# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 15 - Volume and Surface Area of Solids

Choosing the correct option for multiple choice questions in your Maths exam requires practice. For chapter revision, you can refer to TopperLearning’s RS Aggarwal and V Aggarwal Solutions for CBSE Class 9 Maths Chapter 15 Volume and Surface Area of Solids. The solutions by some of the best Maths experts will help you relearn concepts such as the total surface area of a cube, volume of a cube, volume of a cylinder etc.

With our chapter solutions, get clarity on application-type questions such as calculating the concrete mixture for building a pillar. Besides, our CBSE Class 9 Mathematics videos and sample papers will be useful during Maths exam preparation.

## Chapter 15 - Volume and Surface Area of Solids Exercise MCQ

Correct option: (b)

## Chapter 15 - Volume and Surface Area of Solids Exercise Ex. 15C

Radius of a cone, r = 5.25 cm

Slant height of a cone, l = 10 cm

Radius of a cone, r = 12 m

Slant height of a cone, l = 21 cm

Radius of a conical cap, r = 7 cm

Height of a conical cap, h = 24 cm

Thus, 5500 cm^{2}
sheet will be required to make 10 caps.

Let r be the radius of a cone.

Slant height of a cone, l = 14 cm

Curved
surface area of a cone = 308 cm^{2}

Radius of a cone, r = 7 m

Slant height of a cone, l = 25 m

Cost
of whitewashing = Rs. 12 per m^{2}

⇒ Cost of
whitewashing 550 m^{2} area = Rs. (12 ×
550) = Rs. 6600

Radius of a conical tent, r = 24 m

Height of a conical tent, h = 10 m

Radius of a conical heap, r = 4.5 m

Height of a conical tent, h = 3.5 m

Radius of a conical tent, r = 7 m

Area
of canvas used in making conical tent = (551 - 1) m^{2} = 550 m^{2}

⇒ Curved surface
area of a conical tent = 550 m^{2}

_{}

_{ }

_{}

Curved
surface area of the tent = Area of the cloth = 165 m^{2}

## Chapter 15 - Volume and Surface Area of Solids Exercise Ex. 15A

Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.

_{Volume of cuboid= l x b x h}

_{= (24 x 0.25 x 6) m3.}

_{= 36 m3.}

_{Lateral surface area= 2(l + b) x h}

_{= [2(24 +0.25) x 6] m2}

_{= (2 x 24.25 x 6) m2}

_{= 291 m2.}

_{Total surface area =2(lb+ bh + lh)}

_{=2(24 x 0.25+0.25x 6 +24 x 6) m2}

_{= 2(6+1.5+144) m2}

_{= (2 x151.5) m2=303 m2.}

Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m

_{Volume of a cuboid = l x b x h}

_{= (15 x 6 x 0.5) m3=45 m3.}

_{Lateral surface area = 2(l + b) x h }

_{= [2(15 + 6) x 0.5] m2}

_{= (2 x 21x0.5) m2=21 m2}

_{Total surface area =2(lb+ bh + lh)}

_{= 2(15 x 6 +6 x 0.5+ 15 x 0.5) m2}

_{= 2(90+3+7.5) m2}

_{= (2 x 100.5) m2}

_{=201 m2}

Length 26 m, breadth =14 m and height =6.5 m

_{}Volume of a cuboid= l x b x h

= (26 x 14 x 6.5) m^{3}

= 2366 m^{3}

_{}Lateral surface area of a cuboid =2 (l + b) x h

= [2(26+14) x 6.5] m^{2}

= (2 x 40 x 6.5) m^{2}

= 520 m^{2}

_{Total surface area= 2(lb+ bh + lh)}

_{= 2(26 x 14+14 x6.5 +26 x6.5)}

_{= 2 (364+91+169) m2 }

_{= (2 x 624) m2= 1248 m2.}

length =12cm, breadth = 8 cm and height = 4.5 cm

_{}Volume of cuboid = l x b x h

= (12 x 8 x 4.5) cm^{3}= 432 cm^{3}

_{}Lateral surface area of a cuboid = 2(l + b) x h

= [2(12 + 8) x 4.5] cm^{2}

= (2 x 20 x 4.5) cm^{2} = 180 cm^{2}

_{}Total surface area cuboid = 2(lb +b h+ l h)

= 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm^{2}

= 2(96 +36 +54) cm^{2}

= (2 x186) cm^{2}

= 372 cm^{2}

For a matchbox,

Length = 4 cm

Breadth = 2.5 cm

Height = 1.5 cm

Volume of one matchbox = Volume of cuboid

= Length × Breadth × Height

=
(4 × 2.5 × 1.5) cm^{3}

=
15 cm^{3}

Hence,
volume of 12 such matchboxes = 12 × 15 = 180 cm^{3}

For a cuboidal water tank,

Length = 6 m

Breadth = 5 m

Height = 4.5 m

Now,

Volume of a cuboidal water tank = Length × Breadth × Height

=
(6 × 5 × 4.5) m^{3}

=
135 m^{3}

= 135 × 1000 litres

= 135000 litres

Thus, a tank can hold 135000 litres of water.

For a cuboidal water tank,

Length = 10 m

Breadth = 2.5 m

Volume
= 50000 litres = 50 m^{3}

Now,

Volume of a cuboidal tank = Length × Breadth × Height

⇒ 50 = 10 × 2.5 × Height

⇒ Height = 2 m = Depth

Thus, the depth of a tank is 2 m.

For a godown,

Length = 40 m

Breadth = 25 m

Height = 15 m

Volume of a godown = Length × Breadth × Height

= (40 × 25 × 15) m^{3}

For each wooden crate,

Length = 1.5 m

Breadth = 1.25 m

Height = 0.5 m

Volume of each wooden crate = Length × Breadth × Height

= (1.5 × 1.25 × 0.5) m^{3}

_{Length of Cistern = 8 m}

_{ Breadth of Cistern = 6 m }

_{ And Height (depth) of Cistern =2.5 m }

_{ Capacity of the Cistern = Volume of cistern}

_{ Volume of Cistern = (l x b x h) }

_{ = (8 x 6 x2.5) m3}

_{ =120 m3}

_{ }

_{ }

_{Area of the iron sheet required = Total surface area of the cistem.}

_{ Total surface area = 2(lb +bh +lh)}

_{ = 2(8 x 6 + 6x2.5+ 2.5x8) m2 }

_{ = 2(48 + 15 + 20) m2}

_{ = (2 x 83) m2=166 m2}

Area of four walls of the room = 2(length + breadth) × Height

=
[2(9 + 8) × 6.5] m^{2}

=
(34 × 6.5) m^{2}

=
221 m^{2}

Area
of one door = Length × Breadth = (2 × 1.5) m^{2} = 3 m^{2}

Area of two windows = 2 × (Length × Breadth)

=
[2 × (1.5 × 1)] m^{2}

=
(2 × 1.5) m^{2}

=
3 m^{2}

Area to be whitewashed

= Area of four walls of the room - Area of one door - Area of two windows

=
(221 - 3 - 3) m^{2}

=
215 m^{2}

Cost of whitewashing = Rs. 25 per square metre

⇒ Cost of
whitewashing 215 m^{2} = Rs. (25 × 215) = Rs. 5375

_{L}

External length of the box = 36 cm

External breadth of the box = 25 cm

External height of the box = 16.5 cm

∴ External
volume of the box = (36 × 25 × 16.5) cm^{3}
= 14850 cm^{3}

Internal length of the box = [36 - (1.5 × 2)] cm = 33 cm

Internal breadth of the box = [25 - (1.5 × 2)] cm = 22 cm

Internal height of the box = (16.5 - 1.5) cm = 15 cm

∴ Internal
volume of the box = (33 × 22 × 15) cm^{3}
= 10890 cm^{3}

Thus, volume of iron used in the box

= External volume of the box - Internal volume of the box

=
(14850 - 10890) cm^{3}

=
3960 cm^{3}

_{ }

Let the edge of the cube = 'a' cm

Then,
surface area of cube = 6a^{2} cm^{2}

Volume
of a cuboid = (9 × 8 × 2) m^{3}
= 144 m^{3}

Volume
of each cube of edge 2 m = (2 m)^{3} = 8 m^{3}

## Chapter 15 - Volume and Surface Area of Solids Exercise Ex. 15B

_{ }

_{ }

Radius (r) of cylindrical bowl =

Height (h) up to which the bowl is filled with soup = 4 cm

Volume of soup in 1 bowl
= pr^{2}h = 154 cm^{3}

Hence, volume of soup in
250 bowls = (250 × 154) cm^{3 }= 38500 cm^{3
}= 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

Radius (r) of pillar = 20 cm = m

Height (h) of pillar = 10 m

For a tin can of rectangular base,

Length = 5 cm

Breadth = 4 cm

Height = 15 cm

∴ Volume of a tin can = Length × Breadth × Height

=
(5 × 4 × 15) cm^{3}

=
300 cm^{3}

For a cylinder with circular base,

Diameter = 7 ⇒ Radius = r = cm

Height = h = 10 cm

⇒ Volume of plastic cylinder is greater than volume of a tin can.

Difference
in volume = (385 - 300) = 85 cm^{3}

^{ }

Thus,
a plastic cylinder has more capacity that a tin can by 85 cm^{3}.

Radius (r) of 1 pillar =

Height (h) of 1 pillar = 4 m

Curved surface area of a
cylinder = 4.4 m^{2}

Radius (r) of a cylinder = 0.7 m

Lateral surface area of
a cylinder = 94.2 cm^{2}

Height (h) of a cylinder = 5 cm

Volume of a cylinder =
15.4 litres = 15400 cm^{3}

Height (h) of a cylinder = 1 m = 100 cm

Internal diameter of a cylinder = 24 cm

⇒ Internal radius of a cylinder, r = 12 cm

External diameter of a cylinder = 28 cm

⇒ External radius of a cylinder, R = 14 cm

Length of the pipe, i.e height, h = 35 cm

Diameter of a cylindrical pipe = 5 cm

⇒ Radius (r) of a cylindrical pipe = 2.5 cm

Height (h) of a cylindrical pipe = 28 m = 2800 cm

_{ }

_{}

_{}

_{}

_{}

Diameter of a cylinder = 140 cm

⇒ Radius, r = 70 cm

Height (h) of a cylinder = 1 m = 100 cm

Radius (r) of cylindrical vessel = 15 cm

Height (h) of cylindrical vessel = 32 m

Radius of small cylindrical glass = 3 cm

Height of a small cylindrical glass = 8 cm

Radius of the well = 5 m

Depth of the well = 8.4 m

Width of the embankment = 7.5 m

External radius of the embankment, R = (5 + 7.5) m = 12.5 m

Internal radius of the embankment, r = 5 m

Area
of the embankment = π (R^{2} - r^{2})

Volume
of the embankment = Volume of the earth dug out = 660 m^{2}

Speed of water = 30 cm/sec

∴ Volume of water that flows out of the pipe in one second

= Area of cross-section × Length of water flown in one second

=
(5 × 30) cm^{3}

=
150 cm^{3}

Hence, volume of water that flows out of the pipe in 1 minute

=
(150 × 60) cm^{3}

=
9000 cm^{3}

= 9 litres

Suppose the tank is filled in x minutes. Then,

Volume of the water that flows out through the pipe in x minutes

= Volume of the tank

Hence, the tank will be filled in 28 minutes.

Let the rise in the level of water = h cm

Then,

Volume of the cylinder of height h and base radius 28 cm

= Volume of rectangular iron solid

Thus, the rise in the level of water is 4 cm.

Radius, r = 1.5 m

Height, h = 280 m

Let the length of the wire = 'h' metres

Then,

Volume of the wire × 8.4 g = (13.2 × 1000) g

Thus, the length of the wire is 125 m.

Let R cm and r cm be the outer and inner radii of the cylindrical tube.

We have, length of tube = h = 14 cm

Now,

Outside
surface area - Inner surface area = 88 cm^{2}

⇒ 2πRh - 2πrh = 88

⇒ 2π(R - r)h = 88

It
is given that the volume of the tube = 176 cm^{3}

⇒ External
volume - Internal volume = 176 cm^{3}

⇒ πR^{2}h
- πr^{2}h = 176

⇒ π (R^{2}
- r^{2})h = 176

Adding (i) and (ii), we get

2R = 5

⇒ R = 2.5 cm

⇒ 2.5 - r = 1

⇒ r = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm respectively.

When
the sheet is folded along its length, it forms a cylinder of height, h_{1}
= 18 cm and perimeter of base equal to 30 cm.

Let
r_{1} be the radius of the base and V_{1} be is volume.

Then,

Again,
when the sheet is folded along its breadth, it forms a cylinder of height, h_{2}
= 30 cm and perimeter of base equal to 18 cm.

Let
r_{2} be the radius of the base and V_{2} be is volume.

Then,

## Chapter 15 - Volume and Surface Area of Solids Exercise Ex. 15D

Surface
area of sphere = 154 cm^{2}

⇒ 4πr^{2}
= 154

_{}

_{}

_{}

_{}

Inner radius = 5 cm

⇒ Outer radius = 5 + 0.25 = 5.25 cm

Inner diameter of the hemispherical bowl = 10.5 cm

Let the diameter of earth = d

⇒ Radius of the earth =

Then, diameter of moon = .

⇒ Radius of moon =

Volume of moon

Volume of earth

Thus, the volume of moon is of volume of earth.

Volume of a solid hemisphere = Surface area of a solid hemisphere

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