Find the value of 'g' and its uncertainty using T = 2π `sqrt\frac {l}{g}` from the following measurements made during an experiment.
Length of simple pendulum l = 100 cm. Time for 20 vibrations = 40.2 s.
Length was measured by a metre scale of accuracy upto 1 mm and time by stop watch of accuracy upto 0.1 s.
(Ans: 9.76 ± 0.06 ms⁻²)


Given: 

Least count of metre scale = 1 mm = 0.1 cm, So,
Length of simple pendulum = l = 100 cm ± 0.1 cm = 1 m ± 0.001 m . 

Least count of stop watch = 0.1 s. 
Time for 20 vibrations  = 40.2 s, thus
Time for 1 vibration (Time Period) = T = `\frac {40.2}{20}` s = 2.01 s ± 0.1 s

To find: 

Acceleration due to gravity = g  = ?
Uncertainty in Acceleration due to gravity  = ?


Solution:

The formula for the Time Period of the Pendulum
By taking square both side 
and then using basic mathematics rules,  we will get an equation for g


Now by putting values 


by simplifying we get

g = 9.6.7 ms⁻²

Now to find uncertainty in g, we will find the Percentage uncertainty both in Length (l) and in the time period (T) 
As given
[ Length of simple pendulum = l = 100 cm ± 0.1 cm = 1 m ± 0.001 m ]

Absolute uncertainty in Length is = 0.001 m

Percentage uncertainty in Length is = `\frac {0.001 m}{1 m}` ｘ100 %

% uncertainty in length (l) = 0.1 %

Similarly, for time, we are given
Time for 20 vibrations  = 40.2 s, thus
Time for 1 vibration (Time Period) = T = `\frac {40.2}{20}` s = 2.01 s ± 0.1 s

Absolute uncertainty in the time period (T) is = 0.1 s
% uncertainty in the time period (T) = `\frac {0.1 s}{2.o1s ｘ20 vib }` ｘ100 %

% uncertainty in the time period (T) = 0.25 %

% uncertainty in square of time period (T²) = 2 ｘ 0.25 % = 0.5 %

 % uncertainty in g = Percentage uncertainty in Length (l) + % uncertainty in square of time period (T²)

 % uncertainty in g = 0.1 % + 0.5 % = 0.6 %

So,

g = 9.76 ± 0.6 % of 9.76 ms⁻²

g = 9.76 ± `\frac {0.6}{100}` ｘ 9.76 ms⁻²

g = 9.76 ± 0.06 ms⁻² ----- Ans

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Numerical Problem 1.5
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List of 11th Physics Numerical Problems of Chapter - 1
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