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 Volume and Surface Area Questions

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FACTS  AND  FORMULAE  FOR  VOLUME  AND  SURFACE  AREA  QUESTIONS

 

 

I. CUBOID

   Let length=l, breadth =b and height =h units. Then,

1. Volume = (l x b x h)

2. Surface area = 2(lb +bh + lh) sq.units

3. Diagonal =l2+b2+h2 units

 

 

II. CUBE 

Let each edge of a cube be of length a. Then,

1. Volume = a3 cubic units.

2. Surface area = 6a2 sq.units

3. Diagonal = 3a units

 

 

III. CYLINDER 

Let radius of base = r and Height (or Length) = h. Then, 

1.Volume = πr2h cubic units 

2. Curved surface area = 2πrh sq.units

3. Total surface area = 2πrh+2πr2 sq.units

 

 

IV. CONE 

Let radius of base =r and Height = h. Then, 

1. Slant height, l=h2+r2 units

 

2. Volume = 13πr2h cubic units.

 

3. Curved surface area = πrlsq.units 

 

4. Total surface area = πrl+πr2sq.units

 

 

V. SPHERE 

Let the radius of the sphere be r. Then, 

1. Volume =43πr3 cubic units

2. Surface area = 4πr2 sq.units

 

 

VI. HEMISPHERE 

Let the radius of a hemisphere be r. Then, 

1. Volume = 23πr3 cubic units.

2. Curved surface area = 2πr2 sq.units

3. Total surface area = 3πr2 sq.units

 

Q:

A cuboid of dimension 24cm x 9cm x 8 cm is melted and smaller cubers are of side 3 cm is formed.find how many such cubes can be formed ?

A) 64 B) 56
C) 48 D) 40
 
Answer & Explanation Answer: A) 64

Explanation:

Volume of cuboid = (24 x 9 x 8) cm = 1728 cu.cm
Volume of small cube = (3 x 3 x 3) cm = 27 cu.cm


So, No. of small cubes formed = 1728/27 = 64.

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5 9201
Q:

88 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be ?

A) 112 mts B) 84 mts
C) 96 mts D) 108 mts
 
Answer & Explanation Answer: B) 84 mts

Explanation:

Vol of silver = πR2h  = 66 C.C  

Given D= 1mm => R = D/2 = 1/20 cm  

 

=> 227×1202×h = 66

= 66 x 400 x (7/22) = 8400 cm = 84 mtrs.

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11 8645
Q:

The diagonal of the floor of a rectangular closet is 712 feet. The shorter side of the closet is 412 feet. What is the area of the closet in square feet?

A) 5 .5 B) 13.5
C) 27 D) 37
 
Answer & Explanation Answer: C) 27

Explanation:

Other side=1522-922ft  =2254-814ft  =1444= 6ft

 

Area of closet =(6 x 4.5)sq.ft = 27 sq.ft 

 

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5 8475
Q:

A rectangular box measures internally 1.6m long, 1m broad and 60cm deep. The number of cubical box each of edge 20cm that can be packed inside the box is:

A) 30 B) 60
C) 90 D) 120
 
Answer & Explanation Answer: D) 120

Explanation:

Number of blocks =160×100×6020×20×20=120

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10 8410
Q:

A rectangular block has the dimensions 5x6x7 cm it is dropped into a cylindrical vessel of radius 6cm and height 10 cm. If the level of the fluid in the cylinder rises by 4 cm, What portion of the block is immersed in the fluid ?

A) 22/7 x 24/35 B) 22/7 x 36 x 4
C) 22/7 x 36/5 D) 22/7 x 37/21
 
Answer & Explanation Answer: A) 22/7 x 24/35

Explanation:

Since level of water increased in cylinder by height 4. 

This is because of the rectangular block .

 

Therefore , area of rectangular block immersed in water is 

= 227×6×6×4  

Thats why portion of block immersed in water is 

= (22/7 * 36 * 4) / total vol. of rectangle 

= (22/7 * 36 * 4)/ (7*5*6) 

= (22/7 * 24)/ 35.

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7 8324
Q:

A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1cm when a man gets on it. The mass of the man is:

A) 12 kg B) 60 kg
C) 72 kg D) 96 kg
 
Answer & Explanation Answer: B) 60 kg

Explanation:

ke

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11 8277
Q:

How many meshes are there in 1 squrare meter of wire gauge if each mesh is 8mm long and 5mm wide ?

A) 2500 B) 25000
C) 12500 D) 15400
 
Answer & Explanation Answer: B) 25000

Explanation:

We know that,

 

One square meter is equal to 10,00,000 sq mm.
Given, each mesh is 8mm length and 5mm wide
Therefore for each mesh : 8 x 5 = 40 sq mm.
So, number of meshes = 100000040 = 25000.

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6 8052
Q:

The volume of a rectangular solid is to be increased by 50% without altering its base. To what extent the height of the solid must be changed.

A) 50% B) 40%
C) 30% D) 20%
 
Answer & Explanation Answer: A) 50%

Explanation:

Let l, b the sides of the base, so that the area of the base = lb
remains constant.
Let ‘h’ be the height of the solid.
Volume of the solid, V = l× b× h.
Since the volume is to be increased by 50%

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8 7664