The given diagram represents a screw gauge. The circular scale is divided into $50$ divisions and the linear scale is divided into millimetres. If the screw advances by $1 mm$ when the circular scale makes $2$ complete revolutions, find the least count of the instrument and the reading of the instrument in the figure.

Open in App

Solution

Suggest Corrections

Similar questions

Q. The accompanying diagram represents a screw gauge. The circular scale is divided into

50

divisions and the linear scale is divided into millimeters. If the screw advances by

1 m m

when the circular scale makes

2

complete revolutions, the least count of the instrument and the reading of the instrument in figure are:

Find the least count of a screw gauge whose least count of linear scale is 1 mm and the circular scale is divided in 50 divisions and 2 entire rotations makes to 1 division on linear scale.

Q. In a screw gauge,

5

complete rotations of circular scale give \

1.5 m m

reading on linear scale. Circular scale has

50

divisions. The least count of the screw gauge is

x \times 10^{y}

(m m)

. Find

(x + y)

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. Pitch of a screw gauge is 1 mm and there are 50 divisions on circular scale. When screw gauge is completely closed

28^{t h}

division of circular scale coincides with reference line. When a sheet is placed in it, the linear scale reads 2 divisions and circular scale reads 16 divisions. Zero of linear scale is visible when screw gauge is completely closed. Thickness of the sheet is:

Join BYJU'S Learning Program