For the determination of the refractive index of a glass slab,a travelling microscope is used whose main scale contains $300$ equal divisions equal to $15 \ cm$. The vernier scale attached to the microscope has $25$ divisions equal to $24$ divisions of the main scale. The least count $(LC)$ of the travelling microscope is (in $cm$):
- A$0.001$
- B$0.002$
- C$0.0005$
- D$0.0025$
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If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \,mm$, the Vernier constant of travelling microscope is:
$A$ screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: $0 \, mm$
Circular scale reading: $52$ divisions
Given that $1 \, mm$ on the main scale corresponds to $100$ divisions on the circular scale. The diameter of the wire from the above data is ...... $cm$.
Main scale reading: $0 \, mm$
Circular scale reading: $52$ divisions
Given that $1 \, mm$ on the main scale corresponds to $100$ divisions on the circular scale. The diameter of the wire from the above data is ...... $cm$.
MediumNEET 2021
View SolutionThe length of a cylinder measured by a Vernier caliper is given by the following observations: $3.29 \, cm, 3.28 \, cm, 3.29 \, cm, 3.31 \, cm, 3.28 \, cm, 3.27 \, cm, 3.29 \, cm, 3.30 \, cm$. The most accurate length of the cylinder is ........ $cm$.
Easy
View SolutionIn a vernier callipers, $50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division $= 0.05 \ mm$, then the least count of the vernier callipers is . . . . . . $mm$.
An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment,distances are measured by
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