A fixed volume of oil is dispersed in water as identical spherical droplets. If the droplet radius is halved, the number of droplets and the total interfacial area change respectively by factors of:

Answer: C

Total volume

V = n \cdot \frac{4}{3} π r^{3}

is fixed, so

n \propto 1/ r^{3}

and halving

r

gives 8 times as many droplets. Total area

= n \cdot 4 π r^{2} = 3 V / r

, which doubles when

r

is halved.