Question

the sphere of diameter 18cm is dropped into a cylindrical vessel of diameter 36cm partly filled with water if the sphere is completely submerge then the water level rises by?

Answer 1

Plz find the attachment for answers

Answer 2

Answer:

$\red { Rised \:water \:level } \green {= 3\:cm }$

Step-by-step explanation:

$Diameter \: of \: a \: Sphere (d) = 18\:cm$

$Radius \: of \: the \: Sphere (r) = \frac{d}{2}\\= \frac{18}{2} = 9 \:cm$

$Diameter \: of \: a \: cyliderical \: vessel (D)=36\:cm$

$Radius \: the \: vessel (R) = \frac{D}{2} \\= \frac{36}{2} \\= 18 \:cm$

/* According to the problem given

If the sphere dropped into the cylindrical vessel partly filled with water if the Sphere is completely submerged ,

$Let \: the \: water \:level \: rises = H \:cm$

$Volume \:of \: the \: vessel = Volume \:of \: the \: sphere$

$\implies \pi R^{2} H = \frac{4}{3} \pi r^{3}$

$\implies H = \frac{\frac{4}{3} \pi r^{3}}{\pi R^{2}}$

$= \frac{ 4 \times r^{3}}{3\times R^{2}}\\= \frac{ 4 \times 9^{3}}{3\times 18^{2}}\\= \frac{ 4\times 9\times 9 \times 9 }{3 \times 18\times 18 }\\= 3\:cm$

Therefore.,

$\red { Rised \:water \:level } \green {= 3\:cm }$

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