First slide

Accuracy; precision of instruments and errors in measurements

Question

The period of oscillation of a simple pendulum is T $= 2 π \sqrt{\frac{L}{g}}$ Measured value of L is 20 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist-watch of 1 s resolution. The accuracy in the determination of g is

Moderate

A

1%
B

2%
C

3%
D

4%

Solution

Given, $T = 2 π \sqrt{\frac{L}{g}}$
or $g = \frac{4 π^{2} L}{T^{2}}$
Hence $T = \frac{t}{n} and ΔT = \frac{Δt}{n}$
.. $\frac{ΔT}{T} = \frac{Δt}{t}$
The errors in both L and t are the least count errors. Therefore,
$(\frac{Δg}{g}) = (\frac{ΔL}{L}) + 2 (\frac{ΔT}{T}) = \frac{0.1}{20.0} + 2 (\frac{1}{90}) = 0.027$
The percentage error in g is
$(\frac{Δg}{g}) \times 100 = (\frac{ΔL}{L}) \times 100 + 2 \times (\frac{ΔT}{T}) \times 100 = 3 %$

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