Question

1.
If in a Vernier callipers 10 VSD coincides with 8 MSD, then the least count of Vernier calliper is (give
1 MSD = 1 mm)
(1) 1 x 104 m
(2) 2 x 104 m
(3) 1 x 10-3 m
(4) 8 x 104 m​

Answer 1

Answer: 2 x 10^-4

Explanation: 10 vernier division = 8 MSD

                                                       =8 X 1 mm =8 mm

                                             1 VSD = 8/1O mm

                              LC of vernier is =value of MSD - value of VSD

                                                        =    1   -   8/10

                                                        =   2  x 10^-4 m

Answer 2

The least count of Vernier caliper is $2\times10^{-4}\ m$.

Explanation:

The least count of a vernier caliper is given by :

Least count = smallest division of main scale/total no of division of vernier scale

It is also equal to the difference between vernier scale division (VSD) and main scale division (MSD).

It is given that, in a Vernier caliper 10 VSD coincides with 8 MSD

$1\ VSD=\dfrac{8}{10}\ MSD=0.8\ MSD$

Since, 1 MSD = 1 mm

Least count = 1 MSD - 1 VSD

Least count = 1 MSD - 0.8 MSD

Least count = 0.2 MSD

or

Least count = 0.2 mm

Least count = 0.0002 m

$Least\ count=2\times10^{-4}\ m$

So, the least count of Vernier caliper is $2\times10^{-4}\ m$. Hence, this is the required solution.

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