A student determined Young's Modulus of elasticity using the formula Y = M g L 3 4 bd 3 δ . The value of g is taken to be 9 . 8 m s - 2 without any significant error, his observations are as foll

Question

A student determined Young's Modulus of elasticity using the formula $Y = \frac{M g L^{3}}{4 {bd}^{3} δ} .$ The value of $g$ is taken to be $9.8 m s^{- 2}$ without any significant error, his observations are as following.

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 :

TestHub · Accepted Answer

Y = M g L 3 4 bd 3 δ ∆ Y Y = ∆ M M + 3 ∆ l 1 + ∆ b b + 3 ∆ d d + ∆ δ δ ⇒ ∆ Y Y = 0 . 0005 + 3 × 0 . 001 + 0 . 0025 + 3 × 0 . 0025 + 0 . 002 = 0 . 0155

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