The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge?

0.005 cm

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

0.005 mm

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

0.0005 mm

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

0.05 mm

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

Open in App

Solution

The correct option is B
0.005 mm

Pitch of the screw gauge = 0.5 mm

Number of divisions on circular scale = 100

$L . C . = \frac{P i t c h o f t h e s c r e w g a u g e}{N u m b e r o f d i v i s i o n s o n c i r c u l a r s c a l e}$ = $\frac{0.5}{100}$ mm

= 0.005 mm or 0.0005 cm

Suggest Corrections

Similar questions

The pitch of a scrw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge ?

Q. The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge?

Q. The circular scale (head scale) of a screw gauge is divided into

100

equal parts and it moves

0.5 m m

ahead in one rotation. The pitch and the least count are:

Q. What is the least count of a screw gauge if the head scale of a screw gauge contains 100 divisions and its pitch is 1 mm?

It is found that for a screw gauge, the $5^{t h}$ division of the head scale coincides with the pitch scale axis and the zero of the head scale lie below the pitch scale axis. The least count of the screw gauge is 0.01 mm. What is the zero correction?

Join BYJU'S Learning Program

The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge?

The correct option is B 0.005 mm Pitch of the screw gauge = 0.5 mm Number of divisions on circular scale = 100 L.C.=Pitch of the screw gaugeNumber of divisions on circular scale = 0.5100 mm = 0.005 mm or 0.0005 cm

The correct option is B
0.005 mm

Pitch of the screw gauge = 0.5 mm

Number of divisions on circular scale = 100

$L . C . = \frac{P i t c h o f t h e s c r e w g a u g e}{N u m b e r o f d i v i s i o n s o n c i r c u l a r s c a l e}$ = $\frac{0.5}{100}$ mm

= 0.005 mm or 0.0005 cm