JEE Main Physics question
An object of mass m is attached to a light string which passes through a hollow tube. The object is set into rotation in a horizontal circle of radius r₁. If the string is pulled, shortening the radius to r₂, the ratio of the new kinetic energy to the original kinetic energy is
Answer: (B)
Solution
Angular momentum L = mvr is conserved (the tension is a central, radial force), so v ∝ 1/r, and K = ½mv² ∝ 1/r². Hence K₂/K₁ = (r₁/r₂)².