Question #7917e

1 Answer
Jun 22, 2015

Answer:

Here's why the volume of the final drop is equal to the sum of the volumes of the initial drops.

Explanation:

So, you know that you have two water drops, each having a different radius. The first drop has a radius of 3 cm and the second one has a radius of 4 cm.

Now, you can relate mass and volume using density.

#rho = m/V#

Since you're dealing with two water drops, it's clear that the density term will be equal for both drops.

This implies that you have different masses of water in the two drops.

#{(m_1 = rho * V_1), (m_2 = rho * V_2) :}#

Now, when the two drops come together, the mass of the resulting drop will be equal to

#m_"final" = m_1 + m_2#

This means that, if you want to find the volume of the final drop, you can write

#V_"final" = m_"final"/(rho) = (m_1 + m_2)/(rho) = (cancel(rho) * V_1 + cancel(rho) * V_2)/cancel(rho)#

#V_"final" = V_1 + V_2#

Since you can treat the drops as spheres, the two volumes will be equal to

#V_1 = 4/3 pi * 3^3 = 36pi#, and

#V_2 = 4/3 pi * 4^3 = (256pi)/3#

The final volume will thus be

#V_"final" = 36pi + (256pi)/3 = (364pi)/3#

If you take #r_"final"# to be the radius of the final drop, you can write

#V_"final" = 4/3pi * r_"final"^3 = 364/3pi#

This is equivalent to

#4/cancel(3)cancel(pi) * r_"final"^3 = 364/cancel(3)cancel(pi) => r_"final"^3 = 364/4 = 91#

Therefore,

#r_"final" = root(3)(91) = color(green)("4.4979 cm")# #-># without taking into account sig figs