JEE PYQ: Units & Measurements - Question ID 05374317ff82 (JEE Main 2024)

Core Formula & Concept:

To determine the density of the sphere, we use the fundamental definition of density:

$\rho = \frac{m}{V}$

where:

• $\rho$ is the density of the sphere,
• $m$ is the mass of the sphere, and
• $V$ is the volume of the sphere.

The volume of a sphere is given by:

$V = \frac{4}{3} \pi r^3 = \frac{\pi d^3}{6}$

where $d$ is the diameter of the sphere.

The key challenge in this problem is accurately measuring the diameter $d$ using a vernier caliper. A vernier caliper enhances precision by allowing measurements finer than the smallest division on the main scale. The least count (LC) of the vernier caliper is crucial and is calculated as:

$\text{Least Count (LC)} = \frac{\text{Value of 1 main scale division}}{\text{Number of vernier divisions}}$

In this case:

• 9 divisions of the main scale = 10 divisions of the vernier scale.
• 1 main scale division = 1 mm.

Thus, the least count determines the precision of the diameter measurement.

Step-by-Step Derivation:

Step 1: Determine the Least Count (LC) of the Vernier Caliper

Given:

• 9 main scale divisions = 10 vernier scale divisions.
• 1 main scale division = 1 mm.

Let $x$ be the length of 1 vernier division. Then:

$9 \times 1 \text{ mm} = 10 \times x$ $x = \frac{9}{10} \text{ mm} = 0.9 \text{ mm}$

The least count (LC) is the difference between 1 main scale division and 1 vernier division:

$\text{LC} = 1 \text{ mm} - 0.9 \text{ mm} = 0.1 \text{ mm}$

Step 2: Interpret the Vernier Caliper Reading

Given:

• Main scale reading = 2 cm = 20 mm.
• The second division of the vernier scale coincides with a main scale division.

The vernier scale contributes an additional measurement equal to:

$\text{Vernier contribution} = \text{Vernier division number} \times \text{LC} = 2 \times 0.1 \text{ mm} = 0.2 \text{ mm}$

Thus, the total diameter $d$ is:

$d = \text{Main scale reading} + \text{Vernier contribution} = 20 \text{ mm} + 0.2 \text{ mm} = 20.2 \text{ mm}$

Convert $d$ to centimeters:

$d = 20.2 \text{ mm} = 2.02 \text{ cm}$

Step 3: Calculate the Volume of the Sphere

Using the volume formula for a sphere:

$V = \frac{\pi d^3}{6}$

Substitute $d = 2.02 \text{ cm}$:

$V = \frac{\pi (2.02)^3}{6}$

Calculate $(2.02)^3$:

$(2.02)^3 = 2.02 \times 2.02 \times 2.02 = 4.0804 \times 2.02 \approx 8.2424 \text{ cm}^3$

Thus:

$V \approx \frac{3.1416 \times 8.2424}{6} \approx \frac{25.89}{6} \approx 4.315 \text{ cm}^3$

Step 4: Calculate the Density of the Sphere

Given mass $m = 8.635 \text{ g}$, the density $\rho$ is:

$\rho = \frac{m}{V} = \frac{8.635 \text{ g}}{4.315 \text{ cm}^3} \approx 2.0 \text{ g/cm}^3$ Common Traps & Exam Tip:

1. Misinterpreting the Vernier Scale: Students often confuse which scale division coincides. Here, the second vernier division coincides, not the first or any other. This directly affects the vernier contribution.
2. Unit Conversion Errors: Failing to convert millimeters to centimeters before calculating volume leads to incorrect density values. Always ensure consistent units.
3. Incorrect Volume Formula: Some students mistakenly use $V = \frac{4}{3} \pi r^3$ without converting diameter to radius, or forget the $\frac{4}{3}$ factor. Using $V = \frac{\pi d^3}{6}$ avoids this error.
4. Rounding Errors: Premature rounding of intermediate values (e.g., $2.02^3$) can lead to inaccuracies. Retain precision until the final step.
5. Least Count Miscalculation: A common mistake is assuming the least count is 0.01 mm or 0.9 mm. The correct LC is 0.1 mm, derived from the difference between main and vernier divisions.

Exam Tip: Always double-check the vernier coincidence and least count calculation. These are frequent sources of errors in measurement-based questions.