Question

A pan balance has a container of water with an overflow spout on the right-hand pan as shown. It is full of water right up to the overflow spout. A container on the left-hand pan is positioned to catch any water that overflows. The entire apparatus is adjusted so that it's balanced. A brass weight on the end of a string is then lowered into the water, but not allowed to rest on the bottom of the container. What happens next?

• A. Water overflows and the right side of the balance tips down.
• B. Water overflows and the left side of the balance tips down.
• C. Water overflows but the balance remains balanced.
• D. Water overflows but which side of the balance tips down depends on whether the brass weight is partly or completely submerged.

Answer 1

Answer: (A)

Water overflows and the right side of the balance tips down.

Answer 2

When the brass weight is submerged, it displaces a volume of water equal to its submerged volume ($V_{submerged}$). This displaced water overflows into the left pan. The mass of this overflowed water is $m_{overflow} = \rho_w V_{submerged}$.

The brass weight, when submerged, experiences an upward buoyant force $F_B = \rho_w V_{submerged} g$. The tension in the string supporting the weight (and thus the force on the right pan) is the weight of the brass minus the buoyant force: $T = m_{brass}g - F_B$. This is equivalent to an apparent mass of $m_{brass\_apparent} = m_{brass} - \rho_w V_{submerged}$ on the right pan.

The initial state has equal masses on both pans. The final mass on the left pan is $M_{left\_final} = M_{initial} + m_{overflow} = M_{initial} + \rho_w V_{submerged}$. The final mass on the right pan is $M_{right\_final} = M_{initial} - m_{overflow} + m_{brass\_apparent} = M_{initial} - \rho_w V_{submerged} + (m_{brass} - \rho_w V_{submerged}) = M_{initial} + m_{brass} - 2\rho_w V_{submerged}$.

The difference in mass is $M_{left\_final} - M_{right\_final} = (M_{initial} + \rho_w V_{submerged}) - (M_{initial} + m_{brass} - 2\rho_w V_{submerged}) = 3\rho_w V_{submerged} - m_{brass}$.

The right side tips down if $M_{right\_final} > M_{left\_final}$, which means $m_{brass} - 2\rho_w V_{submerged} > \rho_w V_{submerged}$, or $m_{brass} > 3\rho_w V_{submerged}$. Let $\rho_{brass}$ be the density of brass and $V_{brass}$ be the volume of the brass weight. So $m_{brass} = \rho_{brass} V_{brass}$. If the brass is fully submerged, $V_{submerged} = V_{brass}$. The condition becomes $\rho_{brass} V_{brass} > 3\rho_w V_{brass}$, or $\rho_{brass} > 3\rho_w$. Given $\rho_{brass} \approx 8.5 \times 10^3 \, \text{kg/m}^3$ and $\rho_w = 1 \times 10^3 \, \text{kg/m}^3$, $\rho_{brass} > 3\rho_w$ is true ($8.5 > 3$). Thus, the right side tips down when fully submerged.

Even if partly submerged, $V_{submerged} < V_{brass}$, the condition for the right side tipping down is $\rho_{brass} V_{brass} > 3\rho_w V_{submerged}$. This can be written as $\frac{V_{brass}}{V_{submerged}} > \frac{3\rho_w}{\rho_{brass}}$. Since $\frac{V_{brass}}{V_{submerged}} > 1$ and $\frac{3\rho_w}{\rho_{brass}} \approx \frac{3}{8.5} < 1$, this inequality always holds. Therefore, the right side tips down regardless of the degree of submersion.