Question

If in vernier callipers 4 VSD coincides with 2 MSD, then least count of vernier calliper is
0 0.2 mm
0 0.1 mm
0 0.3 mm
O 0.5 mm​

Answer 1

plz check the 4th option is it 0.05mm or 0.5mm I m getting 0.05mm

hope it helps...

Explanation:

$4vsd \: = 2msd \\ vsd = \frac{1}{2} msd \\ least \: count = msd \: - vsd \\ l.c.= msd - \frac{1}{2}msd \: \\ l.c. = \frac{1}{2} msd \\ l.c. = \frac{1}{2} cm \: or \: 0.5cm \: or \: 0.05mm$

Answer 2

Given:

4 VSD coincides with 2 MSD.

To Find:

The least count of the vernier calliper.

Calculation:

- According to the question:

4 VSD = 2 MSD

⇒ 1 VSD = 2/4 MSD

⇒ 1 VSD = 1/2 MSD

- The least count is given as:

LC = 1 MSD - 1 VSD

⇒ LC = 1 MSD - 1/2 MSD

⇒ LC = 1/2 MSD

⇒ LC = 0.5 MSD

- If 1 MSD = 1 mm

⇒ LC = 0.5 mm

- So, the correct answer is option (d) 0.5 mm.