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प्रश्न
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उत्तर
1 MSD = 1 mm
10 VSD = 9 mm
1 VSD =`9/10`mm
Least count = 1 MSD - 1 VSD
=`1/10`MSD
=`1/10` x 1 mm = 0.1 mm
= 0.01 cm
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संबंधित प्रश्न
A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of the circular scale lies below the base line and 4th division of the 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
Is it possible to increase the degree of accuracy by mathematical manipulations? Support your answer by an example.
Who invented vernier callipers?
Up to how many decimal places can a common vernier callipers measure the length in cm?
State the formula for determining a pitch
Which part of vernier callipers is used to measure the internal diameter of a hollow cylinder?
A micrometre screw gauge has a negative zero error of 8 divisions. While measuring the diameter of a wire the reading on the main scale is 3 divisions and the 24th circular scale division coincides with baseline.
If the number of divisions on the main scale are 20 to a centimetre and circular scale has 50 divisions, calculate
- pitch
- observed diameter.
- least count
- corrected diameter.
What is the least count in the case of the following instrument?
Stopwatch
State whether true or false. If false, correct the statement.
With the help of vernier caliper we can have an accuracy of 0.1 mm and with screw gauge we can have an accuracy of 0.01 mm.
