JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

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 $α \times 1 0^{9}$ N/m $^{2}$ . The value of $α$ is ________.

(Ignore the contribution of the load to Young's modulus error calculation)

A
$1.3$
B
$1.65$
C
$0.13$
D
$0.25$

Solution

Young's modulus is $Y = \frac{F ℓ}{A Δ ℓ}$ with $A = \frac{π d ^{2}}{4}$ , so the relative error becomes:

$\frac{Δ Y}{Y} = \frac{Δ F}{F} + \frac{Δ ℓ}{ℓ} + 2 \frac{Δ d}{d} + \frac{Δ ( Δ ℓ )}{Δ ℓ}$

Since the load is taken as exact, $Δ F / F = 0$ . Substituting the least counts:

$\frac{Δ Y}{Y} = \frac{0.1}{150} + 2 \times \frac{0.001}{0.08} + \frac{0.001}{0.5} \approx 0.0277$

The central value of $Y$ :

$Y = \frac{100 \times 1.5}{\frac{π}{4} ( \frac{0.08}{100} ) ^{2} \times ( 0.5 \times 1 0 ^{- 2} )} \approx 5.97 \times 1 0^{10}$ N/m $^{2}$

$∴ Δ Y = 0.0277 \times 5.97 \times 1 0^{10} \approx 1.65 \times 1 0^{9}$ N/m $^{2}$

Hence $α = 1.65$ .

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