JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Units & Measurement

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In an experiment of determine the Young's modulus of wire of a length exactly 1 m , the extension in the length of the wire is measured as 0.4 mm with an uncertainty of

\pm 0.02 mm

when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of

\pm 0.01 mm

. The error in the measurement of Young's modulus (

Δ Y

) is found to be

x \times 1 0^{10} Nm^{- 2}

. The value of x is
[Take

g = 10 m / s^{2}

]

Answer: 2

Solution

Young's modulus:

Y = \frac{F L}{A Δ L} = \frac{m g L}{\frac{π d ^{2}}{4} Δ L}

Substituting values:

Y = \frac{10 \times 1 \times 4}{π ( 0.4 \times 1 0 ^{- 3} ) ^{2} \times 0.4 \times 1 0 ^{- 3}} \approx 0.199 \times 1 0^{12} N/m^{2}

Relative error:

\frac{Δ Y}{Y} = \frac{Δ ( Δ L )}{Δ L} + 2 \frac{Δ d}{d} = \frac{0.02}{0.4} + 2 \times \frac{0.01}{0.4} = 0.05 + 0.05 = 0.1

Absolute error:

Δ Y = 0.1 \times Y = 0.1 \times 0.199 \times 1 0^{12} = 1.99 \times 1 0^{10} \approx 2 \times 1 0^{10}

Therefore,

x = 2

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Units & Measurement

Solution

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