Physics, asked by shivashankarguttula, 7 months ago

A micrometer has 150 equal divisions on the circular
scale, and on one full rotation of circular scale main scale
advances by 0.02 cm on the main scale. The least count
of the micrometer (in cm) is
4x 10-5
6.21 x 10-3
1.33 × 10-4
1.21 x 10-2​

Answers

Answered by amitnrw
7

Given : A micrometer has 150 equal divisions on the circular scale, and on one full rotation of  circular scale main scale advances by 0.02 cm on the main scale.

To FInd : The least count of the  micrometer (in cm) is

Solution:

micrometer has 150 equal divisions on the circular scale, and on one full rotation of  circular scale main scale advances by 0.02 cm on the main scale.

1 main Scale = 150 Vernier scale Divisions

1 main Scale =  0.02 cm

150 Divisions = 0.02 cm

=> 1 Vernier scale Divisions = 0.02/150  cm

=>  1 Vernier scale Divisions = 2/15000  cm

=>  1 Vernier scale  Divisions = 2/(1.5 x 10⁴)   cm

=> 1 Vernier scale  Divisions = 1.33 x 10⁻⁴  cm

The least count of the  micrometer (in cm) is  1.33 x 10⁻⁴  cm

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Answered by nirman95
12

Given:

A micrometer has 150 equal divisions on the circular scale, and on one full rotation of circular scale main scale advances by 0.02 cm on the main scale.

To find:

Least Count of micrometre (screw gauge)?

Calculation:

First of all, what is LEAST COUNT?

  • Least count of a micrometre refers to the distance by which the main scale advances with rotation of the circular scale by 1 unit.

Using, this definition, we can say:

  \boxed{\bold{ \therefore \: LC =  \dfrac{pitch}{total \: number \: of \: circular \: division} }}

  • Pitch refers to the distance by which the main scale moves with full rotation of the circular scale.

Now, putting the values:

 \rm\implies \: LC =  \dfrac{0.2}{150}

 \rm\implies \: LC =  0.000133 \: cm

 \rm\implies \: LC =  1.33 \times  {10}^{ - 4}  \: cm

So, final answer is:

 \boxed{ \bold{ \: LC =  1.33 \times  {10}^{ - 4}  \: cm}}

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