In a vernier callipers,each cm on the main scale is divided into $20$ equal parts. If tenth vernier scale division coincides with nineth main scale division.Then the value of vernier constant will be $\times 10^{- 2} mm$ .

Open in App

Solution

$L C = \frac{1 M S D}{V S D} = \frac{\frac{1}{20} c m}{10}$

$= \frac{1}{200} c m$

$= 5 \times 10^{- 2} m m$

Suggest Corrections

Similar questions

In a vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided in 10 parts. While measuring a length, the zero of the vernier lies just ahead of 1.8 cm mark and 4th division of vernier coincides with a main scale division.

(a) Find the length.

(b) If zero error of vernier callipers is - 0.02 cm, what is the correct length ?

Q. A vernier callipers has its main scale of

10 c m

equally divided into

200

equal parts. Its vernier scale of

25

divisions coincides with

12 m m

on the main scale. The least count of the instrument is -

Q. In a vernier callipers, 20 divisions on the vernier scale coincides with 17 divisions on the main scale. If one cm on the main scale is divided into 20 equal parts, then find the least count of the callipers.

Q. One centimetre on the main scale of vernier callipers is divided into ten equal parts . If 20 division of vernier scale coincide with 19 small division of the main scale then what will be the least count of the callipers

Q. Q. A Vernier callipers has its main scale of 1 cm equally divided into 10 equal parts. 20 vernier scale divisions is equal to 19 main scale divisions. Using this vernier a student measures the diameter of wire. He found that main scale reads 2 small divisions and 4th division of vernier scale coincide with a main scale division. The diameter of the wire is (a) 2.02 mm (b) 2. 2 mm (c) 2.5 mm (d) 2.1 mm

Join BYJU'S Learning Program

In a vernier callipers,each cm on the main scale is divided into 20equal parts. If tenth vernier scale division coincides with nineth main scale division.Then the value of vernier constant will be×10−2 mm.

LC=1MSDVSD=120cm10 =1200cm =5×10−2 mm

In a vernier callipers,each cm on the main scale is divided into $20$ equal parts. If tenth vernier scale division coincides with nineth main scale division.Then the value of vernier constant will be $\times 10^{- 2} mm$ .

$L C = \frac{1 M S D}{V S D} = \frac{\frac{1}{20} c m}{10}$

$= \frac{1}{200} c m$

$= 5 \times 10^{- 2} m m$