Mathematics in Physics

JEE Main level, single correct, medium

62 of 121

Question 62 of 121

Young's modulus is determined by the equation given by Y=49000mldyncm2\mathrm{Y}=49000 \frac{\mathrm{m}}{\mathrm{l}} \frac{\mathrm{dyn}}{\mathrm{cm}^2} where MM is the mass and ll is the extension of wire used in the experiment. Now error in Young modules (Y)(Y) is estimated by taking data from M−lM-l plot in graph paper. The smallest scale divisions are 5 g5 \mathrm{~g} and 0.02 cm0.02 \mathrm{~cm} along load axis and extension axis respectively. If the value of MM and ll are 500 g500 \mathrm{~g} and 2 cm2 \mathrm{~cm} respectively then percentage error of YY is :

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Answer: (B) 2%2 \%

ΔYY=Δmm+Δℓℓ=5500+0.022=0.01+0.01ΔYY=0.02⇒%ΔYY=2%\begin{aligned} \frac{\Delta \mathrm{Y}}{\mathrm{Y}} & =\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \ell}{\ell} \\ & =\frac{5}{500}+\frac{0.02}{2}=0.01+0.01 \\ \frac{\Delta \mathrm{Y}}{\mathrm{Y}} & =0.02 \Rightarrow \% \frac{\Delta \mathrm{Y}}{\mathrm{Y}}=2 \%\end{aligned}

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