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1.Units, Dimensions and Measurement
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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