Home
Class 11
PHYSICS
A spherometer has 100 equal divisions ma...

A spherometer has 100 equal divisions marked along the periphery of its disc, and one full rotation of the disc advances on the main scale by 0.01cm. Find the least count of the system.

Text Solution

Verified by Experts

Given Pitch = 0.01cm
`therefore ` Least count = `("Pitch")/(" Total no. of divisions on the circular scale") = (0.01)/(100)` cm `= 10^(-4)` cm.
Promotional Banner

Topper's Solved these Questions

  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN |Exercise BEGINNER S BOX-1|2 Videos

  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN |Exercise BEGINNER S BOX-2|4 Videos

  • MISCELLANEOUS

    ALLEN |Exercise Question|1 Videos

  • SEMICONDUCTORS

    ALLEN |Exercise Part-3(Exercise-4)|50 Videos

Similar Questions

Explore conceptually related problems

A spherometer has 250 equal divisions marked along the periphery of its disc, and one full rotation of the disc advances by 0.0625 cm on the main scale. What is the least count of the spherometer ?

If the number of divisions on the circular scale is 100 and number of full rotations given to screw is 8 and distance moved by the screw is 4 mm, then what will be least count of the screw gauge ?

A vernier callipers has 20 divisions on the vernier scale which coincide with 19 divisions on the main scale. The least count of the instrument is 0.1 mm. The main scale divisions are of

In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with (n-1) divisions of the main scale. The least count (in cm) of the callipers is :-

One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is

An ammeter has a range (0 - 3A) and there are 30 divisions on its scale. Calculate the least count of the ammeter.

On cm on the main scale of vernier callipers is divided into ten equal parts. If 20 division of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?

If n^(th) division of main scale coincides with (n + 1)^(th) divisions of vernier scale. Given one main scale division is equal to 'a' units. Find the least count of the vernier.

In a given slide callipers 10 division of its vernier coincides with its 9 main scale divisions. If one main scale division is equal to 0.5 mm then find its least count.

Consider a Vernier callipers in which each 1cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions in its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :