Question

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference leveL If screw gauge has a zero error of— 0.004 cm, the correct diameter of the ball is 

(A) 0.521 cm 

(B) 0.525 cm 

(C) 0.053 cm 

(D) 0.529 cm

Answer 1

(D) 0.529 cm

Least count of screw gauge = 0.001 cm = 0.01mm

Main scale reading = 5 mm.

Zero error = – 0.004 cm = -0.04 mm

Zero correction = +0.04 mm

Observed reading = Mainscale reading + (Division × least count)

Observed reading = 5 + (25 × 0.01) = 5.25 mm

Corrected reading = Observed reading + Zero correction

Corrected reading = 5.25 + 0.04

= 5.29 mm = 0.529 cm