Verified Solution • 2025–26 Session moderate

QAIPMT-2009-Ph-2169: NEET 2009 Physics - Class 11 AIPMT-2009-Ph-2169 Solution: If \( {F}\) is the force acting on a particle having position…

If F⃗\vec{F} is the force acting on a particle having position vector r⃗\vec{r} and τ⃗\vec{\tau} be the torque of this force about the origin then r⃗⋅τ⃗>0\vec{r} \cdot \vec{\tau} > 0 and F⃗⋅τ⃗<0\vec{F} \cdot \vec{\tau} < 0 r⃗⋅τ⃗=0\vec{r} \cdot \vec{\tau} = 0 and F⃗⋅τ⃗=0\vec{F} \cdot \vec{\tau} = 0 r⃗⋅τ⃗=0\vec{r} \cdot \vec{\tau} = 0 and F⃗⋅τ⃗≠0\vec{F} \cdot \vec{\tau} \neq 0 r⃗⋅τ⃗≠0\vec{r} \cdot \vec{\tau} \neq 0 and F⃗⋅τ⃗=0\vec{F} \cdot \vec{\tau} = 0

Multiple Choice Options

  • A ωMM+2m{{\omega M} \over {M + 2m}} M + 2 m _ M _
  • B ω(M+2m)M{{\omega \left( {M + 2m} \right)} \over M} M _ ( M + 2 m ) _
  • C ωMM+m{{\omega M} \over {M + m}} M + m _ M _
  • D ω(M−2m)M+2m{{\omega \left( {M - 2m} \right)} \over {M + 2m}} M + 2 m _ ( M _ 2 m ) _

Step-by-Step Verified Solution

As the masses are added to the ring gently, there is no torque and angular momentum is conserved. I _ = I _ _ _ _ M R 2 _ = ( M R 2 + 2 m R 2 ) _ _ _ _ _ = M R 2 _ ( M + 2 m ) R 2 _ _ _ = M _ M + 2 m

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