JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Units & Measurement

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A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ±0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ±0.01 mm. Take

g = 9.8 m / s^{2}

(exact). The Young's modulus obtained from the reading is

(A) $(2.0 \pm 0.3) \times 1 0^{11} N / m^{2}$
(B) $(2.0 \pm 0.2) \times 1 0^{11} N / m^{2}$ ✓ Correct
(C) $(2.0 \pm 0.1) \times 1 0^{11} N / m^{2}$
(D) $(2.0 \pm 0.05) \times 1 0^{11} N / m^{2}$

Answer: (B)

Solution

Young's modulus is given by

Y = \frac{F ℓ}{A Δ ℓ} = \frac{4 F ℓ}{π d ^{2} Δ ℓ}

Given:

ℓ = 2

F = m g = 1.0 \times 9.8 = 9.8

d = 0.4

= 4 \times 1 0^{- 4}

Δ ℓ = 0.8

= 8 \times 1 0^{- 4}

Y = \frac{4 \times 9.8 \times 2}{π \times ( 4 \times 1 0 ^{- 4} ) ^{2} \times 8 \times 1 0 ^{- 4}} \approx 2.0 \times 1 0^{11}

N/m²

For error analysis:

\frac{Δ Y}{Y} = \frac{2Δ d}{d} + \frac{Δ ( Δ ℓ )}{Δ ℓ} = \frac{2 \times 0.01}{0.4} + \frac{0.05}{0.8} = 0.05 + 0.0625 = 0.1125

Δ Y = 2.0 \times 1 0^{11} \times 0.1125 \approx 0.225 \times 1 0^{11} \approx 0.2 \times 1 0^{11}

N/m²

Therefore,

Y = (2.0 \pm 0.2) \times 1 0^{11}

N/m²

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option

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JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Units & Measurement

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