**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.

**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) **πr ^{3} + **

** = (4/3) πr^{3} + πr^{2}h**

**Height of Capsule**

** = 2 (radius of hemisphere) + height of cylinder**

**2(3/2) + h = 12**

**3 + h = 12**

**h = 12 - 3 = 9**

**= ****πr ^{2 }[(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 cm ^{3}**

**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

= 6a^{2} + 2**πr ^{2 }- **

** = 6a ^{2} + πr^{2}**

** = 6(7) ^{2} + (22/7) (7/2)^{2 }**

** = 294 + 38.5**

** = 332.5 cm ^{2}**

**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 = 4**πr ^{2}**

**(ii) the curved surface area of the cylinder**

** = 2****π r h**

** = 2****π**** r(2r)**

= 4**πr ^{2}**

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

= 4**πr^{2 }: **4

** = 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 + 2****π****r ^{2} ----(1) **

Height of shuttle cock = 7

radius of hemisphere + height of frustum cone = 7

1 + h = 7

h = 6

l = √(h^{2} + (R - r)^{2})

l = √(6^{2} + ((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 + 2πr^{2}**

** = ****π[(7/2)(6.18**) + 2 (1)^{2}]

** = (22/7)****[(21.63 + 2**]

= 74.26 cm^{2}

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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