Question

A student got an experiment to find the volume of a glass slab of fixed thickness using a screw gauge. The dimension of the base was found to be 2 cm $\times$ 5 cm. The screw gauge reading for the thickness is shown below. Find the volume of the glass slab. Given the least count of the screw gauge is 0.01 mm.

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Answer 1

Given, area of base = 2 cm $\times$ 5 cm = 10 ${cm}^2$

First part of the measurement, P.S.R = 5 mm

Head scale division that coincides with the pitch scale axis, H.S.C = 29

Second part of the measurement, H.S.C $\times$ L.C = 29 $\times$ 0.01 mm = 0.29 mm

Thickness of the glass slab = P.S.R + H.S.C $\times$ L.C = 5 + 0.29 = 5.29 mm

So, volume of the slab = Area of base $\times$ Thickness = 10 ${cm}^2$ $\times$ 5.29 mm = 1000 ${mm}^2$ $\times$ 5.29 mm = 5290 ${mm}^3$