Q.

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

0.529 cm

 

B

0.521 cm

 

C

0.053 cm

 

D

0.525 cm

 


A.

(a) Measured diameter of the ball = Reading of screw gauge

      = MSR + VSR x LC + zero error

      = 0.5 cm + 25 x 0.001 cm + 0.004 cm = 0.529 cm

 



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