JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Elasticity & SHM

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As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of

4 5^{\circ}

with the load axis. The length of wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be

x \times 1 0^{4} Nm^{- 2}

.
The value of

x

____

Answer: 5

Solution

The straight extension-load line through the origin has slope

\frac{Δ L}{F} = tan 4 5^{\circ} = 1

in SI units, so

F = Δ L

numerically. Young's modulus is

Y = \frac{F L}{A Δ L} = \frac{L}{A} \cdot \frac{F}{Δ L} = \frac{L}{A}

. With

L = 62.8 \times 1 0^{- 2} m

and radius

r = 2 \times 1 0^{- 3} m

, the area is

A = π r^{2} = π (2 \times 1 0^{- 3})^{2} = 1.256 \times 1 0^{- 5} m^{2}

. Hence

Y = \frac{62.8 \times 1 0 ^{- 2}}{1.256 \times 1 0 ^{- 5}} = 5 \times 1 0^{4} N m^{- 2}

, giving

x = 5

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Elasticity & SHM

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