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 $= 1 mm$ )

0.01 mm

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

0.1 mm

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

0.02 mm

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

0.2 mm

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

Open in App

Solution

The correct option is A $0.01 mm$
Least count of a screw gauge is given by,

$L.C. = \frac{Pitch}{Number of circular scale divisions}$

$= \frac{1 mm}{100}$

$= 0.01 mm$

Suggest Corrections

Similar questions

Q. The pitch of a screw gauge is

0.5 mm

and there are

50

divisions on its circular scale. The screw gauge has the least count of

Q. A hypothetical screw gauge has a pitch of

0.2 mm

and

10

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

Consider a Vernier calipers in which each $1 c m$ on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers, $5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then

Q. if the pitch of the screw gauge is 1 mm and there are 200 divisions on the circular scale, the least count of the screw gauge will be

Q. The least count of a screw gauge having 1 mm pitch and 100 circular scale divisions is ____

μ m

Join BYJU'S Learning Program

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)

The correct option is A 0.01 mmLeast count of a screw gauge is given by, L.C.=PitchNumber of circular scale divisions =1 mm100 =0.01 mm

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 $= 1 mm$ )

The correct option is A $0.01 mm$
Least count of a screw gauge is given by,

$L.C. = \frac{Pitch}{Number of circular scale divisions}$

$= \frac{1 mm}{100}$

$= 0.01 mm$