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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Experimental Physics

JEE Main 2024 · 8 April · Morning Shift · Physics · Experimental Physics

The diameter of a sphere is measured using a vernier caliper whose...

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:

Options

  1. A2.0 g/cm32.0 \mathrm{~g} / \mathrm{cm}^3
  2. B1.7 g/cm31.7 \mathrm{~g} / \mathrm{cm}^3
  3. C2.2 g/cm32.2 \mathrm{~g} / \mathrm{cm}^3
  4. D2.5 g/cm32.5 \mathrm{~g} / \mathrm{cm}^3

Correct answer

A. 2.0 g/cm32.0 \mathrm{~g} / \mathrm{cm}^3

Step-by-step solution

Given 9MSD=10VSDmass⁡=8.635 g9 \mathrm{MSD}=10 \mathrm{VSD} \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 $=\mathrm{MSR}+\mathrm{LC} \times \mathrm{VSR}\begin{aligned} & =2 \mathrm{~cm}+(0.01) \times(2) \\ & =2.02 \mathrm{~cm}\end{aligned}$$VolumeofsphereVolume of sphere=\frac{4}{3} \pi\left(\frac{d}{2}\right)^3=\frac{4}{3} \pi\left(\frac{2.02}{2}\right)^3 =4.32 \mathrm{~cm}^3DensityDensity=\frac{\text{ mass }}{\text{ volume }}=\frac{8.635}{4.32}=1.998 \sim 2.00 \mathrm{~g}$
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