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JEE MAIN 2022
28-06-22 (S2)
Question
A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm , $1.23 \mathrm{~mm}, 1.19 \mathrm{~mm}$ and 1.20 mm . The percentage error is $\frac{\mathrm{x}}{121} \%$. The value of x is
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Solution
$\begin{aligned} & X=\frac{1.22 \mathrm{~mm}+1.23 \mathrm{~mm}+1.19 \mathrm{~mm}+1.20 \mathrm{~mm}}{4} \\ & X=1.21 \mathrm{~mm} \\ & \Delta X=\frac{0.01+0.02+0.02+0.01}{4}=\frac{0.06}{4}=0.015 \\ & \text { Percentage error }=\frac{0.015}{1.21} \times 100 \\ & X=150\end{aligned}$
Question Tags
JEE Main
Physics
Easy
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