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1.Units, Dimensions and Measurement
hard
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 $(\delta)$ | $0.01\; {mm}$ | $5 \;{mm}$ |
Then the fractional error in the measurement of ${Y}$ is
A
$0.0083$
B
$0.0155$
C
$0.155$
D
$0.083$
(JEE MAIN-2021)
Solution
${y}=\frac{{MgL}^{3}}{4 {bd}^{3} \delta}$
$\frac{\Delta {y}}{{y}}=\frac{\Delta {M}}{{M}}+\frac{3 \Delta {L}}{{L}}+\frac{\Delta {b}}{{b}}+\frac{3 \Delta {d}}{{d}}+\frac{\Delta \delta}{\delta}$
$\frac{\Delta {y}}{{y}}=\frac{10^{-3}}{2}+\frac{3 \times 10^{-3}}{1}+\frac{10^{-2}}{4}+\frac{3 \times 10^{-2}}{4}+\frac{10^{-2}}{5}$
$=10^{-3}[0.5+3+2.5+7.5+2]=0.0155$
Standard 11
Physics
