Q162 / 170 · Manipal (MET) 2020 · Physics

A wire of resistance $5 \Omega$ is drawn out so that its length is increased by twice its original length, its new resistance is

  • A
    $\dot{4} 5 \Omega$
  • B
    $54 \Omega$
  • Correct answer
    C
    $20 \Omega$
  • D
    $5 \Omega$

Solution

Resistance of a wire is given by $R=\frac{\rho L}{A}$ where, $\rho=$ resistivity, $L=$ length of the wire $A=$ area of cross- section of the wire Equating initial and finial volumes of the wire, we can write $A L=A^{\prime} L^{\prime}=A^{\prime} \times(2 L) \Rightarrow A^{\prime}=\frac{A}{2}$ $\therefore \quad \frac{R^{\prime}}{R}=\frac{\rho L^{\prime} / A^{\prime}}{\rho L / A}=\left(\frac{L^{\prime}}{L}\right)\left(\frac{A}{A^{\prime}}\right)$ $=\left(\frac{2 L}{L}\right)\left(\frac{A}{A / 2}\right)=4$ $\Rightarrow \quad R^{\prime}=4 R=4(5)=20 \Omega$

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

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