In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm . There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main

Correct answer: D. 0.2150 ~cm Least count = Value of 1 part on main scale Number of parts on vernier scale =(0.25)/(5 × 100) ~cm=5 × 10^ -4 ~cm Reading =4 × 0.05 ~cm+30 × 5 × 10^ -4 ~cm =(0.2+0.0150) cm= 0.2150 ~cm (Thickness of wire)

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Physics
Chapter: Experimental Physics

JEE Main 2018 · 15 April · Morning Shift · Online · Physics · Experimental Physics

In a screw gauge, 5 complete rotations of the screw cause it to...

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of

0.25

\mathrm{cm}

. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:

Options

A $0.0430 \mathrm{~cm}$
B $0.3150 \mathrm{~cm}$
C $0.4300 \mathrm{~cm}$
D $0.2150 \mathrm{~cm}$

Correct answer

D. $0.2150 \mathrm{~cm}$

Step-by-step solution

Least count

=\frac{\text{ Value of } 1 \text{ part on main scale }}{\text{ Number of parts on vernier scale }} =\frac{0.25}{5 \times 100} \mathrm{~cm}=5 \times 10^{-4} \mathrm{~cm}

Reading

=4 \times 0.05 \mathrm{~cm}+30 \times 5 \times 10^{-4}

\mathrm{~cm}

=(0.2+0.0150)

\mathrm{cm}=

0.2150

\mathrm{~cm}

(Thickness of wire)

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