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What is least count of the given vernier...

What is least count of the given vernier caliper?
There is one vernier caliper in which 1 main scale division (MSD) is 1mm. It is given that 10 divisions of vernier scale match with 9 divisions of main scale. Given vernier caliper is used to measure the edge of a cube. Main scale reading (MSR) was found to be 10 and first division of the vernier scale was found to be coinciding with some division of the main scale. Mass of the cube is 2.736g.

A

`0.1 mm`

B

`0.1cm`

C

`0.01mm`

D

`1 mu m`

Text Solution

Verified by Experts

The correct Answer is:
A
1 MSD = 1mm.
10 vernier scale divisions match with 9 main scale divisions hence `1 VSD= (9)/(10)=0.9mm`
Least count of the vernier caliper can be written as follows ,
`LC= 1MSD -1VSD= 1-0.9= 0.1mm`.
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Knowledge Check

  • What is the edge length of the cube? There is one vernier caliper in which 1 main scale division (MSD) is 1mm. It is given that 10 divisions of vernier scale match with 9 divisions of main scale. Given vernier caliper is used to measure the edge of a cube. Main scale reading (MSR) was found to be 10 and first division of the vernier scale was found to be coinciding with some division of the main scale. Mass of the cube is 2.736g.

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    `1.01mm`
    B
    `1.01cm`
    C
    `10.1cm`
    D
    `1.001cm`

  • The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n-1) divisions of main scale. The least count of the vernier callipers is,

    A
    `(1)/(n(n+1))cm`
    B
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    B
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    C
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    D
    `(x/(n - 1))`

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