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…

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

Accepted Answer

Correct answer: D. (d) Diameter of the ball = MSR+ CSR (Least count) ‐Zero error =5 mm+25 0.001 cm‐(‐O.004) cm =0.5 cm+25 0.001 cm‐(‐O.004) cm =0.529 cm.

Solution

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