JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 2)
If n main scale divisions coincide with (n + 1) vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into five equal parts, will be :
$${2 \over {n + 1}}$$ mm
$${5 \over {n + 1}}$$ mm
$${1 \over {2n}}$$ mm
$${1 \over {5n}}$$ mm
Explanation
5 parts of main scale division = 1 cm
$$\therefore$$ 1 part of main scale division = $${1 \over 5}$$ cm
$$\therefore$$ 1 M.S.D. = $${1 \over 5}$$ cm
(n + 1) vernier scale division = n main scale division.
$$\therefore$$ 1 V.S.D. = $${n \over n+1}$$ M.S.D.
= $${n \over n+1}$$ $$\times$$ 1 M.S.D.
= $${n \over n + 1}$$ $$\times$$ $${1 \over 5}$$ cm
We know,
L.C. = 1 M.S.D. $$-$$ 1 V,S.D.
= $${1 \over 5}$$ cm $$-$$ $${n \over {5(n + 1)}}$$ cm
= $${{n + 1 - n} \over {5(n + 1)}}$$ cm
= $${1 \over {5(n + 1)}}$$ cm
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