Question

Let main scale division of a vernier callipers is 1 mm. With zero of vernier coinciding with the zero of main scale, the ${50}^{th}$ division of vernier scale coincides with ${49}^{th}$ division of main scale. The least count of the vernier is

• A. 0.1 mm
• B. 0.01 mm
• C. 0.2 mm
• D. 0.02 mm

Answer 1

Answer: (D)

0.02 mm

$1M.S.D=1mm1V.S.D=\frac{49}{50}\times 1mm\therefore L.C=1M.S.D-1V.S.D=-\frac{49}{50}mm=\frac{1}{50}mm=0.02mm$