Error and Uncertainty
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Calculate the least count of the screw gauge.
Main scale reading :0 mm
Circular scale reading :52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
- 0.52 cm
- 0.26 cm
- 0.026 cm
- 0.052 cm
In an experiment of simple pendulum time period measured was 50 sec for 25 vibration when the length of the simple pendulum was taken 100 cm. If the least count of stop watch is 0.1 s and that of meter scale is 0.1 cm. Calculate the maximum percentage error in the measurement of value of g.
none of these
- mean absolute error
- percentage error
- relative error
- arithmetic mean of given measurements
If there is an error of in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is , is
The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42, 2.71 s and 2.80 s respectively. Find the average absolute error.
0.12s
0.11 s
0.108 s
none of these
In a simple pendulum experiment for the determination of acceleration due to gravity , time taken for oscillations is measured by using a watch of second least count. The mean value of time taken comes out to be . The length of the pendulum is measured by using a meter scale of least count and the value obtained is . The percentage error in the determination of is close to
A micrometer screw gauge having a positive zero error of 5 divisions is used to measure the diameter of the wire, when reading on the main scale is the 3rd division, and the 48th circular scale division coincides with the baseline. If the micrometer has 10 divisions to a centimetre on the main scale and 100 divisions on the circular scale, calculate the observed diameter.
- x4
- x2
- 4x
- 2x
- mean absolute error
- percentage error
- relative error
- arithmetic mean of given measurements
The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42, 2.71 s and 2.80 s respectively. Find the average absolute error.
0.12s
0.11 s
0.108 s
none of these
- 0.01 m
- 0.1 m
- 0.008 m
- 0.08 m
- >(2.00+0.05) s
- <(2.00−0.05) s
- in the interval (2.00±0.05) s
- 2.00 s
- Percentage error due to area is more than percentage error due to length.
- Number of significant figures in volume will be two.
- Percentage error in volume will be 1.78%.
- Percentage error in length is 1.6%.
- 0.3 s
- 0.16 s
- 0.2 s
- 0.164 s
Which team's measurements are more reliable?
Team A
Team B
Both are equally reliable
- Both are unreliable
- x4
- x2
- 4x
- 2x
- 1%
- 1/3%
- 3%
- 10%
- 1%
- 2%
- 4%
- No error
- 0.01 m
- 0.1 m
- 0.008 m
- 0.08 m
- >(2.00+0.05) s
- <(2.00−0.05) s
- in the interval (2.00±0.05) s
- 2.00 s
- 4.90 cm
- 4.85 cm
- 4.80 cm
- 4.75 cm
- 1%
- 2%
- 4%
- No error
- 60 V
- 11.4 V
- 20 V
- 40 V