Question

A solid is in the shape of a hemisphere surmounted by a cone of the same radius.The diameter of the cone is 18cm and the height of the cone is 14cm.The solid is completely immersed in a cylindrical tub,full of water.If the diameter of the tub is 26 cm and its height is 21cm,find the quantity of water left in the cylindrical tub in litres.​

Answer 1

Solution :

Volume left in the cylindrical vessel = 8.438 litres

Step by step Explanation

Given:

Radius of a cone, r = 18/2= 9cm

Height of cone, h = 14cm

Radius of hemisphere ,r = 9cm

Radius of base of the cone=Radius of the hemisphere. Because the solid is in the form of a hemisphere surmounted by a cone

To Find :

The quantity of water left in the cylindrical tub in litres.If the solid is completely immersed in a cylindrical tub,full of water.If the diameter of the tub is 26 cm and its height is 21cm.

Solution:

Volume of the solid

= Volume of hemisphere + volume of cone

$\sf=\dfrac{1}{3}\pi\:r^2h+\dfrac{2}{3}\pi\:r^3=$

$\sf=\dfrac{1}{3}\pi\:r^2(h+2r)=$

Now put the given values

$\sf=\dfrac{1}{3}\times\dfrac{22}{7}(9)(14+18)=$

$\sf=\dfrac{22\times9\times32}{7\times3}=$

$\sf=\dfrac{57024}{21}=$

$\sf=2715.42cm^3=2715.42cm$

Now,

Radius of the base of the cylindrical vessel, r=13cm

Height of the cylindrical vessel,h = 21cm

Volume of the water in the cylindrical vessel

$\sf=\pi\:r^2h=πr$

Put the given values

$\sf=\dfrac{22}{7}\:(13)^2(21)=$

$\sf=\dfrac{22\times169\times21}{7}=$

$\sf=\dfrac{78078}{7}=$

7

78078

$\sf=11,154cm^3=11,154cm$

According to the question :

The solid is completely submerged in the cylindrical vessel full of water, then

Volume of the water left in the vessel

=Volume of the water in the cylindrical vessel - volume of solid

=(11,154-2715.42)

=8438.58 cm³

= 8.438 litres

Therefore , the quantity of water left in the cylindrical tub in litres is 8.438 l

Answer 2

Solution :

Volume left in the cylindrical vessel = 8.438 litres

Step by step Explanation

Given:

Radius of a cone, r = 18/2= 9cm

Height of cone, h = 14cm

Radius of hemisphere ,r = 9cm

Radius of base of the cone=Radius of the hemisphere. Because the solid is in the form of a hemisphere surmounted by a cone

To Find :

The quantity of water left in the cylindrical tub in litres.If the solid is completely immersed in a cylindrical tub,full of water.If the diameter of the tub is 26 cm and its height is 21cm.

Solution:

Volume of the solid

= Volume of hemisphere + volume of cone

$\sf=\dfrac{1}{3}\pi\:r^2h+\dfrac{2}{3}\pi\:r^3=$

$\sf=\dfrac{1}{3}\pi\:r^2(h+2r)=$

Now put the given values

$\sf=\dfrac{1}{3}\times\dfrac{22}{7}(9)(14+18)=$

$\sf=\dfrac{22\times9\times32}{7\times3}=$

$\sf=\dfrac{57024}{21}=$

$\sf=2715.42cm^3=2715.42cm$

Now,

Radius of the base of the cylindrical vessel, r=13cm

Height of the cylindrical vessel,h = 21cm

Volume of the water in the cylindrical vessel

$\sf=\pi\:r^2h=πr$

Put the given values

$\sf=\dfrac{22}{7}\:(13)^2(21)=$

$\sf=\dfrac{22\times169\times21}{7}=$

$\sf=\dfrac{78078}{7}=$

7

78078

$\sf=11,154cm^3=11,154cm$

According to the question :

The solid is completely submerged in the cylindrical vessel full of water, then

Volume of the water left in the vessel

=Volume of the water in the cylindrical vessel - volume of solid

=(11,154-2715.42)

=8438.58 cm³

= 8.438 litres

Therefore , the quantity of water left in the cylindrical tub in litres is 8.438 l