The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the least count of screw gauge?

0.01 cm

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0.01 mm

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0.1 mm

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0.1 cm

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Solution

The correct option is B
0.01 mm

Number of divisions on the circular scale = 50.

Pitch = Distance moved ahead in one revolution = $\frac{1 m m}{2}$ = 0.5 mm

L.C. = $\frac{P i t c h}{N u m b e r o f d i v i s i o n s o n t h e c i r c u l a r s c a l e}$ = ( $\frac{0.5}{50}$ ) mm = 0.01 mm

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The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the pitch of screw gauge?

Q. The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the least count of the screw gauge?

The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned two revolutions.

(i) What is the pitch of screw gauge ?

(ii) What is the least count of the screw gauge ?

Q. The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the pitch of screw gauge ?

Q. The thimble of a screw gauge has

50

division. The spindle advances

1 m m

when screw is turned through

2

revaluation. Find:
(1) the pitch of the screw gauge
(2) the least count gauge.

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Standard IX Physics

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The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions. What is the least count of screw gauge?

The correct option is B 0.01 mm Number of divisions on the circular scale = 50. Pitch = Distance moved ahead in one revolution = 1 mm2 = 0.5 mm L.C. = PitchNumber of divisions on the circular scale = (0.550) mm = 0.01 mm

The correct option is B
0.01 mm

Number of divisions on the circular scale = 50.

Pitch = Distance moved ahead in one revolution = $\frac{1 m m}{2}$ = 0.5 mm

L.C. = $\frac{P i t c h}{N u m b e r o f d i v i s i o n s o n t h e c i r c u l a r s c a l e}$ = ( $\frac{0.5}{50}$ ) mm = 0.01 mm