Question

doubt bout fluids: A boat carryinga number of stones is floating in a water tank. If the stones are unloaded into water, then water level will:
FAll
Rise
Remain unchanged
Depends on no. of stones

Answer 1

# Problem:

A boat carrying number of stones is floating in a water tank. If the stones are unloaded into water, then water level will:

- fall

- Rise

- Remain unchanged

- Depends on no. of stones

# Solution:

Everything depends on the density of stones in comparison to water. As usual, density of stones is bigger than the density of water. Let's consider this situation in detail (see figures below).

The equilibrium conditions for boat with stones and without it are:

$$(m + M) g = \rho_ {0} g V$$ $$M g = \rho_ {0} g V _ {2}$$

Thus,

$$\left\{ \begin{array}{l} V = \frac {m + M}{\rho_ {0}} \\ V _ {2} = \frac {M}{\rho_ {0}} \\ V _ {1} = \frac {m}{\rho_ {0}} \\ \rho_ {0} < \rho \end{array} \right.$$

where

- $V$ – volume of displaced water by boat with stones

- $V_{2}$ – volume of displaced water by boat without stones

- $m$ – mass of all stones together

- $M$ – mass of the boat

- $\rho_0$ – density of the water

- $\rho$ – density of stones

$V_{1}$ - volume of displaced water by all stones

In assumption that base area is $S$ ,

$$\left\{ \begin{array}{c} h = \frac {V + V _ {0}}{S} = \frac {M + m}{S \rho_ {0}} + \frac {V _ {0}}{S} \\ H = \frac {V _ {1} + V _ {2} + V _ {0}}{S} = \frac {m}{S \rho} + \frac {M}{S \rho_ {0}} + \frac {V _ {0}}{S} \end{array} \right.$$

Where $h$ and $H$ are initial and final levels of water in tank correspondingly. After comparison of $h$ and $H$ we can conclude that

$$h > H$$

in assumption that $\rho_0 < \rho$ (stones are heavier than water). So that the level of water will fall down if the stones are unloaded into water.

Answer provided by AssignmentExpert.com