Screw Gauge
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Find the least count of a screw gauge whose least count of linear scale is 1 mm and the circular scale is divided in 50 divisions and 2 entire rotations makes to 1 division on linear scale.
0.02 mm
0.1mm
0.2 mm
0.01 mm
In a Searle's experiment, the diameter of the wire as measured by a screw gauge with least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.
y=WA×LX
Where W is the weight suspended from the wire, A is the cross section area, L is the length and X is the extension in the wire.
[IIT JEE 2004]
1.26 %
4.89 %
9.67 %
15.56 %
- Increasing number of divisions on circular scale
- Increasing number of divisions on linear scale
- Increasing pitch of screw gauge
- All of the above
Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum.They use different lenghts of the pendulum and record time for different number of oscillations.The observations are shown in the table.Least count for length = 0.1cm, Least count for time = 0.1s
StudentLength of the pendulum(cm)Number of oscillations(n)Total time for (n) oscillation(s)Time period(s)I64.08128.016.0II64.0464.016.0III20.0436.09.0
If EI, EII and EIII are the percentage errors in g, i.e. (△gg×100) for student I, II and III, respectively, t = 2π√lg , then
= 0
is minimum
and
= 0
is maximum
- 3.67 mm
- 3.38 mm
- 3.73 mm
- 3.32 mm
A screw gauge gives the following reading when used to measure the diameter of a wire.Main scale reading : 0 mm.Circular scale reading : 52 divisionsGiven that 1 mm on main scale corresponds to 100 divisions of the circular scale.The diameter of wire from the above data is :
0.52 cm0.052 cm0.026 cm0.005 cm