JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 16)
One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with (n $$-$$ 1)th division of the main scale. The least count of the callipers in mm is :
$${{10a} \over n}$$
$${{10na} \over {(n - 1)}}$$
$$\left( {{{n - 1} \over {10n}}} \right)a$$
$${{10a} \over {(n - 1)}}$$
Explanation
n VSD = (n $$-$$ 1) MSD
1 VSD = $$\left( {{{n - 1} \over n}} \right)$$MSD
L.C. = 1 MSD $$-$$ 1 VSD
= 1 MSD $$-$$ $$\left( {{{n - 1} \over n}} \right)$$MSD
= 1 MSD $$-$$ 1 MSD + $${{MSD} \over n}$$
= $${{MSD} \over n}$$
= $${a \over n}$$ cm
= $${{10a} \over n}$$ mm
1 VSD = $$\left( {{{n - 1} \over n}} \right)$$MSD
L.C. = 1 MSD $$-$$ 1 VSD
= 1 MSD $$-$$ $$\left( {{{n - 1} \over n}} \right)$$MSD
= 1 MSD $$-$$ 1 MSD + $${{MSD} \over n}$$
= $${{MSD} \over n}$$
= $${a \over n}$$ cm
= $${{10a} \over n}$$ mm
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