Angular Momentum conservation I1ω1=I2ω2ω2=4ω1T=ω2πM= same R2=2r1I∝r2I2=4I1 T1=27 day T2=4 T1=4×27 T2=108 days
The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
Held on 30 Apr 2025 · Verified 9 Jul 2026.
100 days
105 days
115 days
108 days
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A bullet of mass 10 g moving with 400 m/s penetrates a wall and comes to rest. The loss in kinetic energy is:
A body of mass 5 kg is moving with a velocity of 10 m/s. The kinetic energy of the body is:
A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is: 
A ball of mass 0.5 kg is dropped from a height of 40 m . The ball hits the ground and rises to a height of 10 m . The impulse imparted to the ball during its collision with the ground is (Take $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$ )
The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
Work through every NEET UG Mechanics PYQ, year by year.