 \mathrm{R}\)
Now resistance becomes \(1.21\) times of initial and specific resistance is the intrinsic property so remains same.
Now, \(1=1+\frac{1}{10}=\frac{111}{10}\)
and therefore, \(\mathrm{A}=\frac{10 \mathrm{~A}}{11}\)
So \(\mathrm{R}^{\prime}=\frac{\mathrm{p} \times\left(\frac{111}{10}\right)}{\frac{10 \mathrm{~A}}{11}}=\frac{\rho \mathrm{l}}{\mathrm{A}} \times \frac{(11)^{2}}{(10)^{2}}=1.21 \mathrm{R}\)
Now resistance becomes \(1.21\) times of initial and specific resistance is the intrinsic property so remains same.
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