First slide

Measurement

Question

$1 cm$ on the main scale of a vernier callipers is divided into 10 equal parts. If 10 divisions of vernier coincide with 8 small divisions of main scale, then the least count of the calliper is

Moderate

A

0.01 cm
B

0.02 cm
C

0.05 cm
D

0.005 cm

Solution

$1 MSD = \frac{1}{10} (1 cm) = 1 mm$
$10 VSD = 8 MSD$
$1 VSD = \frac{8}{10} MSD$
Least count = $1 MSD - 1 VSD$
$LC = 1 mm - \frac{8}{10} mm$
$LC = \frac{2}{10} mm$
$LC = \frac{2}{100} cm = 0.02 cm$

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