Experimental Physics

JEE Main level, single correct, medium

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Question 2 of 70

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 100100 divisions in circular scale and pitch of screw gauge is 0.10.1 mm. When diameter of a sphere is measured, the reading of main scale is 55 mm and 50th50^{\text{th}} division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.

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Answer: (A) 5.0455.045

Least count of the screw gauge is given by:

LC=PitchNumber of divisions on circular scale\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}

LC=0.1100=0.001 mm\text{LC} = \dfrac{0.1}{100} = 0.001 \text{ mm}

When the two studs are in contact, the 5th5^{\text{th}} division of the circular scale coincides with the reference line. This indicates a positive zero error.

Zero error=+5×LC=+5×0.001=+0.005 mm\text{Zero error} = +5 \times \text{LC} = +5 \times 0.001 = +0.005 \text{ mm}

For the measurement of the sphere's diameter:

Main scale reading (MSR)=5 mm\text{Main scale reading (MSR)} = 5 \text{ mm}

Circular scale reading (CSR)=50\text{Circular scale reading (CSR)} = 50

Measured diameter=MSR+CSR×LC\text{Measured diameter} = \text{MSR} + \text{CSR} \times \text{LC}

Measured diameter=5+50×0.001=5.050 mm\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:

True diameter=Measured diameter−Zero error\text{True diameter} = \text{Measured diameter} - \text{Zero error}

True diameter=5.050−0.005=5.045 mm\text{True diameter} = 5.050 - 0.005 = 5.045 \text{ mm}

Answer: 5.0455.045

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