Question

The mass and diameter of a planet are two times those of earth. If a seconds pendulum is taken to it, the time period of the pendulum in seconds is

• A. $1/\sqrt{2}$
• B. $1/2$
• C. $2$
• D. $2\sqrt{2}$

Answer 1

Answer: (D)

$2\sqrt{2}$

$T=2\pi \sqrt{\frac{l{R}^2}{GM}}$

Replacing R by $2$R and M by $2$M for new planet.
$T_1=2\pi \sqrt{\frac{4l{R}^2}{2GM}}=\sqrt{2}\times 2\pi \sqrt{\frac{l{R}^2}{GM}}$

$T_1=T\sqrt{2}$
$=2\sqrt{2}sec$