Question

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 l 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:

• A. \(2.5 \, \text{g/cm}^3\)
• B. \(1.7 \, \text{g/cm}^3\)
• C. \(2.2 \, \text{g/cm}^3\)
• D. \(2.0 \, \text{g/cm}^3\)

Answer 1

The Correct Option is D

Given:

\[ 9 \text{ MSD} = 10 \text{ VSD}. \]

The least count (LC) is:

\[ \text{LC} = 1 \text{ MSD} - 1 \text{ VSD}. \]

\[ \text{LC} = 1 \text{ MSD} - \frac{9}{10} \text{ MSD} = \frac{1}{10} \text{ MSD}. \]

\[ \text{LC} = 0.01 \, \text{cm}. \]

Reading of the diameter:

\[ \text{Diameter} = \text{MSR} + \text{LC} \times \text{VSR}. \]

\[ \text{Diameter} = 2 \, \text{cm} + (0.01) \times (2) = 2.02 \, \text{cm}. \]

The volume of the sphere is:

\[ V = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 = \frac{4}{3} \pi \left( \frac{2.02}{2} \right)^3. \]

\[ V = \frac{4}{3} \pi (1.01)^3 = 4.32 \, \text{cm}^3. \]

The density is:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{8.635}{4.32}. \]

\[ \text{Density} \approx 1.998 \, \text{g/cm}^3 \approx 2.00 \, \text{g/cm}^3. \]

Final Answer: 2.0 g/cm³.