In a simple pendulum experiment for determination of acceleration due to gravity ( g ) , time taken for 20 oscillations is measures by using a watch of 1 second least count. The mean value of time tak

Question

In a simple pendulum experiment for determination of acceleration due to gravity $(g)$ , time taken for $20$ oscillations is measures by using a watch of $1$ second least count. The mean value of time taken comes out to be $30 s$ . The length of the 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:

B

Solution:

$g = \frac{4 π^{2} l}{T^{2}}$
$\frac{∆ g}{g} = \frac{∆ l}{l} + \frac{2 ∆ T}{T}$
$= \frac{0.1}{55} + 2 \times (\frac{1}{30})$
$= 0.06848$
Percentage error $\approx 6.8 %$

TestHub · Accepted Answer

g = 4 π 2 l T 2 ∆ g g = ∆ l l + 2 ∆ T T = 0.1 55 + 2 × 1 30 = 0.06848 Percentage error ≈ 6.8 %

Physics - Error Question with Solution | TestHub

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