Question

In an instrument, there are 25 divisions on the vernier scale which have length of 24 division of the main scale. 1 cm of main scale is divided into 20 equal parts. Find the least count.

Class 9 ICSE
Ch - Measurements and Experimentation.

Answer 1

Answer:

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Answer 2

Answer: Least count = 0.02 mm

Explanation: Given, No. of divisions on vernier scale (VSD) = 25

No. of divisions on main scale (MSD) = 24

25 VSD = 24 MSD, which means 1 VSD = $\frac{24}{25}$ MSD.

Now, Least count = 1 MSD - 1 VSD = 1 MSD - $\frac{24}{25}$ MSD =$\frac{1}{25}$ MSD = 0.04 MSD.

Also given that 1 cm of main scale is divided into 20 equal parts.

So, 1 MSD =  $\frac{1}{20}$   cm

Therefore, Least count = 0.04 MSD = 0.04 *$\frac{1}{20}$   cm = 0.002 cm = 0.02 mm

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