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The potential energy function of a particle is given by `U(r)=A/(2r^(2))-B/(3r)`, where `A` and `B` are constant and `r` is the radial distance from the centre of the force. Choose the correct option (s)A. The equilibrium distance will be `r_(0)=(2A)/B`B. The equilibrium distance will be `r_(0)=(3A)/B`C. If the total energy of the particle is `(B^(2))/(6A)`, then its radial velocity will vaishh ate `(r_(0))/3`, where `r_(0)` is the equilibrium distanceD. If the total energy of the particle is `(B^(2))/(6A)`, then its radial velocity will vanish at `(r_(0))/2`,where `r_(0)` is the equilibrium distance. |
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Answer» Correct Answer - B::C `F=-(dU)/(dr)=A/(r^(3))-B/(3r^(2))` For equilibrium `F=0, r=(3A)/B` `(B^(2))/(6A)=A/(2r_(0)^(2))-B/(3r_(2))` (for `r_(0)=r/3`) |
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