Question

$\large{\underline{\sf{\red{Question-}}}}$

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

a) 2.25 mm
b) 2.20 mm
c) 1.20 mm
d) 1.25 mm

$\rule{200}2$

Thank You.

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Answer 1

Answer:

(c) 1.20 mm

Explanation:

From figure (i), it is clear that screw gauge has a zero error.

Least count of screw gauge = Reading of one division of main scale/Total number of divisions of circular scale =0.5/50=0.01mm

Therefore, diameter of the ball = (M.S. Reading + Least count (Circular division)− zero error)

                                                    =[1+0.01[25]−5]=1.20 mm

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Answer 2

Answer:

(c) 1.20 mm

Explanation:

Formula :

For LC of S G =

Reading of one div. of Main Scale / Total no. of div. of Cir. Scale.

For S G , Reading = MSR + LC(Circular Div count) - Zero error

Ty !!