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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 × 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.


Solution

Given, area of base = 2 cm × 5 cm = 10 cm2

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 × L.C = 29 × 0.01 mm = 0.29 mm

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

So, volume of the slab = Area of base × Thickness = 10 cm2 × 5.29 mm = 1000 mm2 × 5.29 mm = 5290 mm3

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