JEE Main 2026 — Physics Mechanics

Given:

Length of string A, $L_A = 0.314$LA​=0.314 m

Young's modulus of string A, $Y_A = 2 \times 10^{10}$YA​=2×1010 N/m$^2$2

Radius of both strings, $r = 0.2$r=0.2 mm $= 2 \times 10^{-4}$=2×10−4 m

Length of string B, $L_B = 2L_A = 0.628$LB​=2LA​=0.628 m

Young's modulus of string B, $Y_B = 2Y_A = 4 \times 10^{10}$YB​=2YA​=4×1010 N/m$^2$2

Mass of load, $m = 0.8$m=0.8 kg

Acceleration due to gravity, $g = 10$g=10 m/s$^2$2

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

Cross-sectional area of both strings:

$A = \pi r^2 = 3.14 \times (2 \times 10^{-4})^2 = 3.14 \times 4 \times 10^{-8}$A=πr2=3.14×(2×10−4)2=3.14×4×10−8 m$^2$2

Elongation in string A:

$$\Delta L_A = \dfrac{T L_A}{A Y_A}$$ΔLA​=AYA​TLA​​

$$\Delta L_A = \dfrac{8 \times 0.314}{3.14 \times 4 \times 10^{-8} \times 2 \times 10^{10}}$$ΔLA​=3.14×4×10−8×2×10108×0.314​

$\Delta L_A = \dfrac{8 \times 0.314}{8 \times 3.14 \times 10^2} = \dfrac{0.1}{10^2} = 10^{-3}$ΔLA​=8×3.14×1028×0.314​=1020.1​=10−3 m $= 1$=1 mm

Elongation in string B:

$$\Delta L_B = \dfrac{T L_B}{A Y_B}$$ΔLB​=AYB​TLB​​

$\Delta L_B = \dfrac{8 \times (2 \times 0.314)}{A \times (2 Y_A)} = \dfrac{T L_A}{A Y_A} = 1$ΔLB​=A×(2YA​)8×(2×0.314)​=AYA​TLA​​=1 mm

Net change in length of the combination:

$\Delta L = \Delta L_A + \Delta L_B = 1 + 1 = 2$ΔL=ΔLA​+ΔLB​=1+1=2 mm

Answer: $2$2