Question 73: Units , Dimensions and Error Analysis

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 $$1 \mathrm{~mm}$$. The main scale reading is $$2 \mathrm{~cm}$$ and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 $$\mathrm{g}$$, the density of the sphere is:

Accepted Answer

To find the density of the sphere, we need to calculate its volume using the measured diameter. Then, using the mass of the sphere, we can calculate the density using the formula for density, which is $$	ext{Density} = \frac{	ext{Mass}}{	ext{Volume}}$$. Let's start by finding the accurate measurement of the diameter using the given vernier calipers readings.
The least count (LC) of the vernier calipers can be calculated using the formula:
$$	ext{LC} = \frac{	ext{Value of one main scale division (MSD)}}{	ext{Number of vernier scale divisions (VSD) that match with the main scale}}$$
Given that 9 divisions of the main scale are equal to 10 divisions of the vernier scale and the shortest division on the main scale is equal to $$1\, \mathrm{mm}$$, we find:
$$	ext{LC} = \frac{1\, \mathrm{mm}}{10} = 0.1\, \mathrm{mm} = 0.01\, \mathrm{cm}$$
For the main scale reading (MSR) of $$2\, \mathrm{cm}$$ and the second division of the vernier scale coinciding with a division on the main scale, the vernier scale reading (VSR) can be expressed as $$	ext{VSR} = 2 	imes 	ext{LC}$$.
So, $$	ext{VSR} = 2 	imes 0.01\, \mathrm{cm} = 0.02\, \mathrm{cm}$$.
The total measurement of the diameter (D) can be found by adding MSR and VSR:
$$D = 	ext{MSR} + 	ext{VSR}$$
$$D = 2\, \mathrm{cm} + 0.02\, \mathrm{cm} = 2.02\, \mathrm{cm}$$
Now, we can calculate the volume (V) of the sphere using its diameter with the formula $$V = \frac{4}{3}\pi r^3$$, where $$r$$ is the radius of the sphere. Remembering that the radius is half of the diameter, $$r = \frac{D}{2} = \frac{2.02\, \mathrm{cm}}{2} = 1.01\, \mathrm{cm}$$.
Therefore, $$V = \frac{4}{3}\pi (1.01\, \mathrm{cm})^3$$,
Calculating the volume,
$$V = \frac{4}{3}\pi (1.01)^3\, \mathrm{cm}^3 \approx \frac{4}{3} 	imes 3.1416 	imes 1.0303\, \mathrm{cm}^3 = \frac{4}{3} 	imes 3.1416 	imes 1.0303\, \mathrm{cm}^3 \approx 4.3434\, \mathrm{cm}^3$$
Now to find the density ($$ho$$) using the mass ($$M = 8.635\, \mathrm{g}$$) and volume ($$V = 4.3434\, \mathrm{cm}^3$$) calculated,
$$ho = \frac{M}{V} = \frac{8.635\, \mathrm{g}}{4.3434\, \mathrm{cm}^3} \approx 1.988\, \mathrm{g}/\mathrm{cm}^3$$
Comparing this result to the given options, the closest value is:
Option B $$2.0\, \mathrm{g}/\mathrm{cm}^3$$

Answer

$$2.2 \mathrm{~g} / \mathrm{cm}^3$$

Answer

$$2.0 \mathrm{~g} / \mathrm{cm}^3$$

Answer

$$1.7 \mathrm{~g} / \mathrm{cm}^3$$

Answer

$$2.5 \mathrm{~g} / \mathrm{cm}^3$$

Units , Dimensions and Error Analysis

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