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प्रश्न
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.
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उत्तर
No. of circular scale divisions = 200
Pitch = 1 mm
Least count (L.C.) = `"Pitch"/"No. of circular divisions"`
= `(1 "mm")/200`
= 0.005 mm
L.C. = 0.0005 cm
Main scale reading = 5 mm = 0.5 cm
(C.S.D.) circular scale reading = 34 divisions
Observed diameter of wire = Main scale reading + L.C. × C.S.D.
= 0.5 + 0.0005 × 34
= 0.5 + 0.0170
= 0.5170 cm
Redius of wire = `"diameter"/2`
= `0.5170/2`
= 0.2585 cm
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संबंधित प्रश्न
Define the least count of vernier callipers. How do you determine it?
Define metre according to the old definition.
(a) A vernier scale has 20 divisions. It slides over the main scale, whose pitch is 0.5 mm. If the number of divisions on the left hand of the zero of vernier on the main scale is 38 and the 18th vernier scale division coincides with the main scale, calculate the diameter of the sphere, held in the jaws of vernier callipers.
(b) If the vernier has a negative error of 0.04 cm, calculate the corrected radius of the sphere.
State the formula for calculating length if:
The reading of the main scale is known and the number of vernier scale divisions coinciding with the main scale is known.
State the correction if the negative error is 7 divisions when the least count is 0.01 cm.
State whether the following statement is true or false by writing T/F against it.
The least count of a screw gauge can be lowered by increasing the number of divisions on its thimble.
What is the least count in the case of the following instrument?
Spring balance
