A screw gauge has positive error of 4 divisions. When the screw gauge holds a sphere the main scale reading is 4 mm and the head scale coinciding division is 24. If its least count is 0.01 mm , the volume of the sphere is _____ ${m m}^{3}$

24.456

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38.808

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62.378

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40.567

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Solution

The correct option is B $38.808$
Positive error $= 4 \times 0.01 = 0.04 m m$
Diameter of sphere $d = M . S . R + n \times L . C . - z e r o e r r o r$
Or $d = 4 + 24 \times 0.01 - 0.04 = 4.20 m m$
Volume of sphere $V = \frac{4 π}{3} (\frac{d}{2})^{3}$
$⟹ V = \frac{4 π}{3} \times (\frac{4.20}{2})^{3} = 38.808 m m^{3}$

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A screw gauge has positive error of 4 divisions. When the screw gauge holds a sphere the main scale reading is 4 mm and the head scale coinciding division is 24. If its least count is 0.01 mm , the volume of the sphere is _____ mm3

The correct option is B 38.808Positive error =4×0.01=0.04 mmDiameter of sphere d=M.S.R+n×L.C. − zero errorOr d=4+24×0.01−0.04=4.20 mmVolume of sphere V=4π3(d2)3⟹ V=4π3×(4.202)3=38.808 mm3

A screw gauge has positive error of 4 divisions. When the screw gauge holds a sphere the main scale reading is 4 mm and the head scale coinciding division is 24. If its least count is 0.01 mm , the volume of the sphere is _____ ${m m}^{3}$