A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will…

Question

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be $\frac{x}{100}$ where $x$ is _________.

Aicharya · Accepted Answer

Since least count of the instrument can be calculated as
$\begin{aligned}
& \text { Least count }=\frac{\text { pitch length }}{\text { No. of division on circular scale }} \\ & =\frac{0.75}{15}=0.05 \mathrm{~mm}.
\end{aligned}$
Here we are provided $\mathrm{L}=5 \mathrm{~mm} \& \mathrm{~W}=2.5 \mathrm{~mm}$ $\mathrm{L}=5 \mathrm{~mm} \& \mathrm{~W}=2.5 \mathrm{~mm}$
$\because$ We know that
$\mathrm{A}=\mathrm{L}. \mathrm{W}$
For calculating fractional error, we can write
$\frac{\mathrm{dA}}{\mathrm{~A}}=\frac{\mathrm{dL}}{\mathrm{~L}}+\frac{\mathrm{dW}}{\mathrm{~W}}$
Here $\mathrm{dL}=\mathrm{dW}=0.05 \mathrm{~mm}$
$\begin{aligned}
& \frac{\mathrm{dA}}{\mathrm{~A}}=\frac{0.05}{5}+\frac{0.05}{2.5} \\ & \Rightarrow \frac{\mathrm{dA}}{\mathrm{~A}}=\frac{1}{100}+\frac{2}{100}=\frac{3}{100}
\end{aligned}$
So, $x=3$

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a…

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