JEE PYQ: Units & Measurements - Question ID ac25ca94f42a (JEE Main 2022)

Core Formula & Concept:

To solve this problem, we need to understand the following key concepts and formulas:

1. Screw Gauge Measurement: A screw gauge is a precision instrument used to measure small lengths, such as the diameter of a wire. It consists of:
   • A main scale (linear scale) marked in millimeters.
   • A circular scale (rotating thimble) with divisions that allow for fractional measurements.
   The pitch of the screw gauge is the distance moved by the screw per complete rotation of the circular scale. Here, the pitch is given as $0.5 \text{ mm}$.
2. Least Count (LC): The least count is the smallest measurement that can be made with the screw gauge. It is calculated as: $\text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$ Here, the circular scale has $50$ divisions.
3. Total Reading: The total measurement is the sum of the main scale reading and the product of the circular scale reading and the least count: $\text{Total Reading} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times \text{Least Count})$
4. Curved Surface Area of a Cylinder: The wire is cylindrical in shape. The curved surface area (CSA) of a cylinder is given by: $\text{CSA} = \pi \times \text{Diameter} \times \text{Length}$ Here, the length of the wire is given as $6.8 \text{ cm}$.
5. Significant Figures: The final answer must be reported to the appropriate number of significant figures based on the precision of the measurements.

Step-by-Step Derivation:

Let's solve the problem step-by-step:

1. Calculate the Least Count (LC): Given:
   • Pitch = $0.5 \text{ mm}$
   • Number of divisions on circular scale = $50$
   $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}$


2. Determine the Total Diameter of the Wire: Given:
   • Main scale reading = $1.5 \text{ mm}$
   • Circular scale reading = $7$
   The total diameter is: $\text{Diameter} = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times \text{LC})$ Substituting the values: $\text{Diameter} = 1.5 \text{ mm} + (7 \times 0.01 \text{ mm}) = 1.5 \text{ mm} + 0.07 \text{ mm} = 1.57 \text{ mm}$


3. Convert Diameter to Centimeters: Since the length is given in centimeters, we convert the diameter to centimeters for consistency: $1.57 \text{ mm} = 0.157 \text{ cm}$


4. Calculate the Curved Surface Area (CSA): The formula for CSA is: $\text{CSA} = \pi \times \text{Diameter} \times \text{Length}$ Given:
   • Length = $6.8 \text{ cm}$
   • Diameter = $0.157 \text{ cm}$
   Substituting the values: $\text{CSA} = \pi \times 0.157 \text{ cm} \times 6.8 \text{ cm}$ Using $\pi \approx 3.1416$: $\text{CSA} = 3.1416 \times 0.157 \times 6.8$ First, multiply $0.157 \times 6.8$: $0.157 \times 6.8 = 1.0676 \text{ cm}^2$ Now, multiply by $\pi$: $\text{CSA} = 3.1416 \times 1.0676 \approx 3.354 \text{ cm}^2$


5. Round to Appropriate Significant Figures: The measurements given are:
   • Pitch = $0.5 \text{ mm}$ (1 significant figure, but this is a precise instrument specification, so we treat it as exact).
   • Main scale reading = $1.5 \text{ mm}$ (2 significant figures).
   • Circular scale reading = $7$ (1 significant figure, but this is a count, so it is exact).
   • Length = $6.8 \text{ cm}$ (2 significant figures).
   The least number of significant figures in the given data is 2 (from $1.5 \text{ mm}$ and $6.8 \text{ cm}$). Thus, the final answer should be rounded to 2 significant figures: $\text{CSA} \approx 3.4 \text{ cm}^2$

Common Traps & Exam Tip:

Students often make the following mistakes in this question:

1. Incorrect Least Count Calculation: Some students confuse the pitch with the least count or miscalculate the least count as $\frac{\text{Number of divisions}}{\text{Pitch}}$. Always remember: $\text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions}}$
2. Unit Inconsistency: Failing to convert all measurements to the same unit (e.g., mixing mm and cm) leads to incorrect results. Here, the diameter was in mm, while the length was in cm. Always ensure consistent units before calculations.
3. Significant Figures: Students often overlook the requirement to report the answer to the appropriate number of significant figures. The final answer must reflect the precision of the least precise measurement in the problem.
4. Circular Scale Reading Misinterpretation: Some students forget to multiply the circular scale reading by the least count or miscount the number of divisions. Always verify the circular scale reading and its contribution to the total measurement.
5. Curved Surface Area Formula: Confusing the formula for curved surface area with total surface area (which includes the area of the circular ends) is a common error. For a wire, only the curved surface area is relevant.

Exam Tip: Always double-check the units and significant figures in your final answer. Precision instruments like screw gauges require careful handling of measurements to avoid errors.

The correct answer is Option B: $3.4 \text{ cm}^2$.