 # RS Aggarwal Class 9 Solutions Chapter 13 - Volume And Surface Area Ex 13A(13.1)

## RS Aggarwal Class 9 Chapter 13 – Volume And Surface Area Ex 13A(13.1) Solutions Free PDF

The RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area Ex 13A(13.1) is the best study material that you can refer to while preparing for your Class 9 exam. All the solutions mentioned below are solved in a step-by-step way which helps you to understand the logic behind such questions easily.

These solutions can be referred by students whenever they have any doubt or get stuck while solving any particular question. To excel in the exams, you should practice the RS Aggarwal Class 9 Solutions for Maths regularly. Firstly, practice the difficult topics to clear your concepts and develop your problem-solving skills. It will help you solve questions in quick-time and score good marks.

## Download PDF of RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area Ex 13A(13.1)

Q.1: Find the volume and curved surface area of a right circular cylinder of height 21 cm and base radius 5 cm.

Solution:

Here, r = 5 cm and h = 21 cm

∴ Volume of the cylinder = 𝜋r2h = 227×52×21$\frac{22}{7} \times 5^{2} \times 21$ = 1650 cm3

∴ Curved surface area of the cylinder = 2𝜋rh = 2×227×5×21$2 \times \frac{22}{7} \times 5 \times 21$ = 660 cm2

Q.2: The diameter of a cylinder is 28 cm and its height is 40 cm. Find the curved surface area, total surface area and the volume of the cylinder.

Solution:

Here, diameter = 28 cm, Radius = 14 cm and Height = 40 cm

∴ Curved surface area = 2𝜋rh = 2×227×14×40$2 \times \frac{22}{7} \times 14 \times 40$ = 3520 cm2

∴ Total surface area = 2𝜋rh + 2𝜋r22×227×14×40+2×227×142$2 \times \frac{22}{7} \times 14 \times 40 + 2 \times \frac{22}{7} \times 14^{2}$

= (3520 + 1232) = 4752 cm2

∴ Volume of the cylinder = 𝜋r2h = 227×142×40$\frac{22}{7} \times 14^{2} \times 40$ = 24640 cm3

Q.3: Find the weight of a solid cylinder of radius 10.5 cm and height 60 cm if the material of the cylinder weighs 5 g per cm3.

Solution:

Here, radius(r) = 10.5 cm and height = 60 cm

∴ Volume of the cylinder = 𝜋r2h = 227×10.52×60$\frac{22}{7} \times 10.5^{2} \times 60$ = 20790 cm3

∴ Weight of the solid cylinder if the material of the cylinder weighs 5 g per cm3 = (20790)(5) = 103950 g = 1039501000(Since,1000g=1kg)$\frac{103950}{1000} \: \: \: (Since, 1000 g = 1 kg)$ = 103.95 kg

Q.4: The curved surface area of a cylinder is 1210 cm2 and its diameter is 20 cm. Find its height and volume.

Solution:

Here, curved surface area = 1210 cm2

Diameter = 20 cm $\Rightarrow$ radius = 10 cm

∴ Curved surface area of the cylinder = 2𝜋rh

1210=2×227×10×h$\Rightarrow 1210 = 2 \times \frac{22}{7} \times 10 \times h$
h=1210×72×22×10cm=19.25cm$\Rightarrow h\, = \, \frac{1210 \times 7}{2 \times 22 \times 10} cm \, = \, 19.25 \, cm$

∴ Height = 19.25 cm

∴ Volume of the cylinder = 𝜋r2h = 227×102×19.25$\frac{22}{7} \times 10^{2} \times 19.25$ = 6050 cm3

∴ Volume of the cylinder = 6050 cm3

Q.5: The curved surface area of a cylinder is 4400 cm2 and the circumference of its base is 110 cm. Find the height and volume of the cylinder.

Solution:

Let base radius be ‘r’ and height be ‘h’

Then, 2𝜋rh = 4400 cm2

And, 2𝜋r = 110 cm

2πrh2πr=4400110$\Rightarrow \frac{ 2????rh}{ 2????r}\, = \, \frac{4400}{110}$

$\Rightarrow$ h = 40 cm.

2×227×r×h×40=4400cm$2 \times \frac{22}{7} \times r \times h \times 40 \, = \, 4400 \, cm$

r=4400×744×40=352cm$\Rightarrow \: \: \: \: r \, = \, \frac{4400 \times 7}{44 \times 40} = \frac{35}{2} cm$

∴ Volume of the cylinder = 𝜋r2h = 227×352×352×40$\frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}\times 40$= 38500 cm3

Q.6: The radius of the base and height of a cylinder are in the ratio 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.

Solution:

Let the radius ® = 2x cm and height (h) = 3x cm

Then, Volume of cylinder = 𝜋r2h

Volume = 227×(2x)2×3x$\frac{22}{7} \times (2x)^{2} \times 3x$

Volume = 227×4x×3x$\frac{22}{7} \times 4x \times 3x$

Volume = 227×12x3$\frac{22}{7} \times 12x^{3}$

But, given volume = 1617 cm3

Thus solving the equation for x, we get,

x=72$x = \frac{7}{2}$

∴ Radius = 2x = 7 cm

And height = 3x = 20.5 cm

Total surface area = 2𝜋r(h + r) = 2×227×7212+7$2 \times \frac{22}{7} \times 7{ \frac{21}{2} + 7 }$ cm2 = 44×21+142cm2$44 \times \frac{21 + 14}{2} cm^{2}$ = (22 × 35) cm2 = 770 cm2

Q.7: The total surface area of a cylinder is 462 cm2. Its curved surface area is one – third of its total surface area. Find the volume of the cylinder.

Solution:

Curved surface area = 13×(thetotalsurfacearea)$\frac{1}{3} \times \left ( the\;\;total \;\;surface\;\;area \right )$

= 13×462=154cm2$\frac{1}{3} \times 462 = 154 cm^{2}$

Total surface area – Curved surface area = (462 – 154) cm2 = 308 cm2

2π2=308$\Rightarrow 2\pi^{2} = 308$
2×227×r2=308$\Rightarrow 2 \times \frac{22}{7} \times r^{2} = 308$
r2=308×744=49$\Rightarrow r^{2} = \frac{308 \times 7}{44} = 49$
r=49=7cm$\Rightarrow r = \sqrt{49} = 7\, cm$

Now, curved surface area = 2𝜋rh = 154 cm2 = 2×227×7×h=154cm2$2 \times \frac{22}{7} \times 7 \times h = 154 cm^{2}$

h=15444=3.5cm$\Rightarrow h = \frac{154}{44} = 3.5 cm$

Now, r = 7 cm and h-= 3.5 cm

∴ Volume of the cylinder = 𝜋r2h = 227×7×7×3.5$\frac{22}{7} \times 7 \times 7 \times 3.5$ = 539 cm3

∴ Volume of the cylinder = 539 cm3

Q.8: The total surface area of a solid cylinder is 231 cm2 and its curved surface area is (⅔) of the total surface area. Find the volume of the cylinder.

Solution:

Curved surface area = 23×$\frac{2}{3} \times$ Total surface area

= 23×231=154cm2$\frac{2}{3} \times 231 = 154 cm^{2}$

(Total surface area) – (Curved surface area) = (231 – 154) = 77 cm2

2𝜋r2 = 77 cm2

2×227×r2=77$\Rightarrow 2 \times \frac{22}{7} \times r^{2} = 77$
r2=77×744=494$\Rightarrow r^{2} = \frac{77 \times 7}{44} = \frac{49}{4}$
r=494=72cm$\Rightarrow r = \sqrt{ \frac{49}{4}} = \frac{7}{2} cm$

Now, 2𝜋rh = 154 cm2

2×227×72v×h=154cm2$\Rightarrow 2 \times \frac {22}{7} \times \frac{7}{2}v\times h = 154 cm^{2}$
h=15422=7cm$\Rightarrow h = \frac{154}{22} = 7 cm$

Now, r=72andh=7cm$r = \frac{7}{2}\, and \, h= 7 cm$

Volume of the cylinder = 𝜋r2h = 227×72×72×7$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7$ = 269.5 cm3

Therefore, Volume of the cylinder = 269.5 cm3

Q.9: The sum of the height and radius of the base of a solid cylinder is 37 m. if the total surface area of the cylinder is 1628 m2, find its volume.

Solution:

Here, (r + h) = 37 m [given]

And total surface area = 2𝜋r(r + h) = 1628 m2

$\Rightarrow$ 2𝜋r(37) = 1628 m2

2×227×r×37=1628$\Rightarrow 2 \times \frac{22}{7} \times r \times 37 = 1628$
r=1628×744×37=7m$\Rightarrow r = \frac{1628 \times 7}{44 \times 37} = 7 m$

And, (r + h) = 37 m

$\Rightarrow$ (7 + h) = 37 m

$\Rightarrow$ h = 37 – 7 =30 m

$\Rightarrow$ volume = 𝜋r2h = 227×7×7×30$\frac{22}{7} \times 7 \times 7 \times 30$ = 4620 m3

Q.10: The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm2.

Solution:

Curved surface area = 2𝜋rh

Total surface area = 2𝜋r(h + r)

Since they are in the ratio of 1:2

2πrh2πr(h+r)=12$\frac{2 \pi rh}{2 \pi r(h + r)} = \frac{1}{2}$
hh+r=12$\Rightarrow \frac{h}{h + r} = \frac{1}{2}$

$\Rightarrow$ 2h = h + r

$\Rightarrow$ 2h = h = r

$\Rightarrow$ h = r

2𝜋r(h + r) = 616 cm2

$\Rightarrow$ 4𝜋r2 = 616 cm2 [putting h = r]

4×227×r2=616$\Rightarrow 4 \times \frac{22}{7} \times r^{2} =616$
r2=616×788=49$\Rightarrow r^{2} = \frac{616 \times 7}{88} = 49$
r=49=7cm$\Rightarrow r = \sqrt{49} = 7cm$

Then, r = 7 cm and h = 7 cm

Therefore, Volume of the cylinder = 𝜋r2h =227×7×7×7$\frac{22}{7}\times 7\times 7\times 7$ = 1078 cm3

Volume of the cylinder = 1078 cm3

Q.11: 1 cm3 of gold drawn into a wire 0.1 mm in diameter. Find the length of the wire.

Solution:

1 cm3 = 1 cm × 1 cm × 1 cm

Where, 1 cm = 0.01 m

Therefore, Volume of the gold = 0.01 m × 0.01 m × 0.01 m = 0.000001 m3 . . . . . . . (1)

Diameter of the wire drawn = 0.1 mm

Radius of the wire drawn = 0.05 mm = 0.00005 m . . . . . . (2)

Length of the wire = ‘h’ m . . . . . . (3)

Volume of the wire drawn = Volume of the gold

$\Rightarrow$ 𝜋r2h = 0.000001

$\Rightarrow$ 𝜋 × 0.00005 × 0.00005 × h = 0.000001 [from eqn (1), (2), (3)]

h=0.000001×70.00005×0.00005×22$h = \frac{0.000001 \times 7}{0.00005 \times 0.00005 \times 22}$ = 127.27 m

∴ The length of the wire is 127.27 m

Q.12: The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.

Solution:

Let the radii of the two cylinders be 2R and 3R.

And their heights be 5H and 3H

Then, V1V2=π×(2R)2×5Hπ×(3R)2×3H=π×4H2×5Hπ×9R2×3H=2027$\frac{V_{1}}{V_{2}} = \frac{\pi \times (2R)^{2} \times 5H}{\pi \times (3R)^{2} \times 3H} \, = \, \frac{\pi \times 4H^{2} \times 5H}{\pi \times 9R^{2} \times 3H} \, = \frac{20}{27}$

∴ the ratio of their volumes = 20:27

Now, S1S2=2π(2R)(5H)2π(3R)(3H)=109$\frac{S_{1}}{S_{2}} \, = \, \frac{2 \pi (2R)(5H)}{2 \pi (3R)(3H)} \, = \, \frac{10}{9}$

∴ The ratio of their curved surface = 10:9

Q.13: A powder of tin has a square base with side 12 cm and height 17.5 cm. Another is cylindrical with diameter of its base 12 cm and height 17.5 am. Which has more capacity and by how much?

Solution:

For the tin having square base,

Side = 12 cm and height = 17.5 cm.

∴ Volume = (12 × 12 × 17.5) cm3 = 2520 cm3

Now, diameter of tin with cylindrical base = 12 cm

∴ Radius = 6 cm and height = 17.5 cm.

∴ Volume = (227×6×6×17.5)cm3==1980cm3$(\frac{22}{7} \times 6 \times 6 \times 17.5) cm^{3} \, = \, = 1980 cm^{3}$

Tin with square base has more capacity by (2520 – 1980) cm3 = 540 cm3

Q.14: A cylindrical bucket, 28 cm in diameter and 72 cm high, is full of water. The water is emptied into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level in the tank.

Solution:

Here, cylindrical bucket has diameter = 28 cm

∴ Radius = 14 cm and height = 72 cm

Length of the tank = 66 cm

Breadth of the tank = 28 cm

∴ Volume of the tank = Volume of cylindrical bucket.

$\Rightarrow$ l × b × h = 𝜋r2h

$\Rightarrow$ 66 × 28 × h = 3.14 × 14 × 14 × 72

h=22×2×14×7266×28cm$\Rightarrow h \, = \, \frac{22 \times 2 \times 14 \times 72}{66 \times 28} cm$ = 24 cm.

∴ The height of the water level in the tank = 24 cm.

Q.15: If 1 cm3 of cast iron weighs 21 g, find the weight of a cat iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.

Solution:

Internal radius = (3/2) = 1.5 cm

And, external radius = (1.5 + 1) cm = 2.5 cm

Volume of cast iron = [𝜋 × (2.5)2 × 100 – 𝜋 × (1.5)2 × 100] cm3 = 𝜋 × 100 × [(2.5)2 – (1.5)2] cm3

= 227×100×[6.252.25]$\frac{22}{7} \times 100 \times [6.25 – 2.25]$

= 227×100×4$\frac{22}{7} \times 100 \times 4$

Weight = 227×100×4×211000$\frac{22}{7} \times 100 \times 4 \times \frac{21}{1000}$

(Because, 1kg = 1000g)

Weight = 26.4 kg.

Hence, the weight of the iron pipe = 26.4 kg

Q.16: A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.

Solution:

Internal diameter of the tube = 10.4 cm

Internal radius = 5.2 cm

And length = 25 cm

And external radius = (5.2 + 0.8) = 6 cm

Required volume = [π×(6)2×25π×(5.2)2×25]cm3$[ \pi \times (6)^{2} \times 25 – \pi \times (5.2)^{2} \times 25 ] cm^{3}$

= π×25[(6)2(5.2)2]cm3$\pi \times 25 [(6)^{2} – (5.2)^{2} ] cm^{3}$

= 227×25[3627.04]cm3$\frac{22}{7} \times 25 [36 – 27.04] cm^{3}$

= [227×25×8.96]cm3$[\frac{22}{7} \times 25 \times 8.96] cm^{3}$ = 704 cm3

Therefore, the volume of the metal = 704 cm3

Q.17: The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one – fifth of a liter?

Solution:

Length = 7 cm = height

Diameter = 5 mm $\Rightarrow$ radius = 2.5 mm = 0.25 cm

Volume of the barrel = 𝜋r2h = 227×0.25×0.25×7$\frac{22}{7} \times 0.25 \times 0.25 \times 7$ cm3

= 118cm3$\frac{11}{8} cm^{3}$

Hence, 118cm3$\frac{11}{8} cm^{3}$ is used for writing 330 words.

So, [15×1000]cm3$[ \frac{1}{5} \times 1000 ] cm^{3}$ will be used for writing, [330×811×15×1000]$[330 \times \frac{8}{11} \times \frac{1}{5} \times 1000]$ words = 48000 words.

Q.18: A lead pencil consists of a cylinder of wood with a solid cylinder of graphite fitted into it. The diameter of the pencil is 7 mm, the diameter of the graphite is 1 mm and the length of the pencil is 10 cm. Calculate the weight of the whole pencil, if the specific gravity of the wood is 0.7 g/ cm3 and that of the graphite is 2.1 g/cm3.

Solution:

Weight of the graphite = [227×(0.05)2×10×2.1]g$[\frac{22}{7} \times (0.05)^{2} \times 10 \times 2.1] g$ = 33200g$\frac{33}{200} g$

Weight of wood = [227×10(0.35)2(0.05)2×0.7]$[ \frac{22}{7} \times 10{(0.35)^{2} – (0.05)^{2}} \times 0.7]$ = [227×10(0.12250.0025)×0.7]$[ \frac{22}{7} \times 10(0.1225 – 0.0025) \times 0.7]$

= 6625g$\frac{66}{25} g$

Therefore, total weight of the pencil = [33200+6625]g$[ \frac{33}{200} + \frac{66}{25}] g$

= 33+528200=561200=2.805g$\frac{33 + 528}{200} = \frac{561}{200} = 2.805g$

Therefore, weight of the whole pencil = 2.805 g

### Key Features of RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area Ex 13A(13.1)

• It will improve your speed and accuracy along with overall academic performance.
• RS Aggarwal Maths Solutions helps in practicing different types of questions.
• It gives an idea to solve difficult and tricky questions in a simple way.
• These solutions are based on the recent syllabus of the CBSE Class 9.