Q57 / 75 · JEE Main 2026 (02 April Shift 2) · Physics
In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are $100$ divisions in circular scale and pitch of screw gauge is $0.1$ mm. When diameter of a sphere is measured, the reading of main scale is $5$ mm and $50^{\text{th}}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.
- Correct answerA$5.045$
- B$5.055$
- C$5.450$
- D$5.550$
Solution
Least count of the screw gauge is given by:
$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
$\text{LC} = \dfrac{0.1}{100} = 0.001 \text{ mm}$
When the two studs are in contact, the $5^{\text{th}}$ division of the circular scale coincides with the reference line. This indicates a positive zero error.
$\text{Zero error} = +5 \times \text{LC} = +5 \times 0.001 = +0.005 \text{ mm}$
For the measurement of the sphere's diameter:
$\text{Main scale reading (MSR)} = 5 \text{ mm}$
$\text{Circular scale reading (CSR)} = 50$
$\text{Measured diameter} = \text{MSR} + \text{CSR} \times \text{LC}$
$\text{Measured diameter} = 5 + 50 \times 0.001 = 5.050 \text{ mm}$
The true diameter is obtained by subtracting the zero error from the measured value:
$\text{True diameter} = \text{Measured diameter} - \text{Zero error}$
$\text{True diameter} = 5.050 - 0.005 = 5.045 \text{ mm}$
Answer: $5.045$
$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
$\text{LC} = \dfrac{0.1}{100} = 0.001 \text{ mm}$
When the two studs are in contact, the $5^{\text{th}}$ division of the circular scale coincides with the reference line. This indicates a positive zero error.
$\text{Zero error} = +5 \times \text{LC} = +5 \times 0.001 = +0.005 \text{ mm}$
For the measurement of the sphere's diameter:
$\text{Main scale reading (MSR)} = 5 \text{ mm}$
$\text{Circular scale reading (CSR)} = 50$
$\text{Measured diameter} = \text{MSR} + \text{CSR} \times \text{LC}$
$\text{Measured diameter} = 5 + 50 \times 0.001 = 5.050 \text{ mm}$
The true diameter is obtained by subtracting the zero error from the measured value:
$\text{True diameter} = \text{Measured diameter} - \text{Zero error}$
$\text{True diameter} = 5.050 - 0.005 = 5.045 \text{ mm}$
Answer: $5.045$
Hence, the correct answer is option: 'A'.