First slide

Measurement

Question

A vernier callipers having main scale of $10 cm$ containing 100 divisions. If n be number of divisions on vernier scale of length l. Vernier calliper is designed to have a least count of $0.02 cm$ , then

Easy

A

$l = 0.8 cm, n = 10$
B

$l = 0.5 cm, n = 10$
C

$l = 0.2 cm, n = 10$
D

$l = 0.4 cm, n = 9$

Solution

Least count =1 Main scale division - 1 Vernier scale division
$0.02 = \frac{10}{100} - \frac{l}{n}$
$= 0.1 - \frac{l}{n}$
$\frac{l}{n} = 0.1 - 0.02 = 008$
If n=10, $l = 0.08 cm$

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