Diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal t

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1.Units, Dimensions and Measurement

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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 $1 \mathrm{~s}$ :

A

$2.5 \mathrm{~g} / \mathrm{cm}^3$

B

$1.7 \mathrm{~g} / \mathrm{cm}^3$

C

$2.2 \mathrm{~g} / \mathrm{cm}^3$

D

$2.0 \mathrm{~g} / \mathrm{cm}^3$

(JEE MAIN-2024)

Solution

$\text { Given } 9 \mathrm{MSD}=10 \mathrm{VSD}$

$\text { mass }=8.635 \mathrm{~g}$

$\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}$

$\mathrm{LC}=1 \mathrm{MSD}-\frac{9}{10} \mathrm{MSD}$

$\mathrm{LC}=\frac{1}{10} \mathrm{MSD}$

$\mathrm{LC}=0.01 \mathrm{~cm}$

$\text { Reading of diameter }=\mathrm{MSR}+\mathrm{LC} \times \mathrm{VSR}$

$=2 \mathrm{~cm}+(0.01) \times(2)$

$=2.02 \mathrm{~cm}$

$\text { Volume of sphere }=\frac{4}{3} \pi\left(\frac{d}{2}\right)^3=\frac{4}{3} \pi\left(\frac{2.02}{2}\right)^3$

$=4.32 \mathrm{~cm}^3$

$\text { Density }=\frac{\text { mass }}{\text { volume }}=\frac{8.635}{4.32}=1.998 \sim 2.00 \mathrm{~g}$

Standard 11

Physics

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(IIT-2015)