Question

Seven little spheres of mercury, each with a diameter of 2 mm. When they coalesce to form a single sphere, how big will it be (i.e. what is its diameter)? How does its surface area compare with the total surface area of the previous Seven little spheres?

Answer 1

Answer:

The total surface area of the seven little spheres is 1.91 times the total surface area of the bigger sphere.

Explanation:

Volume of a Sphere

The volume of a sphere of radius r is given by:

$\displaystyle V=\frac{4}{3}\cdot \pi\cdot r^3$

The volume of each little sphere is:

$\displaystyle V_l=\frac{4}{3}\cdot \pi\cdot 2^3$

$V_l=33.51\ mm^3$

When the seven little spheres coalesce, they form a single bigger sphere of volume:

$V_b=7*V_l=234.57\ mm^3$

Knowing the volume, we can find the radius rb by solving the formula for r:

$\displaystyle V_b=\frac{4}{3}\cdot \pi\cdot r_b^3$

Multiplying by 3:

$3V_b=4\cdot \pi\cdot r_b^3$

Dividing by 4π:

$\displaystyle \frac{3V_b}{4\cdot \pi}= r_b^3$

Taking the cubic root:

$\displaystyle r_b=\sqrt[3]{\frac{3V_b}{4\cdot \pi}}$

Substituting:

$\displaystyle r_b=\sqrt[3]{\frac{3*234.57}{4\cdot \pi}}$

$r_b=3.83\ mm$

The surface area of the seven little spheres is:

$A_l=7*(4\pi r^2)=7*(4\pi 2^2)=351.86\ mm^2$

The surface area of the bigger sphere is:

$A_b=4\pi r_b^2=4\pi (3.83)^2=184.33\ mm^2$

The ratio between them is:

$\displaystyle \frac{351.86\ mm^2}{184.33\ mm^2}=1.91$

The total surface area of the seven little spheres is 1.91 times the total surface area of the bigger sphere.