Question

Least count of micrometer screw gauge is 10 X 10-6 m and total 100 divisions on its circular scale. So, pitch distance of M.S.G is _____?​

Answer 1

Explanation:

we know that

pitch

least count = ———-—-——–—-------

no. of division on CS

10 x 10^-6 m = pitch /100

pitch = 10 x 10 ^-4 m

Answer 2

Answer:

Pitch distance of M.S.G is

${10}^{ - 3} m$

Explanation:

The least count of screw gauge is equal to the ratio of pitch of the screw to the number of the division on circular scale

$least \: count = \frac{pitch}{no \: of \: divisions \: on \: cs}$

Given,

Least count of micrometre screw gauge =10 X 10-6 m

Total no. of divisions on its circular scale=100

$10 \times {10}^{ - 6} = \frac{pitch \: distance}{100}$

$pitch \: distance = 10 \times {10}^{ - 6} \times 100$

$pitch \: distance = {10}^{ - 3}$

Pitch distance of M.S.G is

.

${10}^{ - 3} m$

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