JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2026
- Medium
- MCQ
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.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(5.045\)
Step-by-step Solution
Detailed explanation
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: \(5.045\)
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