Experimental Physics

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Question 14 of 70

The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to 1 mm1 \mathrm{~mm}. The main scale reading is 2 cm2 \mathrm{~cm} and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is:

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Answer: (A) 2.0 g/cm32.0 \mathrm{~g} / \mathrm{cm}^3

Given 9MSD=10VSD9 \mathrm{MSD}=10 \mathrm{VSD}
mass⁡=8.635 g\operatorname{mass}=8.635 \mathrm{~g}
LC=1MSD−1VSDLC=1MSD−910MSDLC=110MSDLC=0.01 cm\begin{aligned} & \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \\ & \mathrm{LC}=1 \mathrm{MSD}-\frac{9}{10} \mathrm{MSD} \\ & \mathrm{LC}=\frac{1}{10} \mathrm{MSD} \\ & \mathrm{LC}=0.01 \mathrm{~cm}\end{aligned}
Reading of diameter =MSR+LC×VSR=\mathrm{MSR}+\mathrm{LC} \times \mathrm{VSR}
=2 cm+(0.01)×(2)=2.02 cm\begin{aligned} & =2 \mathrm{~cm}+(0.01) \times(2) \\ & =2.02 \mathrm{~cm}\end{aligned}
Volume of sphere =43π(d2)3=43π(2.022)3=\frac{4}{3} \pi\left(\frac{d}{2}\right)^3=\frac{4}{3} \pi\left(\frac{2.02}{2}\right)^3
=4.32 cm3=4.32 \mathrm{~cm}^3
Density = mass  volume =8.6354.32=1.998∼2.00 g=\frac{\text { mass }}{\text { volume }}=\frac{8.635}{4.32}=1.998 \sim 2.00 \mathrm{~g}

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