CUET Preparation Today
A ball of mass m is dropped from a height h equal to the radius of the earth above the tunnel dug through the earth as shown in the figure. Choose the wrong options. |
Particle will oscillate though the earth to a height h on both sides Particle will execute simple harmonic motion Motion of the particle is periodic Particle passes the centre of earth with a speed v = √(2GM/R) |
Particle will execute simple harmonic motion |
Particle performs SHM inside tunnel but not outside surface, so, if just performs periodic motion with amplitude, (R + h) : \(\frac{1}{2} mv^2 = \frac{3GMm}{2R} - \frac{GMm}{2R}\) \(\frac{1}{2} mv^2 = \frac{GMm}{R}\) \(\Rightarrow v = \sqrt{\frac{2GM}{R}}\) |
