Question

A student measured the diameter of a small steel ball using a screw gauge of least count 0.01 cm . The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauges has zero error of _ 0.04 cm , the correct diameter of the ball is

Answer 1

Answer:

Explanation:

Given -

MSR = 0.5 cm

VSR = 25

LC = 0.001 cm

Zero error = -0.0004 cm

# Solution -

Diameter of steel ball measured by student -

Measured diameter = MSR + VSR × LC

Measured diameter = 0.5 + 25 × 0.001

Measured diameter = 0.525 cm

True diameter of steel ball must be -

True diameter = Measured diamter - Zero error

True diameter = 0.525 - (-0.004)

True diameter = 0.529 cm

Hope this helps you...