A rod of length L has non-uniform linear mass
density given by $$\rho $$(x) = $$a + b{\left( {{x \over L}} \right)^2}$$ , where a
and b are constants and 0 $$ \le $$ x $$ \le $$ L. The value
of x for the centre of mass of the rod is at :

Two particles of equal mass m have respective
initial velocities $$u\widehat i$$ and $$u\left( {{{\widehat i + \widehat j} \over 2}} \right)$$.
They collide completely inelastically. The energy lost in the process is :

A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $$\sqrt {2gh} $$. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of $$\sqrt {{h \over g}} $$ is :

As shown in figure, when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centered at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e, on the surface of the cavity. R can be detemined by the equation :