JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Units & Measurement

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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✓ Correct
(C)0.13
(D)0.25

Answer: (B)

Solution

Young's modulus is

Y = \frac{F ℓ}{A Δ ℓ}

with

A = \frac{π d ^{2}}{4}

, so the fractional error (ignoring the load) is

\frac{Δ Y}{Y} = \frac{Δ ℓ _{L}}{ℓ} + 2 \frac{Δ d}{d} + \frac{Δ ( Δ ℓ )}{Δ ℓ} = \frac{0.1}{150} + \frac{2 ( 0.001 )}{0.08} + \frac{0.001}{0.5} = 0.0277

. The measured value is

Y = \frac{( 100 ) ( 1.5 )}{\frac{π}{4} ( 8 \times 1 0 ^{- 4} ) ^{2} ( 5 \times 1 0 ^{- 3} )} \approx 5.97 \times 1 0^{10} N/ m^{2}

, so

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

. The value of

α

1.65

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Units & Measurement

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