The pitch and the number of circular scale divisions in a screw gauge with least count 0.02 mm are respectively

1 mm and 100

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 mm and 50

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 mm and 50

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.5 mm and 100

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 mm and 200

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 1 mm and 50
Least count $= \frac{p i t c h}{N u m b e r o f c i r c u l a r s c a l e d i v i s i o n s} = \frac{1 m m}{50} = 0.02 m m$
So, $1$ mm and $50$ is a correct answer.

Suggest Corrections

Similar questions

The head scale of a screw gauge has 100 divisions and the screw moves 4 mm ahead in four rotations of the circular head. The pitch and the least count are respectively

Q. The Circular scale of a screw gauge has 80 equal divisions, and its pitch is 2 mm. The least count of screw gauge is

Q. In a screw gauge, the main scale division is

1 mm

and there are

100

divisions on circular scale. The least count of screw gauge is (Pitch

= 1 mm

)

Q. In a screw gauge, the pitch of the screw is 0.5 mm and the number of circular scale divisions is 50, then the least count is __________ mm.

Q. In a screw gauge, the pitch of the screw is 0.5 mm and the number of circular scale divisions are 50, then the least count is ______

\times 10^{- 2}

mm.

Join BYJU'S Learning Program

The pitch and the number of circular scale divisions in a screw gauge with least count 0.02 mm are respectively

The correct option is A 1 mm and 50Least count =pitchNumber of circular scale divisions=1 mm50=0.02mmSo, 1 mm and 50 is a correct answer.

The correct option is A 1 mm and 50
Least count $= \frac{p i t c h}{N u m b e r o f c i r c u l a r s c a l e d i v i s i o n s} = \frac{1 m m}{50} = 0.02 m m$
So, $1$ mm and $50$ is a correct answer.