JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2021
- Hard
- 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 \((\delta)\) | \(0.01\; {mm}\) | \(5 \;{mm}\) |
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.0155\)
Step-by-step Solution
Detailed explanation
\({y}=\frac{{MgL}^{3}}{4 {bd}^{3} \delta}\)
\(\frac{\Delta {y}}{{y}}=\frac{\Delta {M}}{{M}}+\frac{3 \Delta {L}}{{L}}+\frac{\Delta {b}}{{b}}+\frac{3 \Delta {d}}{{d}}+\frac{\Delta \delta}{\delta}\)
\(\frac{\Delta {y}}{{y}}=\frac{10^{-3}}{2}+\frac{3 \times 10^{-3}}{1}+\frac{10^{-2}}{4}+\frac{3 \times 10^{-2}}{4}+\frac{10^{-2}}{5}\)
\(=10^{-3}[0.5+3+2.5+7.5+2]=0.0155\)
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