Question

What will the density be if the radius of the sphere is halved

Answer 1

Answer:

Density would be 8 times.

Step-by-step explanation:

Let r represent radius of original sphere.

We are asked to determine the density of a sphere, when its radius is halved.

We know that density is inversely proportional to cube of radius .

$\text{Density}=\frac{1}{r^3}$

When radius is halved, so new radius would be $\frac{r}{2}$.

Now density would be:

$\text{New density}=\frac{1}{(\frac{r}{2})^3}$

$\text{New density}=\frac{1}{\frac{r^3}{2^3}}$

$\text{New density}=\frac{1}{\frac{r^3}{8}}$

Using property $\frac{a}{\frac{b}{c}}=\frac{a\cdot c}{b}$, we will get:

$\text{New density}=\frac{1*8}{r^3}$

$\text{New density}=\frac{8}{r^3}$

We can see that new density is 8 times the original density. Therefore, the density would be 8 times if the radius of the sphere is halved.