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

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) 0.529 cm (B) 0.521 cm (C) 0.053 cm (D) 0.525 cm

Tardigrade Admin · Accepted Answer

Diameter of the ball = MSR + CSR × (Least count) - Zero error =0.5 cm +25 × 0.001-(-0.004) =0.5+0.025+0.004 =0.529 cm

0.529 cm

0.521 cm

0.053 cm

0.525 cm

Diameter of the ball =MSR+CSR× (Least count) - Zero error =0.5cm+25×0.001−(−0.004) =0.5+0.025+0.004 =0.529cm

Diameter of the ball
$= MSR + CSR \times$ (Least count) - Zero error
$= 0.5 c m + 25 \times 0.001 - (- 0.004)$
$= 0.5 + 0.025 + 0.004$
$= 0.529 c m$