In the determination of Young's modulus (Y=4 MLg / d ^2 ) by usin… - Vidyadip

MCQ

In the determination of Young's modulus $\left(Y=\frac{4 MLg }{\pi / d ^2}\right)$ by using Searle's method, a wire of length $L=2 \ m$ and diameter $d =0.5 \ mm$ is used. For a load $M =2.5 \ kg$, an extension $\ell=0.25 \ mm$ in the length of the wire is observed. Quantities $d$ and $\ell$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $0.5 \ mm$. The number of divisions on their circular scale is $100$ . The contributions to the maximum probable error of the $Y$ measurement

✓
due to the errors in the measurements of $d$ and $\ell$ are the same.
B
due to the error in the measurement of $d$ is twice that due to the error in the measurement of $\ell$.
C
due to the error in the measurement of $\ell$ is twice that due to the error in the measurement of $d$.
D
due to the error in the measurement of $d$ is four time that due to the error in the measurement of $\ell$.

✓

Answer

Correct option: A.

due to the errors in the measurements of $d$ and $\ell$ are the same.

a
$\Delta d=\Delta \ell=\frac{0.5}{100} \ mm$

$y =\frac{4 MLg }{\pi \ell d ^2} $

$\left(\frac{\Delta y }{ y }\right)_{\max }=\frac{\Delta \ell}{\ell}+2 \frac{\Delta d }{ d }$

error due to $\ell$ measurement $\frac{\Delta \ell}{\ell}=\frac{0.5 / 100 mm }{0.25 mm }$

error due to d measurement $2 \frac{\Delta d}{d}=\frac{2 \times \frac{0.5}{100}}{0.5 mm }=\frac{0.5 / 100}{0.25}$

So error in y due to $\ell$ measurement $=$ error in $y$ due to $d$ measurement

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