Find the Binding energy per neucleon for $${}_{50}^{120}Sn$$. Mass of proton m_p = 1.00783 U, mass of neutron m_n = 1.00867 U and mass of tin nucleus m_Sn = 119.902199 U. (take 1U = 931 MeV)

A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force mkv² where v is its speed. The maximum height attained by the ball is :

A quantity x is given by $$\left( {{{IF{v^2}} \over {W{L^4}}}} \right)$$ in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of :

A particle of charge q and mass m is subjected to an electric field
E = E₀ (1 – $$a$$x²) in the x-direction, where $$a$$ and E₀ are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :