Correct Answer - Option 3 : 5%
Concept:
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Error: While doing any experiment due to faulty equipment, carelessness, or any other random cause the final results get affected.
- To resolve the percentage change we use the formula:
\(Y = \frac{{{A^a}\;.\;\;{B^b}}}{{{C^c}}} \Leftrightarrow \frac{{{\bf{Δ }}Y}}{Y} = \; \pm \;\left( {a\;\frac{{{\bf{Δ }}A}}{A} + b\;\frac{{{\bf{Δ }}B}}{B} + c\;\frac{{{\bf{Δ }}C}}{C}\;} \right)\)
[Where ΔY = change in value, Y = original value, a = power of first element, again change in A and follows...]
- When a point mass is attached to an inextensible string and suspended from fixed support then it is called a simple pendulum.
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The time period of a simple pendulum is defined as the time taken by the pendulum to finish one complete oscillation.
\(⇒ T = 2\pi\sqrt{\frac{l}{g}}\)
Calculation:
Given - % change in length = ± 1 % and % change in time period = ± 2%, the time period of simple pendulum is
\( \Rightarrow T = \sqrt {\frac{l}{g}} \; \Leftrightarrow \frac{{{\rm{Δ }}T}}{T}\; = \frac{{\sqrt l }}{{\sqrt g }}\; = \;\frac{{{l^{\frac{1}{2}}}}}{{{g^{\frac{1}{2}}}}}\)
\( \Rightarrow \frac{{{\rm{Δ }}T}}{T} = \pm \left( {\frac{1}{2}\;\frac{{{\rm{Δ }}L}}{L}\; + \;\frac{1}{2}\;\frac{{{\rm{Δ }}g}}{g}} \right)\)
\( \Rightarrow \% \frac{{{\rm{Δ }}T}}{T} = \; \pm \left( {\frac{1}{2}\left( {\frac{{{\rm{Δ }}L}}{L} \times 100} \right) + \frac{1}{2}\left( {\frac{{{\rm{Δ }}g}}{g} \times 100} \right)} \right)\)
\(\implies 2\times\frac{1}{100}= \frac{1}{2}\times 100 + \frac{1}{2}\times\frac{Δ g}{g}\)
Since we have to find the maximum error, we will plus sign.
\(\implies 2\times\frac{1}{100} + \frac{1}{2}\times 100 = \frac{1}{2}\times\frac{Δ g}{g}\)
Since we have to find the maximum error, we will add the percentage error of time period and length
\(\implies \frac{2.5}{100} = \frac{1}{2}\times\frac{Δ g}{g}\)
\(\implies \frac{5}{100} = \frac{Δ g}{g}\)
Now, Δ g / g is the percentage error in acceleration due to gravity. This is equal to 5 /100 which is 5 %.
So, the correct option is 5 %.
