The vernier scale of a travelling microscope has $50$ divisions which coincide with 49 main scale divisions. If each main scale division is $0.5$ mm, then the least count of the microscope is

0.01 c m

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0.5 m m

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0.01 m m

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0.5 c m

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Solution

The correct option is D $0.01 m m$
According to the problem, $50$ divisions of vernier scale which coincide with $49$ main scale divisions.
$50 V S D = 49 M S D$

$\Rightarrow M S D = \frac{50}{49} V S D$ or $1 V S D = \frac{49}{50} M S D$
Where $M S D =$ main scale division and $V S D =$ Vernier scale division.
We know that
$m i n i m u m i n a c c u r a c y = v e r n i e r c o n s t a n t$

$= 1 M S D - 1 V S D$

$= 1 M S D - \frac{49}{50} M S D = \frac{1}{50} M S D$

It is given in the problem that $1 M S D = 0.5 m m$

Hence, minimum inaccuracy $= \frac{1}{50} \times 0.5 m m$

$= \frac{1}{100} = 0.01 m m$

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The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, then the least count of the microscope is

The vernier scale of a travelling microscope has $50$ divisions which coincide with 49 main scale divisions. If each main scale division is $0.5$ mm, then the least count of the microscope is