# SURFACE AREA AND VOLUME OF COMBINATION OF SOLIDS QUESTIONS

Surface Area and Volume of Combination of Solids Questions

Here we are going to see, some practice questions on based on finding surface area and volume of combination of solids.

## Surface Area and Volume of Combination of Solids - Questions

Question 1 :

A capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends. If the length of the entire capsule is 12 mm and the diameter of the capsule is 3 mm, how much medicine it can hold?

Solution : Capacity of capsule

=  2 Volume of hemisphere + volume of cylinder

=  2(2/3) πr3 + πr2h

=  (4/3) πr3 + πr2h

Height of Capsule

=  2 (radius of hemisphere) + height of cylinder

2(3/2) + h  =  12

3 + h =  12

h  =  12 - 3  =  9

=  πr[(4/3) r + h]

=  (22/7) (3/2)2[(4/3)(3/2) + 9]

=  (22/7) (9/4)[2 + 9]

=  (11/7) (9/2)(11)

=  77.78 cm3

Question 2 :

As shown in figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.

Solution : Surface area of solid

=  surface area of cube + curved surface area of hemisphere - area of base  of hemisphere

= 6a2 + 2πrπr

= 6a2 + πr2

=  6(7)2 + (22/7) (7/2)

=  294 + 38.5

=  332.5 cm2

Question 3 :

A right circular cylinder just enclose a sphere of radius r units. Calculate (i) the surface area of the sphere (ii) the curved surface area of the cylinder (iii) the ratio of the areas obtained in (i) and (ii).

Solution : radius of sphere  =  height of cylinder/2

(i) the surface area of the sphere  =  4πr2

(ii) the curved surface area of the cylinder

=  2π r h

=  2π r(2r)

=  4πr2

(iii) the ratio of the areas obtained in (i) and (ii).

=  4πr2  4πr2

=  1 : 1

Question 4 :

A shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.

Solution :

Surface area of shuttle cock  =  curved surface area of frustum cone + curved surface area of hemisphere

=  π (R + r) l + 2πr2  ----(1)

Height of shuttle cock  =  7

radius of hemisphere + height of frustum cone  =  7

1 + h  =  7

h  =  6

l = √(h2 + (R - r)2)

l = √(62 + ((5/2) - 1)2)

l = √(36 + (9/4)

l = √153/2

l = 12.36/2

l = 6.18

By applying the value of l in (1), we get

=  π ((5/2) + 1) l + 2πr2

=  π[(7/2)(6.18) + 2 (1)2]

=  (22/7)[(21.63 + 2]

=  74.26 cm2 After having gone through the stuff given above, we hope that the students would have understood, how to solve problems on surface area and volume of combination of solids.

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