Question

22. One full rotation of the cap of a screw gauge is equivalent to 5mm. The cap has 100 division. Find the least count.
(1) 0.5 m
(2) 0.05 mm (3) 0.005 m (4) 5x10m​

Answer 1

10 rotations of the cap of screw guage is equivalent to 5mm. The cap has 100 divisions. Find the least count. A reading taken for the diameter of wire with the screw guage shows four complete rotations and 35 divisions on the circular scale. Find the diameter of the wire.

Answer

Least count =

100

0.5

=0.005mm

The diameter of the wire = (4×0.5+35×0.005)mm=2.175mm

Answer 2

Answer:

The required least count is $0.05mm$. Thus, the correct answer is option $(2)$ $0.05 mm$.

Explanation:

Concept:

The distance the spindle moves in one revolution is known as the screw's pitch. For a specific number of full screw spins, the distance the head scale advanced over the pitch scale is calculated in order to determine this. Additionally, the least count is the distance covered by the screw tip when it rotates through one division of the head scale.

Given:

A screw gauge top rotates completely once for every $=5mm$.

The division of the cap $=100$

To find:

We have to find out the least count.

Solution:

It is provided that 1 full rotation is equal to $5mm$.

Therefore,

$1$ rotation $=5mm$

least count is, reading in one rotation divided by total divisions. So,

Divisions count$=100$

Least count$=\frac{5}{100}$

$=0.05mm$

The required least count is $0.05mm$.

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