Physics
Two micrometers are shown in figures given below. Give the actual reading shown provided :
(a) The least count of the instrument = 0.001 cm and zero error = +0.05 mm.
(b) The least count of the instrument = 0.001 cm and zero error = -0.021 cm.
Answer
(a) Given,
- Least count = 0.001 cm
- Zero error = +0.05 mm = +0.005 cm
From the figure,
Main scale reading (MSR) = 3 mm = 0.3 cm
And
21st division of the circular scale coincides with the main scale then
Circular scale reading (CSR) = Least count x 21 = 0.001 cm x 21 = 0.021 cm
Now,
Observed reading = MSR + CSR = 0.3 + 0.021 = 0.321 mm
The zero correction = -0.005 cm.
Then,
Actual reading = 0.321 - 0.005 = 0.316 cm
Hence, the actual reading is 0.316 cm.
(b) Given,
- Least count = 0.001 cm
- Zero error = -0.021 cm
From the figure,
Main scale reading (MSR) = 6 mm = 0.6 cm
And
41st division of the circular scale coincides with the main scale then
Circular scale reading (CSR) = Least count x 41 = 0.001 cm x 41 = 0.041 cm
Now,
Observed reading = MSR + CSR = 0.6 + 0.041 = 0.641 cm
The zero correction = +0.021 cm.
Then,
Actual reading = 0.641 + 0.021 = 0.662 cm
Hence, the actual reading is 0.662 cm.
Related Questions
When a screw gauge of least count 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 1 mm and the reading on the thimble is found to be 27 divisions.
What is the diameter of the wire in cm?
If the zero error is +0.005 cm, what is the correct diameter?
A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two rotations. When the flat end of the screw is in contact with the stud, the zero of circular scale lies below the base line and 4th division of circular scale is in line with the base line.
Find —
(i) the pitch,
(ii) the least count and
(iii) the zero error, of the screw gauge.
Figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on 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.
A screw has a pitch equal to 0.5 mm. What should be the number of division on its head so as to read correct up to 0.001mm with its help?