Q99 / 125 · BITSAT 2021 · Physics

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
  • Correct answer
    B
    $1.21$ times, same
  • C
    both remain the same
  • D
    $1.1$ times, $1.1$ times

Solution

\(\mathrm{R}=\frac{\rho \mathrm{l}}{\mathrm{A}}\) 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.

Hence, the correct answer is option: 'B'.

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