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: 
1. 0.7%    
2. 0.2%
3. 3.5%
4. 6.8%

Answer 1

Correct Answer - Option 4 : 6.8%

Concept:

Percentage error:

Percentage error is the difference between a measured and known value, divided by the known value, multiplied by 100%. For many applications, percent error is expressed as a positive value

\({{\rm{\% }}_{{\rm{error}}}} = \left| {\frac{{{\rm{experimental\;error}} - {\rm{theoretical\;error}}}}{{{\rm{theoretical\;error}}}}} \right| \times 100\) 

Calculation:

The time period is given by the formula:

\({\rm{T}} = 2{\rm{\pi }}\sqrt {\frac{l}{{\rm{g}}}} \) 

We need g which is acceleration due to gravity.

So, squaring on both sides,

\( \Rightarrow {{\rm{T}}^2} = \frac{{4{{\rm{\pi }}^2}l}}{{\rm{g}}}\) 

\( \Rightarrow {\rm{g}} = \frac{{4{{\rm{\pi }}^2}l}}{{{{\rm{T}}^2}}}\)      ----(1)              

Now, the percentage error is given by the formula:

\({{\rm{\% }}_{{\rm{error}}}} = \left| {\frac{{{\rm{experimental\;error}} - {\rm{theoretical\;error}}}}{{{\rm{theoretical\;error}}}}} \right| \times 100\) 

\({{\rm{\% }}_{{\rm{error}}}} = \left( {\frac{{{\rm{change\;in\;error}}}}{{{\rm{theoretical\;error}}}}} \right) \times 100\) 

Now, the error for the equation (1) is:

\(\left| {\frac{{{\rm{\Delta g}}}}{{\rm{g}}}} \right| = \left| {\frac{{{\rm{\Delta }}l}}{l}} \right| + 2\left| {\frac{{{\rm{\Delta T}}}}{{\rm{T}}}} \right|\) 

Now, the percentage error for the equation (1) is:

\(\left| {\frac{{{\rm{\Delta g}}}}{{\rm{g}}}} \right|{\rm{\% }} = \left| {\frac{{{\rm{\Delta }}l}}{l}} \right|{\rm{\% }} + 2\left| {\frac{{{\rm{\Delta T}}}}{{\rm{T}}}} \right|{\rm{\% }}\) 

The total error is least count value,

\( \Rightarrow \left| {\frac{{{\rm{\Delta g}}}}{{\rm{g}}}} \right|{\rm{\% }} = \left( {\frac{{0.1}}{{55}} \times 100} \right) + 2\left( {\frac{1}{{30}} \times 100} \right)\) 

∵ 1 mm = 0.1 cm

\( \Rightarrow \left| {\frac{{{\rm{\Delta g}}}}{{\rm{g}}}} \right|{\rm{\% }} = 0.18 + 6.66\) 

\(\therefore \left| {\frac{{{\rm{\Delta g}}}}{{\rm{g}}}} \right|{\rm{\% }} = 6.84{\rm{\% }} \approx 6.8{\rm{\% }}\)