JEE Main 2025 Physics question
A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 Nm for 40s, the speed increases to 2100 rpm. The diameter of the disk is m.
Answer: 40
Solution
τdt = IΔω. , giving I=100 = MR²/4, so R²=400, R=20m, D=40m.