MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Vernier Caliper

A vernier has...

Question

A vernier has 10 divisions and they are equal to 9 divisions of the main scale in length. If the main scale is calibrated in mm, what is its least count?

Open in App

Solution

Step 1, Given data
Value of 1 m.s.d. = 1 mm
10 vernier divisions = 9 m.s.d.
Step 2, Finding the least count
We know,
L.C. = $\frac{V a l u e o f 1 m s d}{N u m b e r o f d i v i s i o n s o n v e r n i e r s c a l e}$
Putting all the values
$L C = \frac{1}{10} = 0.1$
Hence the least count is 0.1 mm or 0.01 cm

Suggest Corrections

thumbs-up

128

Similar questions

Q. The main scale of a vernier callipers is calibrated in mm and 19 divisions of main scale are equal in length to 20 divisions of vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and 4th division of vernier scale coincides with a main scale division. The least count is

5 \times 10^{- x} c m

. Find the value of

x

Q. A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact. the zero of vernier scale is ahead of zero of main scale and 3rd division of vernier scale coincides with a main scale division. The least count is

10^{- x} c m

. Find the value of x.

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Vernier Callipers and Screw Guage

PHYSICS

Watch in App

explore_more_icon

Explore more

Vernier Caliper

Standard IX Physics

Join BYJU'S Learning Program

CrossIcon