1.Units, Dimensions and Measurement
medium

In a vernier callipers, each $cm$ on the main scale is divided into $20$ equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be $\dots \; \times 10^{-2} \;mm$

A

$3$

B

$5$

C

$7$

D

$9$

(JEE MAIN-2022)

Solution

$20 \; MSD =1 \; cm$

$1 \; MSD =\frac{1}{20} \; cm$

$10 \; VSD =9 \; MSD$

$1 \; VSD =\frac{9}{10} \; MSD$

$=\frac{9}{10} \times \frac{1}{20} \; cm$

$1 VSD =\frac{9}{200} \; cm$

VC $=1 MSD -1 \; VSD$

$=\frac{1}{20} \; cm -\frac{9}{200} \; cm$

$=\frac{1}{200} \times 10 \; mm$

VC $=5 \times 10^{-2} \; mm$

Ans. $5$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.