$10$ divisions on the main scale of a Vernier calliper coincide with $11$ divisions on the Vernier scale. If each division on the main scale is of $5$ units, the least count of the instrument is :
A$\frac{1}{2}$
B$\frac{10}{11}$
C$\frac{50}{11}$
D$\frac{5}{11}$
JEE MAIN 2024, Diffcult
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D$\frac{5}{11}$
d $10 \mathrm{MSD}=11 \mathrm{VSD}$
$1 \mathrm{VSD}=\frac{10}{11} \mathrm{MSD}$
$\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}$
$=1 \mathrm{MSD}-\frac{10}{11} \mathrm{MSD}$
$=\frac{1 \mathrm{MSD}}{11}$
$-\frac{5}{11} \text { units }$
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