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प्रश्न
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उत्तर
Pitch of the screw gauge = 0.5mm = 0.05 cm
Circular scale divisions = 100
Least Count of screw gauge = pitch of the gauge/circular scale divisions
= 0.05/100
= 0.0005cm
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संबंधित प्रश्न
Define the least count of vernier callipers. How do you determine it?
Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.
Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.
Define metre in terms of the wavelength of light.
The final result of the calculation in an experiment is 125,347,200. Express the number in term of significant place when accuracy is between 1 and 1000
(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.
Which part of vernier callipers is used to measure the internal length of a hollow cylinder?
State the formula for calculating the least count of screw.
State whether the following statement is true or false by writing T/F against it.
The diameter of a wire can be measured more accurately by using vernier calipers than by a screw gauge.
What is the least count in the case of the following instrument?
Screw gauge
In a vernier calipers, (N + 1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is ______.
