In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular divisions of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is: [2006]

Question

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(3)

$Least count = \frac{Pitch}{no. of divisions on circular scale} = \frac{0.5}{50} = 0.01 mm$

$Zero error = 5 \times L.C. = 5 \times 0.01 mm = 0.05 mm$

$Diameter of ball = [Reading on main scale] + [Reading on circular scale \times L.C.] - Zero error$

$= 0.5 \times 2 + 25 \times 0.01 - 0.05 = 1.20 mm$

Solution

Q.
In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular divisions of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is: [2006]

1 2.25 mm

2 2.20 mm

3 1.20 mm

4 1.25 mm

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