Q98 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics

A conducting wire is stretched by applying a deforming force, so that its diameter decreases to $40 \%$ of original value. The percentage change in its resistance will be

  • A
    $0.9 \%$
  • B
    $0.12 \%$
  • Correct answer
    C
    $1.6 \%$
  • D
    $0.5 \%$

Solution

On stretching, volume of wire remains constant. So, $V=A l$ or $l=\frac{V}{A}$ $\begin{array}{ll} \therefore & R=\rho \cdot \frac{l}{A}=\frac{\rho V}{A^2}=\frac{\rho V}{\frac{\pi^2 D^4}{16}}=\frac{16 \rho V}{\pi^2 D^4} \\ \therefore & \frac{\Delta R}{R}=-4 \frac{\Delta D}{D}=-4(-0.4)=1.6 \% \end{array}$

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

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