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← JEE Advanced 2015 · Paper 1
PhysicsMultiple correct+4 / -2Question 12 of 60

JEE Advanced 2015 · Paper 1, Physics question 12

A ring of mass MM and radius RR is rotating with angular speed ω\omega about a fixed vertical axis passing through its centre O with two point masses each of mass M8\dfrac M8 at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is 89ω\dfrac89\omega and one of the masses is at a distance of 35R\dfrac35R from O. At this instant the distance of the other mass from O is

A horizontal ring rotating with angular speed ω about a vertical axis through its centre O. Two massless rods run from O to the ring, each carrying a small point mass that can slide outward (arrows).
Choose one or more options
Answer and worked solution

Official answer: C

Angular momentum is conserved.
MR2ω=[MR2+M8(925R2+x2)]89ω.MR^2\omega=\left[MR^2+\frac M8\left(\frac9{25}R^2+x^2\right)\right]\frac89\omega.

Solve.
98=1+18(925+x2R2)  ⟹  x2R2=1625  ⟹  x=45R.\frac98=1+\frac18\left(\frac9{25}+\frac{x^2}{R^2}\right)\implies\frac{x^2}{R^2}=\frac{16}{25}\implies x=\frac45R.

Answer: (D) (the official key accepts C, D or CD)

Topics:Angular momentum and its conservation
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