Three bodies A, B and C have equal kinetic energies and their masses are $$400 \mathrm{~g}, 1.2 \mathrm{~kg}$$ and $$1.6 \mathrm{~kg}$$ respectively. The ratio of their linear momenta is :

A body of weight $$200 \mathrm{~N}$$ is suspended from a tree branch through a chain of mass $$10 \mathrm{~kg}$$. The branch pulls the chain by a force equal to (if $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$) :

A light string passing over a smooth light pulley connects two blocks of masses $$m_1$$ and $$m_2\left(\right.$$ where $$\left.m_2>m_1\right)$$. If the acceleration of the system is $$\frac{g}{\sqrt{2}}$$, then the ratio of the masses $$\frac{m_1}{m_2}$$ is:

A particle moves in $$x$$-$$y$$ plane under the influence of a force $$\vec{F}$$ such that its linear momentum is $$\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})$$. If $$\mathrm{k}$$ is constant, the angle between $$\overrightarrow{\mathrm{F}}$$ and $$\overrightarrow{\mathrm{p}}$$ will be :