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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 term 'Vernier constant'.
Draw a neat and labelled diagram of a screw gauge.
Name its main parts and state their functions.
Up to how many decimal places can a common vernier callipers measure the length in cm?
What do you understand by the term zero error?
State whether the following statement is true or false by writing T/F against it.
The metre scale, vernier calipers, and screw gauge are in decreasing order of least count.
Consider the following case where the zero of vernier scale and the zero of the main scale are clearly seen. If L.C. of the vernier calipers is 0.01 cm, write the zero error and zero correction of the following.

