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Physics Question 31 – JEE-MAIN 2026

A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings A to B is 2 and the ratio of their Young's moduli (YA/YB) is 0.5. The ratio of elongations in A to B is _______.

Identify the total force acting on each string due to the suspended masses.

Step 1: Determine the forces on each string✦ Active

String A supports both blocks, so the total force on string A is FA=(M+2M)g=3Mg.

String B supports only the lower block, so the force on string B is FB=2Mg.

Step 2: Recall the formula for elongation○ Expand

The elongation ΔL of a string is given by the formula derived from Young's Modulus: ΔL=FLAY, where F is the force, L is the original length, A is the cross-sectional area, and Y is Young's modulus.

Step 3: Calculate the ratio of elongations○ Expand

For string A, ΔLA=FALAAAYA=3MgLAAYA (since AA=A). For string B, ΔLB=FBLBABYB=2MgLBAYB (since AB=A). The ratio of elongations is:

ΔLAΔLB=3MgLAAYA2MgLBAYB=32LALBYBYA

Given LA/LB=2 and YA/YB=0.5. From YA/YB=0.5, we get YB/YA=1/0.5=2. Substituting these values:

ΔLAΔLB=32(2)(2)=6

Thus, the ratio of elongations in A to B is 6.

💡 Teacher's Secret Hint

Ensure correct identification of forces for each string and careful substitution of ratios.

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