A student determined Young's Modulus of elasticity using the form… - Vidyadip

MCQ

A student determined Young's Modulus of elasticity using the formula $Y=\frac{M g L^{3}}{4 b d^{3} \delta} .$ The value of $g$ is taken to be $9.8 \,{m} / {s}^{2}$, without any significant error, his observation 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 $(\delta)$	$0.01\; {mm}$	$5 \;{mm}$

Then the fractional error in the measurement of ${Y}$ is

A
$0.0083$
✓
$0.0155$
C
$0.155$
D
$0.083$

✓

Answer

Correct option: B.

$0.0155$

b
${y}=\frac{{MgL}^{3}}{4 {bd}^{3} \delta}$

$\frac{\Delta {y}}{{y}}=\frac{\Delta {M}}{{M}}+\frac{3 \Delta {L}}{{L}}+\frac{\Delta {b}}{{b}}+\frac{3 \Delta {d}}{{d}}+\frac{\Delta \delta}{\delta}$

$\frac{\Delta {y}}{{y}}=\frac{10^{-3}}{2}+\frac{3 \times 10^{-3}}{1}+\frac{10^{-2}}{4}+\frac{3 \times 10^{-2}}{4}+\frac{10^{-2}}{5}$

$=10^{-3}[0.5+3+2.5+7.5+2]=0.0155$

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