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.

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 5.045

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

In a screw gauge the zero of main scale reference line coincides...

Options

Correct answer

Step-by-step solution

More from Experimental Physics