A wire of 10 Ω resistance is stretched to thrice its original length.…

The correct answer is Option (3) → 90 Ω

Given:

Initial resistance, $R_1 = 10\,\Omega$

Final length, $L_2 = 3L_1$

Since resistance $R = \rho \frac{L}{A}$ and volume remains constant,

$A_1 L_1 = A_2 L_2 \Rightarrow A_2 = \frac{A_1 L_1}{L_2} = \frac{A_1}{3}$

Therefore,

$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{3L_1}{A_1/3} = 9\rho \frac{L_1}{A_1}$

$R_2 = 9R_1$

Substitute $R_1 = 10\,\Omega$:

$R_2 = 9 \times 10 = 90\,\Omega$

Final Answer: $R_2 = 90\,\Omega$