Question

an object will float at the surface of a liquid if the mass of the object is less than the mess of the liquid that it displaces. A spherical vessel (diameter = 5cm) when empty has a mass of 12g. What is the greatest volume of water that can be placed in the vessel and still have the vessel float at the surface of benzene?

Answer 1

Answer:

45.44 mL

Explanation:

Given:

• m = mass of the empty spherical vessel = 12 g = 0.012 kg
• d = diameter of the spherical vessel = 5 cm

Assume:

• r = radius of the spherical vessel = $\dfrac{d}{2}=\dfrac{5\ cm}{2} = 2.5\ cm = 2.5\times 10^{-2}\ m$
• $\rho$ = density of benzene = $876\ kg/m^3$
• $\rho_w$ = density of water = $997\ kg/m^3$
• V = the greatest volume of water that can be placed in the vessel
• v = volume of the spherical vessel

According to the given question, an object floats at the surface of a liquid if the mass of the object is less than the mass of the liquid displaced by it.

Let the greatest volume of water that can be put in the vessel so that it still floats at the surface of the benzene. Then the sum of the masses of the volume of water in the vessel and the empty vessel is equal to the mass of the volume of benzene displaced by the whole vessel.

i.e., Mass of water in the vessel + Mass of the empty vessel = Mass of the benzene displaced by the whole vessel

$\Rightarrow V\rho_w+m=v \rho\\\Rightarrow V\rho_w+m=\dfrac{4}{3}\pi r^3\rho\\\Rightarrow V\rho_w=\dfrac{4}{3}\pi r^3\times \rho-m\\\Rightarrow V\rho_w=\dfrac{4}{3}\pi (2.5\times 10^{-2})^3\times 876-0.012\\\Rightarrow V\rho_w=0.0573-0.012\\\Rightarrow V\times 997=0.0453\\\Rightarrow V=\dfrac{0.0453}{997}\\\Rightarrow V=4.544\times 10^{-5}\ m^3\\\Rightarrow V=4.544\times 10^{-5}\times 10^6\ mL\\\Rightarrow V=45.44\ mL$

Hence, the greatest volume of water that can be placed in the vessel such that it still floats on the benzene surface is 45.44 mL.