JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2024
- Hard
- MCQ
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 \(1 \mathrm{~mm}\). The main scale reading is \(2 \mathrm{~cm}\) and second division of vernier scale coincides with a division on main scale. If mass of the sphere is \(8.635 \mathrm{~g}\), thedensity of the sphere \(1 \mathrm{~s}\) _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2.0 \mathrm{~g} / \mathrm{cm}^3\)
Step-by-step Solution
Detailed explanation
\(\text { Given } 9 \mathrm{MSD}=10 \mathrm{VSD}\)
\(\text { mass }=8.635 \mathrm{~g}\)
\(\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}\)
\(\mathrm{LC}=1 \mathrm{MSD}-\frac{9}{10} \mathrm{MSD}\)
\(\mathrm{LC}=\frac{1}{10} \mathrm{MSD}\)
\(\mathrm{LC}=0.01 \mathrm{~cm}\)
\(\text { Reading of diameter }=\mathrm{MSR}+\mathrm{LC} \times \mathrm{VSR}\)
\(=2 \mathrm{~cm}+(0.01) \times(2)\)
\(=2.02 \mathrm{~cm}\)
\(\text { Volume of sphere }=\frac{4}{3} \pi\left(\frac{d}{2}\right)^3=\frac{4}{3} \pi\left(\frac{2.02}{2}\right)^3\)
\(=4.32 \mathrm{~cm}^3\)
\(\text { Density }=\frac{\text { mass }}{\text { volume }}=\frac{8.635}{4.32}=1.998 \sim 2.00 \mathrm{~g}\)
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