Question

For a given vernier callipers, 6 divisions of vernier

scale are equal to 4 divisions of main scale. If one

main scale division is of 1 mm, then the least count

of the vernier callipers is

(1) 1

mm

2

(2) 1

mm

6

(3) 1

mm

3

(4) 3 mm​

Answer 1

Answer:

Explanation:

6VSD=4MSD

VSD/MSD=4/6

LC = 1MSD - 1VSD

=MSD(1-VSD/MSD)

=1(1 - 4/6)

=2/6

=1/3

The method is correct I think options or the questions is wrong , bro if you typed the question wrong solve it yourself by using this method .

Answer 2

Answer:

The least count of the given vernier calipers is 1/3mm.

Explanation:

Least count of an instrument is the least measure that a device can be used to measure.

Given the value of 1 main scale division in a vernier calipers, $\mbox{M.S.D}= 1 \mbox{mm}$

6 divisions of vernier scale are equal to 4 divisions of main scale of a vernier calipers

$\implies 6\times 1\mbox{VSD}=4\times 1\mbox{MSD}$

$\implies 1\mbox{VSD}=\dfrac{4}{6} \times 1\mbox{MSD}$

$\implies 1\mbox{VSD}=\dfrac{4}{6} \times 1 \mbox{mm}=\dfrac{2}{3} \mbox{mm}$

Least count of the vernier calipers is given by the difference between the value of one main scale division and vernier scale division.

$\implies \mbox{LC}=1 \mbox{ M.S.D} - 1 \mbox{ V.S.D}$

             $=1 \mbox{ mm} - \mbox{ \dfrac{2}{3} \mbox{mm}}$

             $=\dfrac{3-2}{3} \mbox{mm}}=\dfrac{1}{3} \mbox{mm}}$

Therefore, the least count of the given vernier calipers is 1/3mm.

So, the correct option is 3.

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