Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force, the wire is elongated such that its length becomes $3 L$. Assuming, the resistivity and density of the material is unchanged, the resistance of the elongated wire is

• A

  $40 \Omega$
• Correct answer
  B

  $45 \Omega$
• C

  $50 \Omega$
• D

  $48 \Omega$

Solution

Resistance of wire of length $L, R=5 \Omega$ If area of cross-section be $A$, then $$ R=\rho \frac{L}{A} \quad(\rho=\text { resistivity of the wire }) $$ After applying external force new length becomes, $L^{\prime}=3 L$ $$ \begin{array}{ll} \text { As, density, } \sigma & =\frac{m}{V}=\frac{m}{L \times A} \\ \Rightarrow \quad & A=\frac{m}{L \sigma} \\ \therefore & R^{\prime}=\rho \frac{L^{\prime}}{A^{\prime}}=\rho \times \frac{3 L \times L^{\prime} \sigma}{m}=\frac{3 \rho \sigma L \times 3 L}{m} \\ \Rightarrow & R^{\prime}=\frac{9 L^2 \rho \sigma}{m} \end{array} $$ Also, $\quad R=\rho \cdot \frac{L}{A}=\rho \frac{L L \sigma}{m}$
$=9 \times 5 \Omega=45 \Omega$

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