Q 11-01-112JEE MainJEE Main 2024 (8 Apr, Shift 1)Medium
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 $1\ \text{mm}$. The main scale reading is $2\ \text{cm}$ and second division of vernier scale coincides with a division on main scale. If mass of the sphere is $8.635\ \text{g}$, the density of the sphere is:
Answer: (A) $2.0\ \text{g/cm}^3$
Least count $= 1\ \text{MSD} - 1\ \text{VSD} = 1 - 0.9 = 0.1\ \text{mm} = 0.01\ \text{cm}$.
Diameter $= 2 + 2\times0.01 = 2.02\ \text{cm}$, so $r = 1.01\ \text{cm}$.
$$V = \frac43\pi(1.01)^3 \approx 4.32\ \text{cm}^3$$
$$\rho = \frac{8.635}{4.32} \approx 2.0\ \text{g cm}^{-3}$$
Solution by Sreeraj P, M.Sc Physics