A cylindrical can, whose base is horizontal and of radius 3.5 cm,… - Vidyadip

Question

A cylindrical can, whose base is horizontal and of radius $3.5\ cm,$ contains sufficient water sothat when a sphere is placed in the can, the water just covers the sphere. Given that the spherejust fits into the can, calculate:
(i) the total surface area of the can in contact with water when the sphere is in it;
(ii) the depth of water in the can before the sphere was put into the can.

✓

Answer

Radius of the base of the cylindrical can $= 3.5\ cm$
$(1)$ When the sphere is in can, then total surface area of the can $=$ Base area $+$ curved surface area
$=\pi r^2+2 \pi r h $
$=\left(\frac{22}{7} \times 3.5 \times 3.5\right)+\left(2 \times \frac{22}{7} \times 3.5 \times 7\right) $
$ =\frac{77}{2}+154$
$ =38.5=154 $
$=192.5\ cm ^2$
$(2)$ Let depth of water $= x\ cm$
When sphere is not in the can, then volume of the can $=$ volume of water $+$ volume of sphere
$\Rightarrow \pi r^2 h+\pi^2 x \times+\frac{4}{3} \pi r^3 $
$ \Rightarrow \pi r^2 h+\pi r^2\left(x+\frac{4}{3} r\right) $
$ \Rightarrow h=x+\frac{4}{3} r $
$\Rightarrow x=h-\frac{4}{3} r $
$\Rightarrow x=7-\frac{4}{3} \times \frac{7}{2} $
$ \Rightarrow x=7-\frac{14}{3}$
$ \Rightarrow x=\frac{21-14}{3} $
$\Rightarrow x=\frac{7}{3} $
$\Rightarrow x=2 \frac{1}{3} cm$

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