Physics MCQs
Topic Notes: Physics
MCQs and preparation resources for competitive exams, covering important concepts, past papers, and detailed explanations.
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
1
For a hoop and a disc of identical mass and radius rotating at the same angular velocity, how does their rotational kinetic energy compare?
Answer:
K.E hoop = 2K.E disc
Rotational kinetic energy is given by K = 0.5 * I * omega^2. The moment of inertia (I) for a hoop is MR^2, while for a disc it is 0.5 * MR^2. Since the hoop has twice the moment of inertia of the disc for the same mass and radius, its rotational kinetic energy will be exactly twice that of the disc when rotating at the same angular velocity.
2
Comparing a hoop and a disc of identical mass and radius, how does their rotational kinetic energy relate when rotating at the same angular velocity?
Answer:
K.E hoop = 2K.E disc
Rotational kinetic energy is defined as K = 0.5 * I * omega^2. A hoop has a moment of inertia I = MR^2, while a disc has I = 0.5 * MR^2. Since the hoop's moment of inertia is twice that of the disc, for the same angular velocity, the rotational kinetic energy of the hoop will be exactly twice that of the disc.
3
When a sphere rolls down two different inclined planes that share the same vertical height but have different angles of inclination, which physical quantity remains the same at the bottom?
Answer:
With the same speed
According to the law of conservation of energy, the potential energy at the top is converted into translational and rotational kinetic energy at the bottom. Since the height is identical, the final velocity depends only on the height and the moment of inertia, not the angle of the incline.
4
For a hoop rolling without slipping, how does its rotational kinetic energy compare to its translational kinetic energy?
Answer:
its translational K.E
For a hoop rolling without slipping, the translational kinetic energy is K_trans = 0.5 * M * v^2. The rotational kinetic energy is K_rot = 0.5 * I * omega^2. Substituting I = MR^2 and omega = v/R, we get K_rot = 0.5 * (MR^2) * (v/R)^2 = 0.5 * M * v^2. Thus, the rotational kinetic energy is equal to the translational kinetic energy.