A rectangular block is 5 cm × 5 cm × 10cm in size. The block is floating in water with 5 cm side vertical. If it floats with 10 cm side vertical, what change will occur in the level of water?

Correct Option: No change

Step-by-Step Explanation:

To understand what happens to the water level, we can apply Archimedes' principle, which governs floating objects.

According to Archimedes' principle, when an object floats in a fluid, the buoyant force acting on it is equal to the weight of the fluid displaced by the object. Furthermore, for any floating object in equilibrium, the upward buoyant force must exactly balance the downward gravitational force (the weight of the object).

Let the mass of the rectangular block be represented by $m$, and the acceleration due to gravity be $g$. The weight of the block is:
$W=mg$

The buoyant force ($F_b$) exerted by the water is given by:
$F_b={\rho }_w{V}_{d}g$
where ${\rho }_w$ is the density of water and $V_d$ is the volume of water displaced by the floating block.

Since the block is floating in equilibrium in both cases, the buoyant force must equal the weight of the block:
${\rho }_w{V}_{d}g=mg$

By dividing both sides of the equation by $g$, we can solve for the volume of displaced water ($V_d$):
$V_d=\frac{m}{{\rho }_w}$

From this relation, we can see that the volume of displaced water ($V_d$) depends only on the mass of the block ($m$) and the density of water (${\rho }_w$). Neither of these quantities changes when the orientation of the block is altered.

Whether the block floats with its 5 cm side vertical or its 10 cm side vertical, it will displace the exact same volume of water to support its weight. Because the volume of displaced water remains constant, the overall water level in the container will experience no change.