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(C) Let $M$ be the mass of the boat and $m$ be the total mass of the steel balls. Let $\rho_{\omega}$ be the density of water and $\sigma_{s}$ be the

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It will fall

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(C) Let $M$ be the mass of the boat and $m$ be the total mass of the steel balls. Let $\rho_{\omega}$ be the density of water and $\sigma_{s}$ be the density of steel.When the balls are in the boat,the boat floats. According to the law of floatation,the weight of the displaced water is equal to the total weight of the boat and the balls: $W = (M + m)g$.The volume of water displaced is $V_{1} = \frac{M + m}{\rho_{\omega}}$.When the steel balls are thrown into the water,they sink because $\sigma_{s} > \rho_{\omega}$.The boat now only displaces water equal to its own weight: $V_{boat} = \frac{M}{\rho_{\omega}}$.The steel balls displace water equal to their own volume: $V_{balls} = \frac{m}{\sigma_{s}}$.The total volume of water displaced is $V_{2} = \frac{M}{\rho_{\omega}} + \frac{m}{\sigma_{s}}$.Since $\sigma_{s} > \rho_{\omega}$,it follows that $\frac{m}{\sigma_{s}} Therefore,$V_{2} As the volume of displaced water decreases,the level of water in the tank will fall.

$A$ 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?

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