Question

A cone of radius 9cm and height 12 cm is full of water. A sphere which exactly

fits and touches the sides of the cone is immersed in it. Find: [4]

(i) the radius of the sphere.

(ii) the fraction of the water that overflows​

Answer 1

Step-by-step explanation:

volume of cone=

$\frac{1}{3} \pi {r}^{2} h \\ \frac{1}{3 } \times \frac{22}{7} \times {9}^{2} \times 12 \\ \frac{22}{7} \times 81 \times 4 \\ \frac{22}{7} \times 324 \\ \frac{7128}{7} \\ = 118.28$

cone volume=118.28

1).9cm

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Answer 2

The radius of the sphere is 4.5 cm

The fraction of the water that overflows​ is 3:8

Step-by-step explanation:

radius of a conical vessel (R) = 9 cm

height of the conical vessel (H) = 12 cm

volume of the conical vessel = $\frac{1}{3}\pi R^2H$

= $\frac{1}{3}\pi \times (9)^2\times (12)$

= $324\pi$ cm³

let the radius of the sphere be r cm'

in right triangle PO'R

using pythagoras

l² = r²+h²

$l= \sqrt{9^2+12^2}$

$l= \sqrt{225}$

l = 15 cm

hence

$sin\Theta =\frac{O'P}{PR} =\frac{9}{15} = \frac{3}{5}$...(1)

in triangle MRO

$sin\Theta =\frac{OM}{OR} =\frac{r}{OR}$

$\frac{3}{5} = \frac{r}{12-r}$

3(12-r) = 5r

36- 3r = 5r

36 = 8r

r = 4.5 cm

hence, the radius of the sphere is 4.5 cm

volume of sphere = $\frac{4}{3}\pi r^3$

= $\frac{4}{3} \pi (4.5)^3$

= $121.5 \pi$ cm³

volume of water  = volume of cone

clearly , volume of water that flows out of cone is same as the volume of the sphere

i.e ,

fraction of water that flows out = $\frac{volume of sphere}{volume of cone}$

= $\frac{121.5\pi }{324\pi }$

= $\frac{121.5}{324}$

= $\frac{3}{8}$

hence , the radius of the sphere is 4.5 cm

the fraction of the water that overflows​ is 3:8

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