First slide

Accuracy; precision of instruments and errors in measurements

Question

The time period of oscillation of simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ where l is 100 cm and is known to have 0.1 cm accuracy. The time period is 2s. The time (t) of 100 oscillations is measured by a stop watch of least count 0.1s. The maximum percentage error in measurement of g is

Moderate

A

0.1%
B

0.2%
C

0.8%
D

1%

Solution

Given
The time period of simple pendulum $T = 2 π \sqrt{\frac{l}{g}}$
$\begin{array}{l} \Rightarrow g = 4 π^{2} \frac{l}{T^{2}} \\ \Rightarrow \frac{Δ g}{g} = \frac{Δ l}{l} + 2 \frac{Δ T}{T} \\ t = 100 T = 200 s \\ \Rightarrow \frac{Δ t}{t} = \frac{Δ T}{T} \\ \Rightarrow \frac{Δ g}{g} = \frac{Δ l}{l} + 2 \cdot \frac{Δ t}{t} \\ \Rightarrow \frac{Δ g}{g} = \frac{0.1}{100} + 2 \frac{\frac{0.1}{100}}{2} \\ \Rightarrow \frac{Δ g}{g} \times 100 = 0.2 % \end{array}$

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