In a hypothetical screw,guage pitch of screw guage is `0.5mm` and cap scale is , divided into `250` divisions then the least count of the screw guage is n times`10^(-3)mm` then `(5n)/(2)`=

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 pitch of a screw gauge is 0.5 mm. It's head scale contains 50 divisions. The least count of the screw gauge is

A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale: The pitch of the screw gauge is:

A screw gauge advances by 3mm in 6 rotations. There are 50 divisions on circular scale. Find least count of screw gauge

A screw gauge advances by 3mm in 6 rotations. There are 50 divisions on circular scale. Find least count of screw gauge

The distance advanced by the screw of a screw gauge is 2 mm in 4 rotations. Its cap is divided into 50 divisions. There is no zero error. If the screw reads 3 divisions on the main scale and 32 divisions of the cap, then the diameter (in mm ) of the wire is

In a screw gauge, 5 complete rotations of circular scale give 1.5 mm reading on pitch scale. Consider circular scale has 50 divisions. Least count of the screw gauge is

The circular head of a screw gauge is divided into 50 divisions and the screw moves 1 mm ahead in two revolutions of the circular head. Find its pitch .

In an internal micrometer, main scale division is of 0.5 mm and there are 50 divisions in circular scale. The least count of the instrument is-

1 cm of main scale of a vernier callipers is divided into 10 divisions. The least count of the callipers is 0.005 cm , then the vernier scale must have.