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 scale divisions and \(30\) circular scale divisions . Assuming negligible zero error, the thickness of the wire is

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(D) \(0.2150\, cm\)

Step-by-step Solution

Detailed explanation

In one rotation scale moves \(\frac{0.25}{5}=0.05 cm\)

Least count \(=0.05 \times 10^{-2} cm\)

For 4 main scale division \(=4 \times 0.05=0.2 cm\)

For circular scale divosion \(=30 \times 0.05 \times 10^{-2}=1.5 \times 10^{-2} cm\)

Thickness of wire \(=0.2+0.015=0.2150 cm\)

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