A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass $${m \over 2}$$ collides with A with a velocity which is half $$\left( {{{\overrightarrow v } \over 2}} \right)$$ the instantaneous velocity$${\overrightarrow v }$$ of A. The collision is completely inelastic. Then, the combined body :

The electric fields of two plane electromagnetic plane waves in vacuum are given by
$$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$ and
$$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$$
At t = 0, a particle of charge q is at origin with
a velocity $$\overrightarrow v = 0.8c\widehat j$$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :

A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $${a \over 3}$$ and 2$$a$$, respectively from the axis of the wire is :

Consider a sphere of radius R which carries a uniform charge density $$\rho $$. If a sphere of radius $${{R \over 2}}$$ is carved out of it, as shown, the ratio $${{\left| {\overrightarrow {{E_A}} } \right|} \over {\left| {\overrightarrow {{E_B}} } \right|}}$$ of magnitude of electric field $${\overrightarrow {{E_A}} }$$ and $${\overrightarrow {{E_B}} }$$, respectively, at points A and B due to the remaining portion is :