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Standard XI

Physics

Error and Uncertainty

In an experim...

Question

In an experiment to find acceleration due to gravity $(g)$ using simple pendulum, time period of $0.5 s$ is measured from time of $100$ oscillations with a watchof $1 s$ resolution. If measured value of length is $10 cm$ known to $1 mm$ accuracy, the accuracy in the determination of $g$ is found to be $x$ %. The value of $x$ is _________

5

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5.0

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5.00

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Solution

Time period is given as
$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{d g}{g} \times 100 = \frac{2 d T}{T} \times 100 + \frac{d l}{l} \times 100$

$= 2 \times \frac{1}{50} \times 100 + \frac{1}{100} \times 100 = 5$ %

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Time period is given as T=2π√lg dgg×100=2dTT×100+dll×100 =2×150×100+1100×100=5 %

Time period is given as
$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{d g}{g} \times 100 = \frac{2 d T}{T} \times 100 + \frac{d l}{l} \times 100$

$= 2 \times \frac{1}{50} \times 100 + \frac{1}{100} \times 100 = 5$ %