The mass of and diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be?

The correct option is B 2√3s

Let g be the acceleration due to gravity on Earth and gp be the acceleration due to gravity on the planet.
Let M and R be the mass and radius of Earth, respectively.
Let Mp and Rp be the mass and radius of the planet, respectively.
Given that Mp = 3M and Rp = 3R.
We know that g = GM/R²
Therefore, gp = GMp/Rp² = G(3M)/(3R)² = G(3M)/(9R²) = (1/3)GM/R² = (1/3)g
Also, the period of oscillation of a simple pendulum is given by T = 2π√(L/g), where L is the length of the pendulum.
Since the length of the pendulum remains the same, we can write:
T ∝ 1/√g
Therefore, Tp/To = √(go/gp) = √(g/(g/3)) = √3
Given that To = 2s, we have Tp = To√3 = 2√3 s.