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

To determine the density of the sphere, we need to start by calculating its volume using the measurements given. We'll use the following steps:

1. Calculate the least count of the vernier caliper.
2. Determine the total reading from the vernier caliper.
3. Calculate the volume of the sphere using the obtained diameter.
4. Finally, calculate the density using the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \).

Step 1: Calculate the Least Count of the Vernier Caliper

Given:

• \(9\) divisions of main scale \(= 10\) divisions of vernier scale.
• Length of one main scale division \(= l \, \text{mm} = 0.1 \, \text{cm}\) (since 1 cm = 10 mm).

The least count (LC) of the vernier caliper is given by:

\(\text{LC} = \frac{\text{Value of one main scale division}}{\text{Number of divisions on vernier scale}} = \frac{l}{9} \, \text{cm} = \frac{0.1}{9} \, \text{cm}\approx 0.0111 \, \text{cm}\)

Step 2: Determine the Total Reading

Given:

• Main scale reading \(= 2 \, \text{cm}\)
• The second division of the vernier scale coincides with a division on the main scale.

Vernier scale reading \(= 2 \times \text{Least Count} = 2 \times 0.0111 \, \text{cm} = 0.0222 \, \text{cm}\).

Total reading (diameter of the sphere) is:

\(\text{Total reading} = \text{Main Scale Reading} + \text{Vernier Scale Reading} = 2 + 0.0222 \approx 2.0222 \, \text{cm}\)

Step 3: Calculate the Volume of the Sphere

The formula for the volume of a sphere is given by:

\(V = \frac{4}{3} \pi r^3\)

where \( r \) is the radius of the sphere.

Diameter \(= 2.0222 \, \text{cm}\), hence radius \( r = \frac{2.0222}{2} \, \text{cm} = 1.0111 \, \text{cm}\).

Plugging the radius into the volume formula:

\(V = \frac{4}{3} \pi (1.0111)^3 \approx \frac{4}{3} \times 3.1416 \times 1.033 \approx 4.32 \, \text{cm}^3\)

Step 4: Calculate the Density

Now, using the formula for density:

\(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{8.635 \, \text{g}}{4.32 \, \text{cm}^3} \approx 2.0 \, \text{g/cm}^3\)

Therefore, the density of the sphere is \(2.0 \, \text{g/cm}^3\), which is the correct answer.

Answer 2

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.