JEE Mains · Physics · STD 11 - 13. oscillations
The length of a seconds pendulum at a height \(h=2 R\) from earth surface will be _______. (Given: \(R =\) Radius of earth and acceleration due to gravity at the surface of earth \(g =\pi^{2}\,m / s ^{-2}\) )
- A \(\frac{2}{9}\,m\)
- B \(\frac{4}{9}\,m\)
- C \(\frac{8}{9}\,m\)
- D \(\frac{1}{9}\,m\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{9}\,m\)
Step-by-step Solution
Detailed explanation
\(T =2 \pi \sqrt{\frac{ L }{ g }}, \quad g ^{\prime}=\frac{ GM }{9 R ^{2}}=\frac{ g }{9}=\frac{\pi^{2}}{9}\) \(2=2 \pi \sqrt{\frac{ L }{\pi^{2}} \times 9}\) \(1=\pi \sqrt{ L } \times \frac{3}{\pi} \Rightarrow L =\frac{1}{9}\,m\)
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