Question

An ice cube contains a glass ball. The cube is floating on the surface of water contained in a trough. What will happen to the water level, when the cube melts?

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

Answer 1

The correct option is B

It will fall

Let the volume of the cube be V and that of the ball be V'

Now when the cube floats

Mass of water displaced = mass of ice + mass of ball

$=(V-V^{\prime }){\rho }_i+V^{\prime }{\rho }_G$

Volume of water displaced =$=(V-V^{\prime })\frac{{\rho }_i}{{\rho }_{\omega }}+V^{\prime }\frac{{\rho }_G}{{\rho }_{\omega }}$

After melting, mass of water melted is equal to mass of ice

$=(V-V^{\prime }){\rho }_i$

Volume of water melted $=\frac{(V-V^{\prime }){\rho }_i}{{\rho }_{\omega }}$

But after the water has melted, rise in the level of water will be equal to the sum of the volume of water melted and the volume of the glass ball as the ball would sink into the water.

$\therefore$ Rise in water $=(V-V^{\prime })\frac{{\rho }_i}{{\rho }_{\omega }}+V^{\prime }$

$=(V-V^{\prime })\frac{{\rho }_i}{{\rho }_{\omega }}+V^{\prime }\frac{{\rho }_G}{{\rho }_{\omega }}$

Volume of water displaced

As ${\rho }_G>{\rho }_{\omega }$

We can see that volume of water displaced is greater

$\therefore$The water level will fall