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Mechanical Properties of Solids

JEE Main level, single correct, medium

14 of 118

Question 14 of 118

A string $A$ of length $0.314$ m and Young's modulus $2 \times 10^{10}$ 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 $= 10$ m/s $^2$ ) (Mass of both strings is to be neglected as compared to the mass of load)

Show answer and solution

Answer: (B) $2$

Given:
Length of string A,

L_A = 0.314

m
Young's modulus of string A,

Y_A = 2 \times 10^{10}

N/m

^2

Radius of both strings,

r = 0.2

= 2 \times 10^{-4}

m
Length of string B,

L_B = 2L_A = 0.628

m
Young's modulus of string B,

Y_B = 2Y_A = 4 \times 10^{10}

N/m

^2

Mass of load,

m = 0.8

kg
Acceleration due to gravity,

g = 10

m/s

^2

Tension in both strings,

T = mg = 0.8 \times 10 = 8

N

Cross-sectional area of both strings:

A = \pi r^2 = 3.14 \times (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8}

^2

Elongation in string A:

\Delta L_A = \dfrac{T L_A}{A Y_A}

\Delta L_A = \dfrac{8 \times 0.314}{3.14 \times 4 \times 10^{-8} \times 2 \times 10^{10}}

\Delta L_A = \dfrac{8 \times 0.314}{8 \times 3.14 \times 10^2} = \dfrac{0.1}{10^2} = 10^{-3}

= 1

mm

Elongation in string B:

\Delta L_B = \dfrac{T L_B}{A Y_B}

\Delta L_B = \dfrac{8 \times (2 \times 0.314)}{A \times (2 Y_A)} = \dfrac{T L_A}{A Y_A} = 1

mm

Net change in length of the combination:

\Delta L = \Delta L_A + \Delta L_B = 1 + 1 = 2

mm

Answer:

2

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JEE Main 2025 Apr 2 Shift 1 question paper

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