First slide

Accuracy; precision of instruments and errors in measurements

Question

A simple pendulum is being used to determine the value of gravitational acceleration ‘g’ at a certain place. The length of the pendulum is 20.0 cm known to 1mm accuracy and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50s. The accuracy in g is

Moderate

A

3.5%
B

5.5%
C

4.5%
D

2.4%

Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{l}{g}} \\ o n s q u a r i n g b o t h s i d e s \\ g = 4 π^{2} \frac{l}{T^{2}} \\ o n d i f f e r e n t i a t i n g, \\ \frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2 Δ T}{T} = \frac{0.1}{20} + \frac{2 \times 1}{50} = 0.045 \\ \Rightarrow (\frac{Δ g}{g}) \times 100 = (0.045) \times 100 \\ ∴ % \frac{Δ g}{g} = 4.5 % \end{array}$

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