Q59 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
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)
- A$3$
- Correct answerB$2$
- C$1.9$
- D$1$
Solution
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, $r = 0.2$ mm $= 2 \times 10^{-4}$ 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:
$A = \pi r^2 = 3.14 \times (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8}$ m$^2$
Elongation in string A:
$\Delta L_A = \dfrac{T L_A}{A Y_A}$
$\Delta L_A = \dfrac{8 \times 0.314}{3.14 \times 4 \times 10^{-8} \times 2 \times 10^{10}}$
$\Delta L_A = \dfrac{8 \times 0.314}{8 \times 3.14 \times 10^2} = \dfrac{0.1}{10^2} = 10^{-3}$ m $= 1$ 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: $2$
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, $r = 0.2$ mm $= 2 \times 10^{-4}$ 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:
$A = \pi r^2 = 3.14 \times (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8}$ m$^2$
Elongation in string A:
$\Delta L_A = \dfrac{T L_A}{A Y_A}$
$\Delta L_A = \dfrac{8 \times 0.314}{3.14 \times 4 \times 10^{-8} \times 2 \times 10^{10}}$
$\Delta L_A = \dfrac{8 \times 0.314}{8 \times 3.14 \times 10^2} = \dfrac{0.1}{10^2} = 10^{-3}$ m $= 1$ 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: $2$
Hence, the correct answer is option: 'B'.