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  • A student determined Young's Modulus of elasticity using the formula <img alt="\mathrm{Y}=\frac{\mathrm{MgL}^{3}}{4 \mathrm{bd}^{3} \delta}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cmathrm%7BY%7D%3D%5Cfrac%7B%5Cmathrm%7BMgL%7D%5E%7B3%7D%7

A student determined Young's Modulus of elasticity using the formula \mathrm{Y}=\frac{\mathrm{MgL}^{3}}{4 \mathrm{bd}^{3} \delta}, The
value of \mathrm{g} is taken to be 9.8 \mathrm{~m} / \mathrm{s}^{2}, without any significant error, his observation are as following,

Physical Quantity Least count of the Equipment Observed Value
Mass (M) 1g 2kg
Length of bar(L) 1mm 1m
Breadth of bar(b) 0.1mm 4cm
Thickness of bar(d) 0.01mm 0.4cm
Depression \left ( \delta\right ) 0.01mm 5mm

Then the fractional error in the measurement of Y is :
Option: 1 0.155
Option: 2 0.0083
Option: 3 0.083
Option: 4 0.0155

Answers (1)

best_answer

Y= \frac{MgL^{3}}{4bd^{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{10^{-3}}{2}+\frac{3\times 10^{-3}}{1}+\frac{10^{-4}}{4\times 10^{-2}}+\frac{3\times 10^{-5}}{4\times 10^{-3}}+\frac{10^{-5}}{5\times 10^{-3}}
=\left ( 0\cdot 5+3+2\cdot 5+7\cdot 5+2 \right )\times 10^{-3}
=15\cdot 5\times 10^{-3}
\Delta Y=0\cdot 0155

Posted by

Ritika Jonwal

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