Question

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

• A. $0.01cm$
• B. $0.5mm$
• C. $0.01mm$
• D. $0.5cm$

Answer 1

Answer: (C)

$0.01mm$

According to the problem, $50$ divisions of vernier scale which coincide with $49$ main scale divisions.

$50VSD=49MSD$

$\Rightarrow MSD=\frac{50}{49}VSD$ or $1VSD=\frac{49}{50}MSD$

Where $MSD=$ main scale division and $VSD=$ Vernier scale division.

We know that

$minimuminaccuracy=vernierconstant$

$=1MSD-1VSD$

$=1MSD-\frac{49}{50}MSD=\frac{1}{50}MSD$

It is given in the problem that $1MSD=0.5mm$

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

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