Least count of a vernier caliper is 1/20N cm . The value of one division on the main scale is 1 mm. Then the number of divisions ​

Question

Least count of a vernier caliper is \(\frac{1}{20\text{N}}\ \text{cm}\) . The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is : 

(1) \(\left(\frac{2 \mathrm{N}-1}{20 \mathrm{N}}\right)\)

(2) \(\left(\frac{2 \mathrm{N}-1}{2}\right)\)

(3) \((2 \mathrm{N}-1)\)

(4) \(\left(\frac{2 \mathrm{N}-1}{2 \mathrm{N}}\right)\)    

Answer 1

Correct option is :  (2) \( \left(\frac{2 \mathrm{N}-1}{2}\right)\) 

Least count of vernier calipers \(=\frac{1}{20 \mathrm{N}} \mathrm{~cm}\)

\(\because\) Least count\( =1 \ \mathrm{MSD}-1 \ \mathrm{VSD}\)

let x no. of divisions of main scale coincides with N division of vernier scale, then

\(1 \ \mathrm{VSD}=\frac{\mathrm{x} \times 1 \mathrm{mm}}{\mathrm{~N}}\)

\(\therefore \frac{1}{20 \mathrm{N}} \mathrm{cm}=1 \ \mathrm{mm}-\frac{\mathrm{x} \times 1 \mathrm{mm}}{\mathrm{~N}}\)

\(\frac{1}{2 \mathrm{N}} \mathrm{mm}=1 \mathrm{mm}-\frac{\mathrm{x}}{\mathrm{N}} \mathrm{mm}\)

\(\mathrm{x}=\left(1-\frac{1}{2 \mathrm{N}}\right) \mathrm{N}\)

\(\mathrm{x}=\frac{2 \mathrm{N}-1}{2}\)