Question

Time period of simple pendulum of infinite length oscillating near the surface of the earth is: [$M_e=$ mass of the earth, R = radius ]

• A. $2\pi \sqrt{R^{2}G{M}_e}$
• B. $2\pi \sqrt{R^3}G{M}_e$
• C. $2\pi \sqrt{RG{M}_e}$
• D. $2\pi \sqrt{R^4}G{M}_e$

Answer 1

Answer: (A)

$2\pi \sqrt{R^{2}G{M}_e}$

The general formula of a time period of a pendulum of length comparable to radius of earth

$T=2\pi \sqrt{\frac{1}{(\frac{1}{l}+\frac{1}{R})g}}$

Putting $l=\infty$

$T=2\pi \sqrt{\frac{R}{g}}T=2\pi \sqrt{R^{2}G{M}_e}[g=\frac{G{M}_e}{R}]$