A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0 c m$ and a stopwatch with $1 s$ resolution measures the time taken for $40$ oscillation to be $50 s$ . The accuracy in $g$ is:

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Answer 1

For simple pendulum

$T = 2 π \frac{l}{g}$

so, $\frac{Δ T}{T} = \frac{1}{2} (\frac{Δ l}{l} + \frac{Δ g}{g})$

Givn $Δ T = 1 s e c$ $T = 50 s e c$

$Δ l = 0.1 c m$

$l = 25 c m$

$\frac{Δ g}{g} = \frac{2 Δ T}{T} + \frac{Δ l}{l}$

$\frac{Δ g}{g} = 2 \times \frac{1}{5 0} + \frac{0 . 1}{2 5} = \frac{1 . 1}{2 5}$

$\frac{Δ g}{g} = \frac{1 . 1}{2 5}$

$% \frac{Δ g}{g} = \frac{1 . 1}{2 5} \times 100 = 4.4 %$

Answer 2

For simple pendulum

T = 2 π \frac{l}{g}

so,

\frac{Δ T}{T} = \frac{1}{2} (\frac{Δ l}{l} + \frac{Δ g}{g})

Givn

Δ T = 1 s e c

T = 50 s e c

Δ l = 0.1 c m

l = 25 c m

\frac{Δ g}{g} = \frac{2 Δ T}{T} + \frac{Δ l}{l}

\frac{Δ g}{g} = 2 \times \frac{1}{5 0} + \frac{0 . 1}{2 5} = \frac{1 . 1}{2 5}

\frac{Δ g}{g} = \frac{1 . 1}{2 5}

% \frac{Δ g}{g} = \frac{1 . 1}{2 5} \times 100 = 4.4 %