Mensuration Practice Test 10

According to the question,
Volume of the can upto the length of sphere placed + Volume of the sphere = volume of the can upto the depth of water that was in the can before the sphered is placed.
$\mathit{\pi }{\mathit{r}}^2\mathit{h}+\frac{4}{3}\mathit{\pi }{\mathit{r}}^3$ = $\mathit{\pi }{\mathit{r}}^2\mathit{H}$
Where H is the Height of the depth of water
before the sphered is place
Dividing by $\mathit{\pi }{\mathit{r}}^2$ on both sides.
$\mathit{h}+\frac{4}{3}\left(3.5\right)$ = 7
h = $\frac{7}{3}$ cm.