A screw gauge gives the following readings when used to measure the diameter of a wire Main scale reading: 0 mm Circular scale reading: 52 divisions Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is :

Question

A screw gauge gives the following readings when used to measure the diameter of a wire Main scale reading: 0 mm Circular scale reading: 52 divisions Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is : (A) 0.52 cm (B) 0.026 cm (C) 0.26 cm (D) 0.052 cm

Tardigrade Admin · Accepted Answer

Here, pitch of the screw gauge, p =1 mm Number of circular divisions, n =100 Thus, least count LC = p / n =1 / 100=0.01 mm =0.001 cm Diameter of the wire = MSR +( CSR × LC )=0+(52 × 0.001 cm )=0.052 cm

Q. A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading : $0 mm$
Circular scale reading : $52$ divisions
Given that $1 mm$ on main scale corresponds to $100$ divisions on the circular scale. The diameter of the wire from the above data is :

0.52 cm

0.026 cm

0.26 cm

0.052 cm

Here, pitch of the screw gauge, $p = 1 mm$
Number of circular divisions, $n = 100$
Thus, least count $L C = p / n = 1/100 = 0.01 mm = 0.001 c m$
Diameter of the wire $= MSR + (CSR \times L C) = 0 + (52 \times 0.001 c m) = 0.052 c m$

Q. A screw gauge gives the following readings when used to measure the diameter of a wire Main scale reading : 0mm Circular scale reading : 52 divisions Given that 1mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is :