JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 24)
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 _________ $$\times$$ 10$$-$$2 mm.
답변
5
설명
1 MSD = $${1 \over {20}}$$ cm
$$\because$$ 10 VSD = 9 MSD
1 VSD = $${9 \over {10}}$$ $$\times$$ $${1 \over {20}}$$ cm = $${9 \over {200}}$$ $$\times$$ 10 mm = 0.45 mm
Now, 1 MSD = $${1 \over {20}}$$ $$\times$$ 10 mm = 0.50 mm
LC = (0.50 $$-$$ 0.45) mm = 0.05 mm
= 5 $$\times$$ 10$$-$$2 mm