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← JEE Main 2022 · 26 Jul · Shift 1
PhysicsInteger+4 / -1Question 52 of 90

JEE Main 2022 · 26 Jul · Shift 1, Physics question 52

A disc of mass 1 kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be 4x3R4\sqrt{\dfrac x{3R}} rad s−1^{-1} where x=x= ______.

(g=10g=10 ms−2^{-2})

Answer and worked solution

Official answer: 5

Energy. The mass drops 2R2R: mg 2R=12(12mR2+mR2)ω2mg\,2R=\frac12\left(\frac12mR^2+mR^2\right)\omega^2, so ω2=8g3R\omega^2=\frac{8g}{3R}, ω=453R\omega=4\sqrt{\frac{5}{3R}} with g=10g=10.

Answer: 5