First slide

Accuracy; precision of instruments and errors in measurements

Question

The period of oscillation of a simple pendulum is $T = 2 π \sqrt{L / g}$ . Measured value of L is $20.0 cm$ known to $1 mm$ accuracy and time for 100 oscillations of the pendulum is found to be $90 s$ using a wrist watch of 1s resolution. What is the accuracy in the determination of g?

Moderate

A

2.1%
B

2.7%
C

3.1%
D

3.7%

Solution

$g = 4 π^{2} L / T^{2}$
Here, $T = \frac{t}{n}$ and $Δ T = \frac{Δ t}{n}$
Therefore, $\frac{Δ T}{T} = \frac{Δ t}{t}$
The error in both L and t are least count error,
Therefore,
$(Δ g / g) = (Δ L / L) + 2 (Δ T / T)$
$= (Δ L / L) + 2 (Δ t / t)$
$= \frac{0.1}{20.0} + 2 (\frac{1}{90}) = 0.027$
Thus, the percentage error in g is,
$(Δ g / g) \times 100 = 2.7 %$

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