In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for $20$ oscillations is measured by using a watch of $1$ second least count. The mean value of time taken comes out to be $30 s$ . The length of pendulum is measured by using a meter scale of least count $1 m m$ and the value obtained is $55.0 c m$ . The percentage error in the determination of g is close to :-

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Answer 1

$T = \frac{3 0 s e c}{2 0}$ $Δ T = \frac{1}{2 0} s e c$
$L = 55 c m$ $Δ L = 1 m m = 0.1 c m$
$g = \frac{4 π ^{2} L}{T ^{2}}$
percentage error in g is
$\frac{Δ g}{g} \times 100 % = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) 100 %$
$= ⎝ ⎜ ⎜ ⎜ ⎛ \frac{0 . 1}{5 5} + \frac{2 ( \frac{1}{2 0} )}{\frac{3 0}{2 0}} ⎠ ⎟ ⎟ ⎟ ⎞ 100 % = 6.8 %$

Answer 2

T = \frac{3 0 s e c}{2 0}

Δ T = \frac{1}{2 0} s e c

L = 55 c m

Δ L = 1 m m = 0.1 c m

g = \frac{4 π ^{2} L}{T ^{2}}

percentage error in g is

\frac{Δ g}{g} \times 100 % = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) 100 %

= ⎝ ⎜ ⎜ ⎜ ⎛ \frac{0 . 1}{5 5} + \frac{2 ( \frac{1}{2 0} )}{\frac{3 0}{2 0}} ⎠ ⎟ ⎟ ⎟ ⎞ 100 % = 6.8 %