Question 184: Units , Dimensions and Error Analysis

Question

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 mm and the
value obtained is 55.0 cm. The percentage
error in the determination of g is close to :-

Accepted Answer

Time period of a pendulum (T) = $$2\pi \sqrt {{l \over g}} $$
$$ \Rightarrow $$ T
2
= $$4{\pi ^2}{l \over g}$$
$$ \Rightarrow $$ $$g = {{4{\pi ^2}l} \over {{T^2}}}$$
Fractional change
$$\left( {{{dg} \over g}} ight) 	imes 100 = \left( {{{dl} \over l}} ight) 	imes 100 - \left( {2{{dT} \over T}} ight) 	imes 100$$
$$ 	herefore $$ Maximum possible percentage error,
$$\left( {{{dg} \over g}} ight) 	imes 100 = \left( {{{dl} \over l}} ight) 	imes 100 + \left( {2{{dT} \over T}} ight) 	imes 100$$
Error in time period(dT) = least count of time = 1 second
and T = 30 second
Error in length(dl) = least count of length = 1 mm
and $$l$$ = 55.0 cm
$$ 	herefore $$ $$\left( {{{dg} \over g}} ight) 	imes 100 =$$ $$\left( {{{0.1} \over {55}}} ight) 	imes 100 + 2\left( {{1 \over {30}}} ight) 	imes 100$$ = 6.8%

Answer

0.2%

Answer

3.5%

Answer

0.7%

Answer

6.8%

Units , Dimensions and Error Analysis

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