A student uses a screw gauge to measure the thickness of a glass plate. The pitch of the screw gauge is 0.5 mm and it has 50 divisions on its circular scale. The main scale reading is 1.5 mm and the 24^ th division of

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Quantrex Academy · Accepted Answer

Correct answer: B. 0.174 cm

Least count of the screw gauge is given by the ratio of pitch to the number of divisions on the circular scale.

\text{Least count (LC)} = \dfrac{0.5\text{ mm}}{50} = 0.01\text{ mm}

The thickness of the glass plate is the sum of the main scale reading and the circular scale reading multiplied by the least count.

\text{Thickness} = \text{MSR} + (\text{CSR} \times \text{LC})

\text{Thickness} = 1.5\text{ mm} + (24 \times 0.01\text{ mm})

\text{Thickness} = 1.5\text{ mm} + 0.24\text{ mm} = 1.74\text{ mm}

Converting the thickness into centimeters:

\text{Thickness} = 0.174\text{ cm}

Answer:

0.174\text{ cm}

A student uses a screw gauge to measure the thickness of a glass...

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