Q132 / 180 · NEET 2026 (Cancelled) · Physics
In a vernier callipers, $20$ VSD coincide with $16$ MSD (each division of length $1$ mm). The least count of the vernier callipers is :
- A$0.2$ cm
- B$0.01$ cm
- Correct answerC$0.02$ cm
- D$0.1$ cm
Solution
Given $20$ VSD $= 16$ MSD.
$1$ VSD $= \dfrac{16}{20}$ MSD $= 0.8$ MSD.
Least count is given by $1$ MSD $- 1$ VSD.
Least count $= 1$ MSD $- 0.8$ MSD $= 0.2$ MSD.
Since $1$ MSD $= 1$ mm, we get Least count $= 0.2$ mm.
Converting to cm, Least count $= 0.02$ cm.
Answer: $0.02$ cm
$1$ VSD $= \dfrac{16}{20}$ MSD $= 0.8$ MSD.
Least count is given by $1$ MSD $- 1$ VSD.
Least count $= 1$ MSD $- 0.8$ MSD $= 0.2$ MSD.
Since $1$ MSD $= 1$ mm, we get Least count $= 0.2$ mm.
Converting to cm, Least count $= 0.02$ cm.
Answer: $0.02$ cm
Hence, the correct answer is option: 'C'.