Question

A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water?

• A. It will remain unchanged
• B. It will rise
• C. It will fall
• D. First it will rise and then fall

Answer 1

The correct option is C

It will fall

Let us take the simple case of when a single steel ball is removed from a container floating in water.

Let the volume submerged initially be $V_1$ and after the ball is removed be $V_2$

Let the mass of container be M

Density of steel ${\rho }_s$ and volume of the ball be V

Then,

From, Free body diagram case (i)

$Mg+{\rho }_{s}Vg={\rho }_{\omega }{V}_{1}g$

case (ii)

$Mg={\rho }_{\omega }{V}_{2}g$

$\Rightarrow {\rho }_{\omega }{V}_{2}g+{\rho }_{s}Vg={\rho }_{\omega }{V}_{1}g$

$\Rightarrow {\rho }_{s}V={\rho }_{\omega }(V_1-V_2)$

We know ${\rho }_s>{\rho }_{\omega }$

$\therefore V_1-V_2>V$

Now $V_1-V_2$ will be the volume of fall in the level of water.

So when the ball is removed, the level will fall by $V_1-V_2$ and when he ball is thrown into the water, the level will rise by V

As $V<V_1-V_2$, the final level will be lower than the initial.

Now if there are multiple balls. The level will decrease continuously after each ball is thrown.

$\therefore$ Final level will be lower.