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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Experimental Physics

JEE Main 2025 · 28 January · Morning Shift · Physics · Experimental Physics

A tiny metallic rectangular sheet has length and breadth of 5 mm...

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 x100\frac{x}{100} where xx is _________.A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a specially designed

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Step-by-step solution

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a specially designed
Since least count of the instrument can be calculated as  Least count = pitch length  No. of division on circular scale =0.7515=0.05 mm.\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 L=5 mm& W=2.5 mmL=5 mm& W=2.5 mm∵\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 A=L.W\mathrm{A}=\mathrm{L} . \mathrm{W} For calculating fractional error, we can write dA A=dL L+dW W\frac{\mathrm{dA}}{\mathrm{~A}}=\frac{\mathrm{dL}}{\mathrm{~L}}+\frac{\mathrm{dW}}{\mathrm{~W}} Here dL=dW=0.05 mm\mathrm{dL}=\mathrm{dW}=0.05 \mathrm{~mm}dA A=0.055+0.052.5⇒dA A=1100+2100=3100\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}$So,So,x=3$
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