AP EAMCET · PHYSICS · Gravitation
The time period of a simple pendulum on the surface of the earth is \(T\). If the pendulum is taken to a height equal to half of the radius of the earth, then its time period is
- A \(\frac{\mathrm{T}}{2}\)
- B \(\frac{3 \mathrm{~T}}{2}\)
- C 2 T
- D 3 T
Answer & Solution
Correct Answer
(B) \(\frac{3 \mathrm{~T}}{2}\)
Step-by-step Solution
Detailed explanation
\( g' = g \left( \frac{R}{R+h} \right)^2 \) \( g' = g \left( \frac{R}{R+\frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4g}{9} \) \( T' = 2\pi\sqrt{\frac{L}{g'}} \)…
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