JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Units & Measurement

In order to determine the young's modulus of a wire of radius 0.2 cm (measured using a scale of least count

= 0.001 cm

) and length 1 m (measured using a scale of least count

= 1 mm

), a weight of mass 1 kg (measured using a scale of least count

= 1 g

) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm .) What will be the fractional error in the value of young's modulus determined by this experiment?

(A) $9%$
(B) $0.9%$
(C) $0.14%$
(D) $1.4%$ ✓ Correct

Answer: (D)

Solution

From

Y = \frac{F L}{A Δ L} = \frac{m g L}{π r ^{2} Δ L}

, the fractional errors of the measured quantities add as

\frac{Δ Y}{Y} = \frac{Δ m}{m} + \frac{Δ L}{L} + 2 \frac{Δ r}{r} + \frac{Δ ( Δ L )}{Δ L}

each uncertainty being the least count of the instrument used for that quantity:

\frac{1 g}{1 kg} = 0.001, \frac{1 mm}{1 m} = 0.001, 2 \times \frac{0.001 cm}{0.2 cm} = 0.01, \frac{0.001 cm}{0.5 cm} = 0.002

Adding them,

\frac{Δ Y}{Y} = 0.001 + 0.001 + 0.01 + 0.002 = 0.014

so the fractional error in the measured Young's modulus is

1.4%

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Units & Measurement

Solution

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