A string A of length 0.314 m and Young's modulus 2 × 10¹⁰ N/m² is… | JEE Main 2026 (4 Apr, Shift 1)

Q2JEE Main 2026 · 4 APR · SHIFT 1Medium

Wire Combinations & Special Cases

A string $A$ of length 0.314 m and Young's modulus $2 \times 10^{10}\text{ N/m}^2$ is connected to another string $B$ of length and Young's modulus both twice of those of $A$ . This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg. The net change in length of the combination is ______ mm.

(radius of both the strings is 0.2 mm and acceleration due to gravity $= 10\text{ m/s}^2$ ) (Mass of both strings is to be neglected as compared to the mass of load)

Show answer and solution

Answer: (b)

Both strings carry $F = 0.8 \times 10 = 8\text{ N}$ ; $A = \pi (0.2 \times 10^{-3})^2 = 1.256 \times 10^{-7}\text{ m}^2$
$\Delta l_A = \frac{FL}{AY} = \frac{8 \times 0.314}{1.256 \times 10^{-7} \times 2 \times 10^{10}} = 10^{-3}\text{ m} = 1\text{ mm}$
For B, $L$ and $Y$ are both doubled, so $\frac{L}{Y}$ is unchanged and $\Delta l_B = 1\text{ mm}$ .
Total $= 2\text{ mm}$ , so the answer is (b).

Report a mistake

Wire Combinations & Special Cases question from JEE Main 2026 (4 Apr, Shift 1)

More questions on Wire Combinations & Special Cases