Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Four Kashmiri girls defy the odds, score 99+ percentile in JEE Main 2025
anam-khan

JEE, NEET, CUET coaching centres: Assam, Rajasthan have framed laws, but will they work?
anam-khan

JEE Main 2025 withheld results declared; NTA issues show cause notices ahead of action

Latest Question