Q132 / 180 · JIPMER 2016 · Physics

The length of a given cylindrical wire is increased by $150 \%$. Due to the consequent decrease in diameter the change in the resistance of the wire will be

  • A
    200%
  • Correct answer
    B
    525%
  • C
    300%
  • D
    400%

Solution

If suppose initial length, $I_1=100$, then $I_2=100+150=250$ $\because \quad \frac{R_1}{R_2}=\left(\frac{I_1}{I_2}\right)^2=\left(\frac{100}{250}\right)^2$ $\Rightarrow \quad \frac{R_1}{R_2}=\frac{4}{25} \Rightarrow R_2=\frac{25}{4} R_1$ $\begin{aligned} \Rightarrow \frac{\Delta R}{R} \times 100 & =\frac{R_2-R_1}{R_1} \times 100 \\ & =\frac{\frac{25}{4} R_1-R_1}{R_1} \times 100=525 \%\end{aligned}$

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

Relevant Questions