Test Index

JEE Main 4 April 2026 Shift 1 Paper

Section: Physics

© examsnet.com

Question : 9 of 75

Marks: +1, -0

A string A of length 0.314 m of 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)

[JEE Main 4 Apr 2026 Shift 1]

3
2
1.9
1

Solution:

\Delta \ell \Rightarrow \Delta \ell_{1} + \Delta \ell_{2}

\Rightarrow \; \frac{mg \ell}{AY} + \; \frac{mg (2 \ell)}{A (2 Y)} = \; \frac{2 mg \ell}{AY}

= \; \frac{2 \times 0.8 \times 10 \times (0.314)}{3.14 \times (2 \times 10^{-4})^{2} \times 2 \times 10^{10}} = 2 \text{mm}

© examsnet.com

Go to Question:

Prev Question

Next Question