The main scale reading of a screw gauge is 8 mm and the circular scale coincides with 25th division of the main scale. If the pitch of the screw gauge is 0.5 mm and the number of circular scale division is 100, then the observed reading will be

(A) 8.005 mm
(B) 8.025 mm
(C) 8.125 mm
(D) 8.000 mm

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Solution

$D e a r s t u d e n t, L . C = \frac{p i t c h}{n o o f d i v i s i o n s o n m a i n s c a l e} = \frac{0.5}{100} = 0.005 m m R e a d i n g = 8 + 25 \times 0.005 = 8.125 R e g a r d s$

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The main scale reading of a screw gauge is 8 mm and the circular scale coincides with 25th division of the main scale. If the pitch of the screw gauge is 0.5 mm and the number of circular scale division is 100, then the observed reading will be (A) 8.005 mm (B) 8.025 mm (C) 8.125 mm (D) 8.000 mm

Dear student,L.C=pitchno of divisions on main scale=0.5100=0.005mmReading=8+25×0.005=8.125Regards

The main scale reading of a screw gauge is 8 mm and the circular scale coincides with 25th division of the main scale. If the pitch of the screw gauge is 0.5 mm and the number of circular scale division is 100, then the observed reading will be

(A) 8.005 mm
(B) 8.025 mm
(C) 8.125 mm
(D) 8.000 mm

$D e a r s t u d e n t, L . C = \frac{p i t c h}{n o o f d i v i s i o n s o n m a i n s c a l e} = \frac{0.5}{100} = 0.005 m m R e a d i n g = 8 + 25 \times 0.005 = 8.125 R e g a r d s$