Exams › JEE Advanced › Physics

A thin circular metal disc of radius 500.0 mm is rotating about its central axis perpendicular to its plane. When its temperature is raised slowly, the radius grows to 507.5 mm. Assuming angular momentum is conserved, what is the approximate percentage change in its rotational kinetic energy?

1. 1.5%
2. -1.5%
3. 3%
4. -3%

Correct answer: -3%

Solution

With L conserved, KE = L²/(2I) and I proportional to r², so delta_KE/KE = -2*(delta_r/r) = -2*(1.5%) = -3%.

Related JEE Advanced Physics questions

• Seven identical birds are flying north in a flock at constant velocity v. A hunter shoots one bird, which instantly becomes motionless and falls straight to the ground. The remaining six birds continue flying north at the same original speed v. After the dead bird has landed, what is the velocity of the center of mass of all seven birds?
• A semicircular ring has mass 2 kg and radius 2 m. Let I_O be the moment of inertia about an axis perpendicular to the plane of the ring and passing through the geometric center O of the full circle (midpoint of the diameter), and let I_C be the moment of inertia about a parallel axis through the midpoint C of the arc (the point on the arc that lies on the axis of symmetry, at distance R from O). Which of the following statements is/are correct?
• A uniform ring of mass M and radius R starts from rest at the top of a rough inclined plane and rolls down without slipping through a vertical height h. What is the magnitude of the angular momentum of the ring about its own center of mass when it reaches the bottom of the incline?
• A particle of mass m moves with constant velocity v along the straight line 3y = 2x + 6. Observer A is at the origin (0, 0) and observer B is at the point (x1, 0) on the x-axis. If the angular momentum of the particle with respect to B is twice the angular momentum with respect to A, find x1.
• A uniform rod AB of mass M is pivoted at end A and released from rest in the horizontal position. When it swings down to the vertical position, it strikes a sphere of mass m (initially at rest on a frictionless surface) with a perfectly elastic, horizontal impact. After the collision, the sphere moves horizontally and the rod swings back upward to a maximum angle of 60 degrees from the vertical. Given that the length of the rod is sqrt(2)*r (where r = 6*sqrt(2)/10 m), what is the ratio M/m?
• A uniform disc rolls without slipping such that the velocity of its centre of mass is 6v0 and its angular velocity is 2v0/R, where R is the radius of the disc. The distance of the instantaneous axis of rotation (IAR) from the centre of mass is

⚔️ Practice JEE Advanced Physics free + battle 1v1 →