JEE PYQ: Units & Measurements - Question ID fbac5f451cfd (JEE Main 2011)

In precision measurements using a screw gauge, the total reading is the sum of two parts:

• The main scale reading (MSR), which gives the whole millimeter value.
• The circular scale reading (CSR), which gives the fractional part in hundredths of a millimeter.

The key relationship is:

Here, 1 mm on the main scale corresponds to 100 divisions on the circular scale. Hence, each circular scale division represents:

1. Identify given values:
   Main scale reading (MSR) = 0 mm
   Circular scale reading (CSR) = 52 divisions
   Number of circular scale divisions per mm = 100
2. Compute the least count (LC):
   $\text{LC} = \frac{1\ \text{mm}}{100} = 0.01\ \text{mm}$
3. Calculate the circular scale contribution:
   $\text{Circular scale contribution} = \text{CSR} \times \text{LC} = 52 \times 0.01\ \text{mm} = 0.52\ \text{mm}$
4. Total diameter in millimeters:
   $\text{Diameter} = \text{MSR} + \text{Circular scale contribution} = 0\ \text{mm} + 0.52\ \text{mm} = 0.52\ \text{mm}$
5. Convert millimeters to centimeters:
   $1\ \text{mm} = 0.1\ \text{cm}$
   $0.52\ \text{mm} = 0.52 \times 0.1\ \text{cm} = 0.052\ \text{cm}$
6. Match with the given options:
   The computed diameter is $0.052\ \text{cm}$, which corresponds to option A.

• Unit confusion: Students often forget to convert millimeters to centimeters, leading to incorrect options like $0.52\ \text{cm}$ (option D).
• Least count misinterpretation: Some assume 1 circular division equals 1 mm instead of 0.01 mm, causing errors in fractional calculations.
• Zero error oversight: Although not applicable here, always check if the screw gauge has a zero error before taking readings.

Exam Tip: Always double-check unit conversions and ensure the least count is correctly calculated before proceeding with the measurement.