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# A Buoy is Made in the Form of Hemisphere Surmounted by a Right Cone Whose Circular Base Coincides with the Plane Surface of Hemisphere. - ICSE Class 10 - Mathematics

ConceptArea and Volume of Solids - Cone

#### Question

A buoy is made in the form of hemisphere surmounted by a right cone whose circular base coincides with the plane surface of hemisphere. The radius of the base of the cone is 3.5 metres and its volume is two third of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal

#### Solution

Radius of hemispherical part (r)=3.5 m=7/2 m

Therefore, Volume of hemisphere=2/3pir^3

=2/3xx22/7xx7/2xx7/2xx7/2

=539/6 m^3

Volume of conical part2/3xx539/6m^3(2/3 "of hemisphere")

Let height of the cone = h
Then,

1/3pir^2h=(2xx539)/(3xx6)

⇒ 1/3xx22/7xx7/2xx7/2xxh=(2xx539)/(3xx6)

⇒ =(2xx539xx2xx2xx7xx3)/(3xx6xx22xx7xx7)

⇒h=14/3m=4 2/3m =4.67 m

Height of the cone = 4.67 m

Surface area of buoy=2pir^2+pirl

But l=sqrt(r^2+h^2)

l=sqrt((7/2)^2+(14/3)^2

=sqrt(49/4+196/9)=sqrt1225/36=35/6m

Therefore, Surface area =

=(2xx22/7xx7/2xx7/2)+(22/7xx7/2xx35/6)m^2

=77/1+385/6=847/6

=141.17 m^2

Surface Area=141.17 m^2

Is there an error in this question or solution?

#### APPEARS IN

Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2020 to Current)
Chapter 20: Cylinder, Cone and Sphere
Exercise 20(E) | Q: 2 | Page no. 311
Solution A Buoy is Made in the Form of Hemisphere Surmounted by a Right Cone Whose Circular Base Coincides with the Plane Surface of Hemisphere. Concept: Area and Volume of Solids - Cone.
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