JEE Main 2026 — Physics Mechanics

The formula for Young's modulus is given by:

$$Y = \dfrac{F L_0}{A \Delta L} = \dfrac{4 F L_0}{\pi d^2 \Delta L}$$Y=AΔLFL0​​=πd2ΔL4FL0​​

Substituting the given values into the formula:

$$Y = \dfrac{4 \times 100 \times 1.5}{\pi \times (0.08 \times 10^{-2})^2 \times (0.5 \times 10^{-2})}$$Y=π×(0.08×10−2)2×(0.5×10−2)4×100×1.5​

$$Y = \dfrac{600}{\pi \times 64 \times 10^{-8} \times 5 \times 10^{-3}} = \dfrac{15}{8\pi} \times 10^{11} \text{ N/m}^2$$Y=π×64×10−8×5×10−3600​=8π15​×1011 N/m2

The relative error in Young's modulus is:

$$\dfrac{\Delta Y}{Y} = \dfrac{\Delta L_0}{L_0} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta (\Delta L)}{\Delta L}$$YΔY​=L0​ΔL0​​+2dΔd​+ΔLΔ(ΔL)​

Substituting the least counts as absolute errors and the measured values:

$$\dfrac{\Delta Y}{Y} = \dfrac{0.1}{150} + 2 \times \dfrac{0.001}{0.08} + \dfrac{0.001}{0.5}$$YΔY​=1500.1​+2×0.080.001​+0.50.001​

$$\dfrac{\Delta Y}{Y} = \dfrac{1}{1500} + \dfrac{1}{40} + \dfrac{1}{500}$$YΔY​=15001​+401​+5001​

$$\dfrac{\Delta Y}{Y} = \dfrac{2 + 75 + 6}{3000} = \dfrac{83}{3000}$$YΔY​=30002+75+6​=300083​

The absolute error in Young's modulus is:

$$\Delta Y = Y \times \dfrac{\Delta Y}{Y} = \left(\dfrac{15}{8\pi} \times 10^{11}\right) \times \dfrac{83}{3000}$$ΔY=Y×YΔY​=(8π15​×1011)×300083​

$$\Delta Y = \dfrac{15 \times 83}{24000 \pi} \times 10^{11} = \dfrac{83}{1600 \pi} \times 10^{11} = \dfrac{83}{16\pi} \times 10^9 \text{ N/m}^2$$ΔY=24000π15×83​×1011=1600π83​×1011=16π83​×109 N/m2

Using $\pi \approx 3.14$π≈3.14:

$$\Delta Y = \dfrac{83}{16 \times 3.14} \times 10^9 = \dfrac{83}{50.24} \times 10^9 \approx 1.65 \times 10^9 \text{ N/m}^2$$ΔY=16×3.1483​×109=50.2483​×109≈1.65×109 N/m2

Comparing this with $\alpha \times 10^9 \text{ N/m}^2$α×109 N/m2, we get $\alpha = 1.65$α=1.65.

Answer: $1.65$1.65