Question

The pitch of a screw gauge is 0.1 cm. The number of divisions on its circular scale is 100. In the measurement of diameter of a wire with this screw gauge the linear scale reading is 'N' cm and the number of division on the reference line is n. Then the radius of the wire in cm will be –

• A. N + 0.01 n
• B. N + 0.001 n
• C. 0.5 N + 0.001 n
• D. 5( 0.1 N + 0.0001 n)

Answer 1

Answer: (D)

5( 0.1 N + 0.0001 n)

Answer 2

Least count = $\frac { 0.1 } { 100 }$

2r = N + n ( $\frac { 0.1 } { 100 }$ )

r = $\frac { \mathrm { N } } { 2 }$ + $\frac { \mathrm { n } } { 2 }$ (0.001)

r = 0.5 N + n (0.0005)

r = 5 (0.1 N + 0.0001 n)