Please help with this one

Dear Student,


Given : A cylindrical can with radius = 3.5 cm

And when a sphere is placed in a can, the water just covers the sphere . Given that the sphere just fits into the can , So

Radius of sphere  =  3.5 cm 

And

Height of can  =  Height of water after sphere in the can  =  Diameter of sphere =  7  cm 

And can is open from one side , So

Total surface area of can in contact with water  = $2 \mathrm{\pi } r h + \mathrm{\pi } r^2 = 2 \times \frac{22}{7}\times 3.5 \times 7 + \frac{22}{7}\times 3.5^2 = 22 \times 7 + 22 \times .5\times 3.5 = 154 + 38.5 = 192.5 = \frac{\mathbf{385}}{\mathbf{2}}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$
Here ,

Volume of sphere =  Volume of water rise

Let level of water rise = h m 

So,

$$\begin{gathered} \frac{ 4 \mathrm{\pi } r^3}{3} = \mathrm{\pi } r^{2}h \\ \Rightarrow \frac{ 4 \times \mathrm{\pi } \times 3.5^3}{3} = \mathrm{\pi } \times 3.5^2 \times h \\ \Rightarrow \frac{ 4 \times 3.5}{3} = h \\ \mathbf{\Rightarrow }\mathit{h}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{14}}{\mathbf{3}} \end{gathered}$$
So,

Level of water rise = $\frac{14}{3}$ cm

Therefore ,

The depth of the water in the can before the sphere was put into the can =  7 - $\frac{14}{3}$  = $\frac{21 - 14}{3} = \frac{\mathbf{7}}{\mathbf{3}}$cm                        ( Ans )




.

Regards