Consider the arrangement shown. Light of intensity $l$ coming from an infinite sheet source falls on a perfectly black cone of base diameter $l$  and semi-vertex angle  $\theta ={\tan }^{-1}\frac{1}{2}$. The same light is incident on a perfectly reflecting square mirror of side  $l$ and is keeping it in equilibrium as shown. If mass of the mirror is same as that of the cone, acceleration of the cone is

        

Force experienced by cone, as it is perfectly absorbing will be 
F = radiation pressure $\times$ projected area As projected area of cone is triangle as shown in Fig. 1, we get
 
  Fig.1
$$\begin{gathered} F=\frac{l}{C}\times \frac{1}{2}\times b\times h\Rightarrow F=\frac{l}{C}\times \frac{1}{2}\times l\times l=\frac{l{l}^2}{2c}\mathrm{..........}\left(i\right) \\ (ash=\frac{l}{2}\cot \theta =\frac{l}{2}\times 2)=l \end{gathered}$$
Let radiation force experience by square mirror be  $F^{\text{'}}$.
As radiation pressure in case of perfectly reflecting surface is given by
(where , i is angle of incidence)
This force will acting at centre of the mirror, as force is uniformly distributed as shown in FBD of mirror Fig.2
 
       Fig.2
For equilibrium of mirror, net torque is zero.
$$\begin{gathered} \Rightarrow Torqueof{F}^{\text{'}}=Torqueofmg\Rightarrow F^{\text{'}}\frac{1}{2}=mg\frac{1}{2\sqrt{2}} \\ \Rightarrow F^{\text{'}}=\frac{mg}{\sqrt{2}}\mathrm{......}\left(iii\right) \end{gathered}$$
From Eqs.  (ii) and (iii), we get$\Rightarrow \frac{l{l}^2}{c}=\frac{mg}{\sqrt{2}}\mathrm{......}\left(iv\right)$ 
From Eqs. (i) and (iv), we get 
 $F=\frac{mg}{2\sqrt{2}}$
$\Rightarrow$Acceleration of cone $=\frac{F}{m}=\frac{g}{2\sqrt{2}}$