Advertisements
Advertisements
प्रश्न
The least count of a vernier callipers is 0.0025 cm and it has an error of + 0.0125 cm. While measuring the length of a cylinder, the reading on the main scale is 7.55 cm, and the 12th vernier scale division coincides with the main scale. Calculate the corrected length.
Advertisements
उत्तर
Least count (L.C.) = 0.0025 cm
Error = +0.0125 cm
Correction = – (Error)
= – (+0.0125)
= – 0.0125 cm
Main scale reading = 7.55 cm
Vernier scale division (V.S.D.) coinciding with main scale = 12th
Length recorded = Main scale reading + L.C. × V.S.D.
= 7.55 + 0.0025 × 12
= 7.55 + 0.0300
= 7.58 cm
Correct length = Length recorded + Correction
= 7.58 + (−0.0125)
= 7.58 – 0.0125
= 7.5675 cm
= 7.567 cm
APPEARS IN
संबंधित प्रश्न
Define the least count of vernier callipers. How do you determine it?
A vernier callipers has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.
Name the part of the vernier callipers which is used to measure the following
Thickness of a pencil.
When does a vernier callipers has the positive error?
Which part of vernier callipers is used to measure the internal diameter of a hollow cylinder?
How the backlash error is avoided?
Consider the following case where the zero of vernier scale and the zero of the main scale are clearly seen. If L.C. of the vernier calipers is 0.01 cm, write the zero error and zero correction of the following.
An instrument that is used to measure the diameter of a cricket ball is ______.
