AP EAMCET · Maths · Application of Derivatives
The time period \(T\) of a simple pendulum of length \(l\) is given by \(T=2 \pi \sqrt{\frac{l}{g}}\), where \(g\) denotes the acceleration due to gravity. If the length of the pendulum is increased by \(1 \%\), then the approximate change in its time period is
- A \(0.5 \%\)
- B \(2 \%\)
- C \(1 \%\)
- D \(4 \%\)
Answer & Solution
Correct Answer
(A) \(0.5 \%\)
Step-by-step Solution
Detailed explanation
Given, \(T=2 \pi \sqrt{\frac{l}{g}}\) Then, \(\frac{d T}{d l}=\frac{2 \pi}{\sqrt{g}} \cdot \frac{1}{2 \sqrt{l}}=\frac{T}{\sqrt{l}} \cdot \frac{1}{2 \sqrt{l}} \quad\left[\because T=\frac{2 \pi}{\sqrt{g}} \cdot \sqrt{l}\right]\)…
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