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JEE Main 2026 · 2 Apr · Shift 1, Physics question 27

The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm. The length measured by a scale of least count 0.1 cm is 150 cm. When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm, measured by a micrometer of least count 0.001 cm. The error in the measured Young's modulus is α×109\alpha\times10^9 N/m². The value of α\alpha is ________.
(Ignore the contribution of the load to Young's modulus error calculation)

Choose one option
Answer and worked solution

Official answer: B

Young's modulus. Y=4FLπd2 ΔL=4(100)(1.5)π(8×10−4)2(5×10−3)≈5.97×1010Y=\dfrac{4FL}{\pi d^2\,\Delta L}=\dfrac{4(100)(1.5)}{\pi(8\times10^{-4})^2(5\times10^{-3})}\approx5.97\times10^{10} N/m².

Relative error.
ΔYY=2Δdd+ΔLL+Δ(ΔL)ΔL=2⋅0.0010.08+0.1150+0.0010.5=0.0277.\frac{\Delta Y}Y=2\frac{\Delta d}d+\frac{\Delta L}L+\frac{\Delta(\Delta L)}{\Delta L}=2\cdot\frac{0.001}{0.08}+\frac{0.1}{150}+\frac{0.001}{0.5}=0.0277.

Error. ΔY≈0.0277×5.97×1010=1.65×109\Delta Y\approx0.0277\times5.97\times10^{10}=1.65\times10^9 N/m², so α=1.65\alpha=1.65.

(B)

Topics:Errors in measurementHooke's law and elastic moduli
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