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:
- \(2.5 \, \text{g/cm}^3\)
- \(1.7 \, \text{g/cm}^3\)
- \(2.2 \, \text{g/cm}^3\)
- \(2.0 \, \text{g/cm}^3\)
The Correct Option is D
Approach Solution - 1
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:
- Calculate the least count of the vernier caliper.
- Determine the total reading from the vernier caliper.
- Calculate the volume of the sphere using the obtained diameter.
- 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.
Approach Solution -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.
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