A student determined Young's Modulus of elasticity using the formula $Y=\frac{\mathrm{MgL}^{3}}{4 \mathrm{bd}^{3} \delta}$. The value of g is taken to be $9.8 \mathrm{~m} / \mathrm{s}^{2}$, without any significant error, his observation are as follows: | Physical Quantity | Least count of the Equipment used for measurement | Observed value | |---|---|---| | Mass (M) | 1 g | 2 kg | | Length of bar (L) | 1 mm | 1 m | | Breadth of bar (b) | 0.1 mm | 4 cm | | Thickness of bar (d) | 0.01 mm | 0.4 cm | | Depression ($\delta$) | 0.01 mm | 5 mm | Then the fractional error in the measurement of Y is:

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Units & Measurement

A student determined Young's Modulus of elasticity using the formula

Y = \frac{MgL ^{3}}{4 bd ^{3} δ}

. The value of g is taken to be

9.8 m / s^{2}

, without any significant error, his observation are as follows:

Physical Quantity	Least count of the Equipment used for measurement	Observed value
Mass (M)	1 g	2 kg
Length of bar (L)	1 mm	1 m
Breadth of bar (b)	0.1 mm	4 cm
Thickness of bar (d)	0.01 mm	0.4 cm
Depression ( $δ$ )	0.01 mm	5 mm

Then the fractional error in the measurement of Y is:

Answer: (B)

The fractional error follows the exponents in

Y = \frac{M g L ^{3}}{4 b d ^{3} δ}

\frac{Δ Y}{Y} = \frac{Δ M}{M} + 3 \frac{Δ L}{L} + \frac{Δ b}{b} + 3 \frac{Δ d}{d} + \frac{Δ δ}{δ}

. Taking each least count over its reading,

\frac{1}{2000} + 3 \cdot \frac{1}{1000} + \frac{0.1}{40} + 3 \cdot \frac{0.01}{4} + \frac{0.01}{5} = 0.0005 + 0.003 + 0.0025 + 0.0075 + 0.002 = 0.0155

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option