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 ~mm . The main scale reading is 2 ~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:

Units and Measurements PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q74 with Solution

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Correct answer: B

Given:

So, $9\,\text{MSD} = 10\,\text{VSD}$ $10\,\text{VSD} = 9\text{ mm}$ $1\,\text{VSD} = 0.9\text{ mm}$

Least count: $\text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = 1.0 - 0.9 = 0.1\text{ mm}$

Main scale reading: $2\text{ cm} = 20\text{ mm}$

Second vernier division coincides, so vernier reading: $2 \times \text{LC} = 2 \times 0.1 = 0.2\text{ mm}$

Hence diameter: $d = 20 + 0.2 = 20.2\text{ mm} = 2.02\text{ cm}$

$r = \frac{d}{2} = \frac{2.02}{2} = 1.01\text{ cm}$

$V = \frac{4}{3}\pi r^3$

Now, $r^3 = (1.01)^3 = 1.030301$

So, $V = \frac{4}{3}\pi (1.030301) \approx 4.32\text{ cm}^3$

Given mass: $m = 8.635\text{ g}$

Density: $\rho = \frac{m}{V} = \frac{8.635}{4.32} \approx 2.0\text{ g/cm}^3$

Thus the density is $\boxed{2.0\text{ g/cm}^3}$

So the correct option is B.