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JEE Main 20264 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2026 · 4 April · Morning Shift · Physics · Mechanical Properties of Solids

A string A of length 0.314 m and Young's modulus 2 × 10^ 10 N/m ^2...

A string AA of length 0.3140.314 m and Young's modulus 2×10102 \times 10^{10} N/m2^2 is connected to another string BB of length and Young's modulus both twice of those of AA. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.80.8 kg. The net change in length of the combination is _____ mm. (radius of both the strings is 0.20.2 mm and acceleration due to gravity =10= 10 m/s2^2) (Mass of both strings is to be neglected as compared to the mass of load)

Options

  1. A33
  2. B22
  3. C1.91.9
  4. D11

Correct answer

B. 22

Step-by-step solution

Given:
Length of string A, LA=0.314L_A = 0.314 m
Young's modulus of string A, YA=2×1010Y_A = 2 \times 10^{10} N/m2^2
Radius of both strings, r=0.2r = 0.2 mm =2×10−4= 2 \times 10^{-4} m
Length of string B, LB=2LA=0.628L_B = 2L_A = 0.628 m
Young's modulus of string B, YB=2YA=4×1010Y_B = 2Y_A = 4 \times 10^{10} N/m2^2
Mass of load, m=0.8m = 0.8 kg
Acceleration due to gravity, g=10g = 10 m/s2^2

Tension in both strings, T=mg=0.8×10=8T = mg = 0.8 \times 10 = 8 N

Cross-sectional area of both strings:
A=πr2=3.14×(2×10−4)2=3.14×4×10−8A = \pi r^2 = 3.14 \times (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8} m2^2

Elongation in string A:
ΔLA=TLAAYA\Delta L_A = \dfrac{T L_A}{A Y_A}
ΔLA=8×0.3143.14×4×10−8×2×1010\Delta L_A = \dfrac{8 \times 0.314}{3.14 \times 4 \times 10^{-8} \times 2 \times 10^{10}}
ΔLA=8×0.3148×3.14×102=0.1102=10−3\Delta L_A = \dfrac{8 \times 0.314}{8 \times 3.14 \times 10^2} = \dfrac{0.1}{10^2} = 10^{-3} m =1= 1 mm

Elongation in string B:
ΔLB=TLBAYB\Delta L_B = \dfrac{T L_B}{A Y_B}
ΔLB=8×(2×0.314)A×(2YA)=TLAAYA=1\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:
ΔL=ΔLA+ΔLB=1+1=2\Delta L = \Delta L_A + \Delta L_B = 1 + 1 = 2 mm

Answer: 22
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