- #1

brotherbobby

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- Homework Statement
- The figure I put below shows three identical open-top containers filled to the brim with water - toy ducks float in two of them. If each of the containers were put on a weighing scale, rank the containers (a), (b) and (c) with the heaviest of them first.

- Relevant Equations
- 1. ##\text{Law of floatation : Weight (or mass) of a body = Weight (or mass) of liquid displaced}##. 2. Weight of a body ##w = mg = \rho V g##

We understand that the crucial thing about the problem is that the volume of water present in the three containers

**are not**the same. Also, we note that in each case the weight of the container is the total weight of its contents. (A student might be confused as to why should be so - after all, isn't a floating body weightless?! Yes it is - but in order for it to be weightless, the liquid must apply a buoyant force equal to the body's weight on it. The floating body in turn applies the same force back on to the liquid by the third law and this increases the weight of the liquid by an amount equal to that of the floating body).

If we denote the volume of the container itself as ##V##, then the volume of water in each container can be ranked : ##(V_W)_a [= V] > (V_W)_b > (V_W)_c##. Far as the densities are concerned, the density of water is more than that of the duck (##\rho_W > \rho_D##) and for the ducks in (b) and (c), we have ##(V_D)_b < (V_D)_c##.

(a) The weight of the container in case (a) is simple : ##\boxed{w_a = \rho_W Vg}##

(b) The weight of the container in case (b) : ##w_b = (w_W)_b + (w_D)b##. The weight of the duck is ##w_D = \rho_D (V_D)_b g##. As for the volume of water in (b), ##(V_W)_b = V- (V_D)_b##. Hence the weight of the water : ##(w_W)_b = \rho_W [V - (V_D)_b]g##. Opening brackets and adding the two weights, we find that the weight of the container in (b) ##w_b = \rho_D (V_D)_b g + \rho_W Vg - \rho_W (V_D)_b g## or we have ##\boxed{w_b = \rho_W Vg - (V_D)_b g\underbrace{(\rho_W - \rho_D)}_{>\;0}}##. Given that the density of the duck is less than that of water, we have ##\boxed{\color {green} {w_a > w_b}}##.

(c) This is identical to (b) above, except that the volume of the duck is more : ##(V_D)_c > (V_D)_b##. Hence the weight of the container for c is ##\boxed{w_c = \rho_W Vg - (V_D)_c g(\rho_W - \rho_D)}##. Subtracting a larger amount from the weight in (a) here than in (b) implies that ##\boxed{\color {green} {w_b > w_c}}##.

Hence, finally : ##\boxed{\color {green} {\mathbf{w_a > w_b>w_c}}}##.

**Is my solution correct?**