Math, asked by vijay9356, 5 months ago

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.

Answers

Answered by Anonymous

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

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

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

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

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

Answered by vicky6616