Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Here, zero of the vernier scale lies on the right of the zero of the main scale.
Also, the 5th vernier division is coinciding with some main scale division
Thus, zero error = 0 + 5 x L.C. = 0.05 cm
Hence zero correction = -0.05 cm
APPEARS IN
संबंधित प्रश्न
A screw gauge has a least count 0.001 cm and zero error +0.007 cm. Draw a neat diagram to represent it.
Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.
Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.
Define metre in terms of the wavelength of light.
The least count of a vernier callipers is 0.01 cm and it has an error of + 0.07 cm. While measuring the radius of a sphere, the main scale reading is 2.90 cm and the 5th vernier scale division coincides with the main scale. Calculate the correct radius.
State the formula for calculating length if:
The reading of the main scale is known and the number of vernier scale divisions coinciding with the main scale is known.
A micrometre screw gauge having a positive zero error of 5 divisions is used to measure diameter of a wire, when reading on the main scale is 3rd division and the 48th circular scale division coincides with baseline. If the micrometer has 10 divisions to a centimetre on the main scale and 100 divisions on a circular scale, calculate
- Pitch of screw
- Least count of screw
- Observed diameter
- Corrected diameter.
What do you understand by the following term as applied to micrometre screw gauge?
Sleeve cylinder
How do you account for negative zero error, for calculating the correct diameter of wires?
