Two bodies of masses $$m_1 = 5$$ kg and $$m_2 = 3$$ kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass $$m_1$$ will be: [Take $$g = 10$$ m s$$^{-2}$$]

Now here we have 2 bodies of masses $$m_1$$ and $$m_2$$

Let us consider the forces acting on these 

On $$m_2$$ the forces acting are $$T\ and\ F=\ m_2g$$

As they are in opposite directions and for the system to be in equilibrium, they have to be equal

$$T\ =\ m_2g$$

coming to the first body of mass $$m_1$$

Forces along the direction of incline are $$T\ and\ F_{ }=\ m_1g\sin\theta\ $$

These are also equal as the system is in equilibrium

So , $$T\ =m_1g\sin\theta\ $$

We know  , $$T\ =\ m_2g$$

After Substituting T

$$m_1g\sin\theta\ =\ m_2g$$

$$5g\sin\theta\ =\ 3g$$

$$5\sin\theta\ =\ 3$$

$$\sin\theta\ =\ \frac{3}{5}$$

Now coming to Forces which are  perpendicular to the incline

 $$N\ and\ m_1g\cos\theta\ $$

Even they have to equal

so , $$N\ =\ m_1g\cos\theta\ $$

As $$\sin\theta\ =\frac{3}{5}\ then\ \cos\ \theta\ =\frac{4}{5}\ $$

Then $$N\ =\ m_1g\times\ \frac{4}{5}$$

$$N\ =5\times\ g\times\ \frac{4}{5}$$

$$N\ =5\times\ 10\times\ \frac{4}{5}$$

$$N\ =\ 10\times\ 4$$

$$N\ =\ 40N$$

This is the force exerted by the inclined plane on the body of mass $$m_1$$