The side of a cube is measured by vernier callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). [2005]

Question

The side of a cube is measured by vernier callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). The main scale reads 10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g. Find the density of the cube in appropriate significant figures. [2005]

Smart Achievers · Accepted Answer

(2.66)

$L.C. = 1 MSD - 1 VSD$

$= 1 MSD - \frac{9}{10} MSD (∵ 9 MSD = 10 VSD)$

$= (1 - \frac{9}{10}) MSD = \frac{1}{10} MSD = (\frac{1}{10} \times 1) mm = 0.1 mm (∵ 1 MSD = 1 mm)$

$Reading of side, l = MSR + VSR (LC) = 10 mm + 1 \times 0.1 mm = 10.1 mm = 1.01 cm$

$Now, density = \frac{mass}{volume} = \frac{m}{l^{3}} = \frac{2.736 g}{{(1.01)}^{3}} = 2.66 {g/cm}^{3}$

$(Rounding off to 3 significant figures)$

Solution

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Solution

Ans. (2.66) L.C.=1 MSD-1 VSD =1 MSD-910 MSD (∵ 9 MSD=10 VSD) =(1-910) MSD=110 MSD=(110×1) mm=0.1 mm (∵ 1 MSD=1 mm) Reading of side, l=MSR+VSR (LC)=10 mm+1×0.1 mm=10.1 mm=1.01 cm Now, density=massvolume=ml3=2.736 g(1.01)3=2.66 g/cm3 (Rounding off to 3 significant figures)

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