Question

A sphere of diameter 6 cm , is droped in a right circular cylindrical vessel partly filled with water. the diameter of the cylindrical vessel is 12 cm . if the sphere completely submarged in water , by how much will the level of water rise in the cylindrical vessel ?

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Answer 1

Answer:

1 cm

Step-by-step explanation:

Given, diameter of speher = 6 cm

⇒ radius of sphere R= 6/2 =3cm

Diameter of cylindrical vessel = 12 cm

⇒ radius of cylindrical vessel = r=

2/12

=6cm

volume of sphere =

$\frac{4}{3} \pi \: r {}^{3 }$

=

$\frac{4}{3} \pi \:(3 {}^{3} )$

Volume of cylindrical vessel =

$\pi \: r {}^{2} \: h$

=

$\pi \: 6 {}^{2} h$

Here, the volume of sphere = volume increased in the cylindrical vessel.

∴

$\frac{4}{3} \pi(3) {}^{2} \: = \: \pi(6 ){}^{2} h$

h =

$\frac{4}{3} \times \frac{3}{6} \: = \: \frac{4 \times 3 \times 3}{6 \times 6} \: = \: 1cm$

=1cm

∴ 1 cm level of water rise in the cylindrical vessel.

Answer 2

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$☃ ƛƝƧƜЄƦ ☃ \\ \small \red{{Diameter \: of \: the \: sphere \: = \: 6 cm}} \\ \small\green{{ \therefore Radius ( r ) = 3 cm}}$

$\small \blue{{{Volume \: of \: the \: sphere = \: \frac{4}{3}\pi \: r^{3} = \frac{4}{3} \times \pi \times (3)^{2} \: cm^{3}}}}$

$\small \pink{{ = 36\pi \: cm^{3}}}$

$\small \orange{{Let \: the \: water \: level \: in \: the \: cylindrical \: vessel \: rises \: h \: cm}} \\ \small \red{{volume \: of \: sphere = volume \: of \: water \: rises}} \\ \small \blue{ 36\pi = 36\pi \: h = 1 \: cm \: } \\ \small \pink{\therefore \:level \: of \: water \: rise \: is \}} \: \small\red{{1 \: cm}}$

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