Question

Two containers of equal mass and containing equal volumes of water are placed on a balance as shown. The ball on the left is a ping pong ball and the ball on the right is made of steel. Both have equal volume. Towards which side will the balance tilt?

• A. Towards the side of the steel ball
• B. Towards the side of the ping pong ball
• C. It will not it will not tilt toward any side
• D. None of these

Answer 1

The correct option is A

Towards the side of the steel ball

Let the mass of the ping pong ball be m1 and that of the steel ball be m2 and mass of the container and the water be M

Left container,

$\Rightarrow F_B=M_{1}g+T$ - - - - - - (1)

$\Rightarrow T+N_1=Mg+F_B$ - - - - - - (2)

$\Rightarrow F_B-M_{1}g+N_1=Mg+F_B$

$\Rightarrow N_1=M_{1}g+Mg$ - - - - - - (3)

Now for the container on the right side

The mass of the container and water will be the same and the magnitude of the buoyant force will be the same as the volume of the balls are the same.

$\Rightarrow T_2+F_B=M_{2}g$

$N_2=Mg+F_B$ - - - - - - (4)

We know $F_B={\rho }_{\omega }Vg$ $\forall$ V is the volume of the ball

${\rho }_{\omega }$ = density of water

From (3) and (4)

$N_1=Mg+M_{1}g$

$N_2=Mg+{\rho }_{\omega }Vg$

$M_{1}g<{\rho }_{\omega }Vg$ as density of ping pong ball is less than that of water

$\therefore N_1<N_2$

$\therefore$ It will tilt toward the right