An Application of Archimedes' Principle

Initially the buoyant force B supports both blocks, but after m₂ is placed on the bottom of the bucket, the buoyant force on the system of blocks is less than the sum of the weights of the blocks. This is so because m₂ is now partially supported by the normal force N exerted on it by the bottom of the bucket. The buoyant force is therefore decreased and the change is

$\Delta B = m_1 g + (m_2 g - N) - (m_1 g + m_2 g) = - N,$

where $g$ is the acceleration of gravity. The normal force $N$ is the difference in the weight of m₂ and the buoyant force exerted by the water on $m_2$ . By Archimedes' principle that buoyant force is $\rho_0 V_2g$ , so

$\begin{aligned} N &=& m_2 g - \rho_0 V_2 g \\ &=& \rho_2 g V_2 - \rho_0 V_2 g \\ &=& (\rho_2 - \rho_0) gV_2. \end{aligned}$

Using Archimedes' principle again, with the fact that $m_1$ is initially fully submerged, as stated in the problem, one can determine $\rho_2$ in terms of $\rho_0$ .

$\rho_1 V_1 g + \rho_2 gV_2 = \rho_0 g V_1 .$

Remembering that $V_1=2V_2$ and solving for $\rho_2$ , one finds

$\rho_2 = 2(\rho_0 - \rho_1) .$

It is given in the statement of the problem that $\rho_1=\rho_0/3$ . Therefore, $\rho_2=4\rho_0/3$ . Substituting above gives

$N = \frac13 \rho_0 g V_2.$ Thus, $\Delta B = -\frac13 \rho_0 g V_2 .$

By Archimedes' principle $\Delta B=\rho_0 g\Delta V$ , where $\Delta V$ is the change in the volume of displaced water. Using $\Delta V=A \Delta H$ in the preceding equations gives

$\Delta H = -\frac{V_2}{3A} .$

By comparing with the form

$\Delta H = \dfrac{V_2}{KA}$

given in the statement of the problem, one finds $K=\boxed{-3}$ .

When $m_{2}$ is placed over $m_{1}$ , $m_{1}$ is submerged completely.So the amount of water displaced is equal to the buoyant force acting on $m_{1}$ .Let the rise in the level of water be $h_{1}$ in this case.

$v_{1}$ =volume of $m_{1}$

$v_{2}$ =volume of $m_{2}$

A =area of the cross section of the bucket

Weight of water displaced = $ρ_{0}$ $v_{1}$ $g$

$ρ_{0}$ A $h_{1}$ $g$ = $ρ_{0}$ $v_{1}$ $g$

A $h_{1}$ = $v_{1}$

Now, after we remove $m_{2}$ and place it at the bottom of the bucket, $m_{1}$ will rise by a certain amount until its weight is balanced by the buoyant force acting on it.Let $v_{3}$ be the volume of $m_{1}$ immersed.

So,

$m_{1}$ $g$ = $v_{3}$ $ρ_{0}$ $g$

$v_{1}$ $ρ_{1}$ = $v_{3}$ $ρ_{0}$

$v_{3}$ = $\frac{v_{1}}{3}$

So,weight of water displaced in the second case is due to both $m_{1}$ & $m_{2}$ .Let the rise in the level of water be $h_{2}$ in this case.

Weight of water displaced = $ρ_{0}$ $g$ ( $\frac{v_{1}}{3}$ $+$ $v_{2}$ )

$ρ_{0}$ A $h_{2}$ $g$ = $ρ_{0}$ $g$ ( $\frac{v_{1}}{3}$ $+$ $v_{2}$ )

A $h_{2}$ = $\frac{v_{1}}{3}$ $+$ $v_{2}$

A $h_{2}$ = $\frac{2v_{2}}{3}$ $+$ $v_{2}$

A $h_{2}$ = $\frac{5v_{2}}{3}$

Therefore change in height is,

$ΔH$ = $h_{2}$ $-$ $h_{1}$

$AΔH$ = $\frac{5v_{2}}{3}$ $-$ $v_{1}$

$AΔH$ = $\frac{5v_{2}}{3}$ $-$ $2v_{2}$

$ΔH$ = $\frac{-v_{2}}{3A}$

On comparing $ΔH$ = $\frac{v_{2}}{KA}$

$\boxed{K= -3}$

We know that the change in volume , $A . \Delta H$ is given by :

$A . \Delta H = V_{2} - V''_{1}$ (1),

with $V_{2}$ given and $V''_{1}$ the part of block 1 that comes above water after removing block 2.

Now because of equilibrium, the force of buoyancy is just as large as the force of gravity on block 1, after we have removed block 2. This gives us :

$\rho_{0} . V'_{1} . g = m_{1} . g$ or $\rho_{0} . V'_{1} = \rho_{1} . V_{1}$ and $V'_{1} = \dfrac{V_{1}}{3}$ ,

with $V'_{1}$ the volume under water after removing block 2.

Now $V''_{1} = V_{1} - V'_{1} = \dfrac{2V_{1}}{3} = \dfrac{4V_{2}}{3}$ .

Plugging this into (1), we get $A . \Delta H = V_{2} - \dfrac{4V_{2}}{3} = \dfrac{V_{2}}{-3}$ , and K = -3.

Let the volume of water in the tube before anything is added be $V_{w}$ . Then notice that when the initial combination of blocks is added, one on top of the other, the volume of water displaced is clearly equal to $V_{1}$ . The total volume of the cylinder formed by the water and anything submerged in it is going to be

$V_{w}+V_{1}$

By the volume of a cylinder, this makes the original height

$H_{1}=\frac{V_{w}+V_{1}}{A}$

Next, consider the second state of affairs. Since $\rho_{2} > \rho_{0}$ , we know that the block $m_{2}$ will sit at the bottom of the tube and hence displace a volume of water equal to $V_{2}$ . Additionally the block $m_{1}$ will sit at the surface of the water, partially submerged. Since the weight of water it displaces is equal to the weight of the block, paraphrasing Archimedes' Principle where the object is in equilibrium, we can say that (letting $V_{D1}$ represent the volume displaced by block $m_{1}$ in this second case)

$m_{1}g = \rho_{0} V_{D1}g$

$V_{D1}=\frac{m_{1}}{\rho_{0}}$

Using the substitution that $\rho_{0} = 3\rho_{1}$

$V_{D1}=\frac{m_{1}}{3\rho_{1}}$

Since, in general, $V = \frac{m}{\rho}$

$V_{D1}=\frac{V_{1}}{3}$

The total volume of the new cylinder of water and anything submerged in it is going to be

$V_{W}+V_{2}+V_{D1} = V_{W}+V_{2}+\frac{V_{1}}{3}$

The new height is therefore

$H_{2} = \frac{V_{W}+V_{2}+\frac{V_{1}}{3}}{A}$

And the change in height

$\Delta H = H_{2} - H_{1} = \frac{V_{W}+V_{2}+\frac{V_{1}}{3}}{A} - \frac{V_{w}+V_{1}}{A} = \frac{V_{2}+\frac{V_{1}}{3}-V_{1}}{A} = \frac{V_{2}-\frac{2V_{1}}{3}}{A}$

Using the substitution that $V_{1}=2V_{2}$

$\Delta H = \frac{V_{2}-\frac{4V_{2}}{3}}{A} = \frac{V_{2}}{-3A}$

Thus we conclude that $\boxed{K = -3}$