A student measured the diameter of a wire using a screw gauge with least count 0.001 cm and listed the measurements. The correct measurement is-
(1) 5.3 cm
(2) 5.32 cm
(3) 5.320 cm
(4) 5.3200 cm

Hint: Here only the least count of the screw gauge is given. The least count of the screw gauge is the distance through which the screw advances when it is rotated through one division of the head scale. Hence the diameter of the wire can be written as the multiples of 0.001cm. Thus we will get the solution.
Formula used:
The least count of the screw gauge is,
$L.C=\dfrac{Pitch}{n}$
where, n is the no. of divisions on the head scale.
Pitch of a screw is the distance through which it advances in one complete rotation of the head scale.
$Total\operatorname{Re}ading=P.S.R+\left( correctedH.S.R\times L.C \right)$
where, PSR is the pitch scale reading
HSR is the head scale reading
LC is the least count

Complete answer:
The least count of the screw gauge is the distance through which the screw advances when it is rotated through one division of the head scale.
The least count of the screw gauge is,
$L.C=\dfrac{Pitch}{n}$
where, n is the no. of divisions on the head scale.
Pitch of a screw is the distance through which it advances in one complete rotation of the head scale.
Given that the least count is 0.001 cm. Hence diameter of the wire can be written as the multiples of 0.001cm. Thus 5.320 cm is 5320 times 0.001cm.

Therefore option (C ) is correct.

Note:
The screw gauge is also known as the micrometer. The principle employed in the screw gauge is that the linear distance moved by the screw is directly proportional to the rotation given to the head. This is the principle of the micrometer screw. Screw gauge is a device generally used to measure the diameter of wires. When the screw tip is at the zero line of the head scale and does not coincide with the reference line of the pitch scale, then there is a zero error.