First slide

Errors

Question

The time period of oscillation of simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ ; The length of the pendulum is measured as l = $(10 \pm 0.1)$ cm and time period as T = $(0.5 \pm 0.02)$ s. The percentage error in measuring ‘g’ is

Moderate

A

9
B

5
C

8
D

7

Solution

$T = 2 π \sqrt{\frac{l}{g}} g = \frac{4 π^{2} l}{T^{2}}$
$\frac{Δg}{g} = \frac{Δl}{l} + 2 \frac{ΔT}{T} = \frac{0.1}{10} + \frac{2 \times 0.02}{0.5} = 0.01 + 0.08 = 0.09$
$\frac{∆ g}{g} % = 9 %$

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