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प्रश्न
The circular scale of a screw gauge has 50 divisions. Its spindle moves by 2 mm on the sleeve, when given four complete rotations calculate
- pitch
- least count
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उत्तर
Number of circular scale divisions (C.S.D.) = 50
Distance moved by screw (spindle) on sleeve = 2 mm
Number of complete rotations given = 4
(1) Pitch = `"Distance moved by screw on sleeve"/"No. of complete rotations"`
= `(2 "mm")/4`
= 0.5 mm
(2) Pitch = 0.05 cm
Least count = `"Pitch"/"No. of circular scale divisions"`
= `0.05/50`
= 0.001 cm
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संबंधित प्रश्न
In the vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided into 10 parts. While measuring the length, the zero of the vernier lies just ahead of the 1.8 cm mark and the 4th division of vernier coincides with a main scale division.
(a) Find the length.
(b) If zero error of vernier callipers is -0.02 cm,
What is the correct length ?
Is it possible to increase the degree of accuracy by mathematical manipulations? Support your answer by an example.
Up to how many decimal places can a common vernier callipers measure the length in cm?
The thimble of a screw gauge has 50 divisions for one rotation. The spindle advances 1 mm when the screw is turned through two rotations.
- What is the pitch of the screw?
- What is the least count of screw gauge?
- When the screw gauge is used to measure the diameter of wire the reading on the sleeve is found to be 0.5 mm and reading on thimble is found, 27 divisions. What is the diameter of the wire in centimetres?
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The ratchet of a screw gauge is used to measure the depth of a beaker.
State whether the following statement is true or false by writing T/F against it.
The least count of a screw gauge can be lowered by increasing the number of divisions on its thimble.
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.
Least count of a vernier caliper is ______ cm.
