ML Aggarwal Solution Class 10 Chapter 17 Mensuration Exercise 17.1

 Exercise 17.1

Question 1

Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of π.

Sol :

Radius of the cylinder (r) = 5 cm

Height (h) = 10 cm

Total surface area = 2πr (h + r)

= 2π x 5(10 + 5) $\mathrm{cm}^{2}$

= 10 x 15π

= 150π $\mathrm{cm}^{2}$

Question 2

An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2m. Neglecting the thickness of its walls, calculate

(i) its outer lateral surface area,

(ii) its capacity in litres.

Sol :

Diameter of cylindrical geyser = 35 cm

Radius $(r)=\frac{35}{2} \mathrm{~cm}$

Height = 1.2 m = 120 cm

(i) Outer lateral surface area $=2 \pi r h$

$=2 \times \frac{22}{7} \times \frac{35}{2} \times 120 \mathrm{~cm}^{2}=13200 \mathrm{~cm}^{2}$

(ii) Capacity $=\pi r^{2} h$

$=\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 120 \mathrm{~cm}^{3}=115500 \mathrm{~cm}^{3}$

$=\frac{115500}{1000}$ litres $=115.5$ litres

Question 3

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto a height of 12 cm, find how many litres of milk is needed to serve 1600 students.

Sol :

Number of students = 1600

Diameter of cylindrical glasses = 7 cm

Radius $(r)=\frac{7}{2} \mathrm{~cm}$

Height of milk filled in it $(h)=12 \mathrm{~cm}$

$\therefore$ Volume of one glass $=\pi r^{2} h$

$=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12 \mathrm{~cm}^{3}=462 \mathrm{~cm}^{3}$

Volume of milk for 1600 students

$\therefore$ Total milk needed

$=462 \times 1600 \mathrm{~cm}^{3}$

$=\frac{462 \times 1600}{1000}$ litres

$=\frac{7392}{10}=739.2$ litres

Question 4

In the given figure, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.

Sol :

Length of rectangular tin foil (l) = 22 cm

and breadth (b) = 16 cm

By folding lengthwise, the radius of the cylinder

$(r)=\frac{22}{2 \pi}=\frac{11 \times 7}{22}=\frac{7}{2}$

and height $(h)=16 \mathrm{~cm}$

$\therefore$ Volume of the cylinder so formed

$=\pi r^{2} h=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 16=616 \mathrm{~cm}^{3}$

Question 5

(i) How many cubic metres of soil must be dug out to make a well 20 metres deep and 2 metres in diameter?

(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 50 per $m^{2}$, find the cost of plastering.

Sol :

(i) Depth of well (h) = 20 m

and diameter = 2 m

$\therefore$ Radius $=\frac{2}{2}=1 \mathrm{~m}$

$\therefore$ Volume of earth dug out

$=\pi r^{2} h=\frac{22}{7} \times 1 \times 1 \times 20 \mathrm{~m}^{3}$

$=\frac{440}{7}=62 \frac{6}{7} \mathrm{~m}^{3}$

(ii) Inner curved surface area $=2 \pi r h$

$=2 \times \frac{22}{7} \times 1 \times 20=\frac{880}{7} \mathrm{~m}^{2}$

Rate of plastering $=₹ 50$ per $\mathrm{m}^{2}$

$\therefore$ Total $\operatorname{cost}=\frac{880}{7} \times 50=₹ \frac{44000}{7}$

=₹ 6285.70

Question 6

A roadroller (in the shape of a cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.

Sol :

Diameter of a road roller = 0.7 m

Radius $(r)=\frac{0.7}{2}=0.35 \mathrm{~m}$

and width (h) = 1.2 m

$\therefore$ Curved surface area $=2 \pi r h $

$=2 \times \frac{22}{7} \times 0.35 \times 1.2 \mathrm{~m}^{2}$

$=44 \times 0.05 \times 1.2=2.64 \mathrm{~m}^{2}$

Area of playground $=120 \times 44=5280 \mathrm{~m}^{2}$

$\therefore$ Number of revolutions $=\frac{5280}{2.64}=2000$

Question 7

If the volume of a cylinder of height 7 cm is 448 π $\mathrm{cm}^{3}$, find its lateral surface area and total surface area.

Sol :

Volume of a cylinder = 448 π $\mathrm{cm}^{3}$

Height (h) = 7 cm

$\therefore$ Radius $(r)=\sqrt{\frac{\text { Volume }}{\pi h}}=\sqrt{\frac{448 \pi}{\pi \times 7}}$

$=\sqrt{64}=8 \mathrm{~cm}$

(i) Now lateral surface area $=2 \pi r h$ $=2 \pi \times 8 \times 7 \mathrm{~cm}^{2}=112 \pi \mathrm{cm}^{2}$

(ii) Total surface area $=2 \pi r(h+r)$

$=2 \pi \times 8(7+8)$

$=16 \pi \times 15=240 \pi \mathrm{cm}^{2}$

Question 8

A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per $m^{3}$.

Sol :

Height of a wooden pole (h) = 7 m

Diameter = 20 cm

$\therefore$ Radius $(r)=\frac{20}{2}=10 \mathrm{~cm}$

$\therefore$ Volume $=\pi r^{2} h$

$=\frac{22}{7} \times \frac{10 \times 10}{100 \times 100} \times 7 \mathrm{~m}^{3}=\frac{22}{100} \mathrm{~m}^{3}$

Weight of wood used $=225 \mathrm{~kg}$ per $\mathrm{m}^{3}$

$\therefore$ Total weight $=\frac{22}{100} \times 225 \mathrm{~kg}$

$=\frac{99}{2}=49.5 \mathrm{~kg}$

Question 9

The area of the curved surface of a cylinder is 4400 $\mathrm{cm}^{2}$, and the circumference of its base is 110 cm. Find

(i) the height of the cylinder.

(ii) the volume of the cylinder.

Sol :

Area of the curved surface of a cylinder = 4400 $\mathrm{cm}^{2}$

Circumference of base = 110 cm

Radius $=\frac{\text { Circumference }}{2 \pi}=\frac{110 \times 7}{2 \times 22} \mathrm{~cm}$

$=\frac{35}{2} \mathrm{~cm}$

(i) $\therefore$ Now height $=\frac{\text { Curved surface area }}{2 \pi r}$

$=\frac{4400 \times 7 \times 2}{2 \times 22 \times 35}=40 \mathrm{~cm}$

(ii) Volume of $=\pi r^{2} h$

$=\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 40 \mathrm{~cm}^{3}$

$=38500 \mathrm{~cm}^{3}$

Question 10

A cylinder has a diameter of 20 cm. The area of curved surface is 1000 $\mathrm{cm}^{2}$. Find

(i) the height of the cylinder correct to one decimal place.

(ii) the volume of the cylinder correct to one decimal place. (Take π = 3.14)

Sol :

Diameter of a cylinder = 20 cm

Radius $(r)=\frac{20}{2}=10 \mathrm{~cm}$

Curved surface area $=1000 \mathrm{~cm}^{2}$

(i) $\therefore$ Height $(h)=\frac{\text { Area }}{2 \pi r}=\frac{1000}{2 \times 3.14 \times 10}$

$=\frac{1000 \times 100}{2 \times 314 \times 10}=\frac{5000}{314}=15.9 \mathrm{~cm}$

(ii) Volume $=\pi r^{2} h=3.14 \times 10 \times 10 \times 15.9 \mathrm{~cm}^{3}$

$=\frac{314 \times 10 \times 10 \times 159}{100 \times 10}=\frac{49916}{10}$

$=4992.6 \mathrm{~cm}^{3}$

Question 11

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 when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

Answer correct to the nearest. 100 words.

Sol :

Height of cylindrical barrel of a pen (h) = 7 cm

Diameter = 5 mm

$\therefore$ Radius $(r)=\frac{5}{2} \mathrm{~mm}=\frac{5}{2 \times 10}=\frac{1}{4} \mathrm{~cm}$

$\therefore$ Volume of the ink used in it $=\pi r^{2} h$

$=\frac{22}{7} \times 7 \times \frac{1}{4} \times \frac{1}{4} \mathrm{~cm}^{3}=\frac{11}{8} \mathrm{~cm}^{3}$

Ink in the bottle $=\frac{1}{5} l=200 \mathrm{ml}$

Volume $=200 \mathrm{~cm}^{3}$

$\therefore$ Number of work written by this ink

$=\frac{200 \times 8}{11} \times 310=\frac{496000}{11}$

=45090.9=45100 words

Question 12

Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.

Sol :

Radius of a cylinder (r) = 3.5 cm

and height (h) = 7.5 cm

Total surface area = 2πr(r + h)

$=2 \pi \times 3.5(3.5+7.5) \mathrm{cm}^{2}$

$=7 \pi \times 11=77 \pi \mathrm{cm}^{2}$

and curved surface area $=2 \pi r h$

$=2 \pi \times 3.5 \times 7.5 \mathrm{~cm}^{2}$

$=7 \pi 7.5 \mathrm{~cm}^{2}$

$\therefore$ Ratio $77 \pi: 7 \pi \times 7.5=11: 7.5$

$=11: \frac{15}{2}=22: 15$

Question 13

The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?

Sol :

Let the radius of the base of a right circular cylinder = r

and height (h) = h

Volume = $\pi r^{2} h$

Radius of new cylinder $=\frac{r}{2}$

and height =2h

$\therefore$ Volume $\pi\left(\frac{r}{2}\right)^{2} \times 2 h=\pi \times \frac{r^{2}}{4} \times 2 h=\frac{\pi r^{2} h}{2}$

Ratio between the two cylinder (new + old)

$=\frac{\pi r^{2} h}{2}: \pi r^{2} h$

$=\frac{1}{2}: 1=1: 2$

Question 14

(i) The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 $\mathrm{cm}^{2}$. Find the height and the volume of the cylinder.

(ii) The total surface area of a cylinder is 352 $\mathrm{cm}^{2}$. If its height is 10 cm, then find the diameter of the base.

Sol :

Sum of radius and height of a cylinder = 37 cm

Total surface area = 1628 $\mathrm{cm}^{2}$

Let r be radius and h be height, then r × h = 37

and 2πr(r + h) = 1628

$2 \pi r \times 37=1628 $

$\Rightarrow 2 \pi r=\frac{1628}{37}$

$\Rightarrow \frac{2 \times 22}{7} r=\frac{1628}{37} $

$\Rightarrow r=\frac{1628}{37} \times \frac{7}{44}=7 \mathrm{~cm}$

$\therefore$ Radius $=7 \mathrm{~cm}$

and height $=37-7=30 \mathrm{~cm}$

$\therefore$ Volume $=\pi r^{2} h=\frac{22}{7} \times 7 \times 7 \times 30 \mathrm{~cm}^{3}$

$=4620 \mathrm{~cm}^{3}$

(ii) Total surface area of a cylinder $=352 \mathrm{~cm}^{2}$ 

Height =10 cm 

Let radius r, then

$2 \pi r(h+r)=352 $

$\Rightarrow \frac{2 \times 22}{7} r(10+r)=352$

$\Rightarrow r(10+r)=\frac{352 \times 7}{44}=56$

$10 r+r^{2}-56=0$

$\Rightarrow r^{2}+10 r-56=0 $

$\Rightarrow r^{2}+14 r-4 r-56=0$

$\Rightarrow r(r+14)-4(r+14)=0$

$\Rightarrow(r+14)(r-4)=0$

Either r+14=0, then r=-14 which is not possible being negative. or r-4=0, then r=4

$\therefore$ Diameter $=2 r=2 \times 4=8 \mathrm{~cm}$

Question 15

The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 $\mathrm{cm}^{2}$

Sol :

Ratio between curved surface area and total surface area = 1 : 2

Total surface area = 616 $\mathrm{cm}^{2}$

$\therefore$ Curved surface area $=\frac{616 \times 1}{2}=308 \mathrm{~cm}^{2}$

$=2 \pi r h=308$

$r h=\frac{308}{2 \pi}=\frac{308 \times 7}{2 \times 22}=49$

and $2 \pi r^{2}=308$

$r^{2}=\frac{308 \times 7}{2 \times 22}=49=(7)^{2}$

$\therefore r=7 \mathrm{~cm}$

and $h=\frac{49}{7}=7 \mathrm{~cm}$

Volume $=\pi r^{2} h=\frac{22}{7} \times 7 \times 7 \times 7=1078 \mathrm{~cm}^{3}$

Question 16

Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.

Sol :

Volume of two cylinders is the same

Diameter of both cylinder are in the ratio = 3 : 4

Let radius of the first $=\frac{3 x}{2}$

and second $=\frac{4 x}{2}=2 x$

Let $h_{1}$ and $h_{2}$ be the height of the two Then, $\pi r^{2} h_{1}=\pi r^{2} h_{2}$

$\pi \times \frac{3 x}{2} \times \frac{3 x}{2} \times h_{1}=\pi \times 2 \ddot{x} \times 2 x \times h_{2}$

$\therefore \frac{h_{1}}{h_{2}}=\frac{\pi \times 2 x \times 2 x \times 2 \times 2}{\pi \times 3 x \times 3 x}=\frac{16}{9}$

$\therefore$ Ratio in their heights $=16: 9$

Question 17

A rectangular sheet of tin foil of size 30 cm x 18 cm can be rolled to form a cylinder in two ways along length and along breadth. Find the ratio of volumes of the two cylinders thus formed.

Sol :

Size of the sheet = 30 cm × 18 cm

(i) By rolling lengthwise,

The circumference of the cylinder = 2πr = 30

$\therefore$ Radius $=\frac{30}{2 \pi}=\frac{15}{\pi} \mathrm{cm}=\frac{15 \times 7}{22}=\frac{105}{22} \mathrm{~cm}$

and height =18 cm

Volume $=\pi r_{1}^{2} h_{1}=\frac{22}{7} \times \frac{105}{22} \times \frac{105}{22} \times 18$

$=\frac{15 \times 105 \times 9}{11} \mathrm{~cm}^{3}$

and by rolling breadthwise circumference=18 cm

$\therefore$ Radius $\left(r_{2}\right)=\frac{18}{2 \pi}=\frac{18 \times 7}{2 \times 22}=\frac{63}{22} \mathrm{~cm}^{3}$

and $h_{2}=30 \mathrm{~cm}$

Volume $=\pi r_{2}^{2} h$

$=\frac{22}{7} \times \frac{63}{22} \times \frac{63}{22} \times 30$

$=\frac{9 \times 63 \times 15}{11} \mathrm{~cm}^{3}$

$\therefore$ Ratio $=\frac{15 \times 105 \times 9}{11}: \frac{9 \times 63 \times 15}{11}$

=105: 63

=15: 9 

= 5: 3

Question 18

A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.

Sol :

Internal diameter of a metal tube = 11.2 cm

and radius $(r)=\frac{11.2}{2}=5.6 \mathrm{~cm}$

Length (h) = 21 cm

Thickness of metal = 0.4 cm

External radius (R) = 5.6 + 0.4 = 6.0 cm

$\therefore$ Volume of metal used

$=\pi h\left(R^{2}-r^{2}\right)=\frac{22}{7} \times 21\left[6^{2}-5.6^{2}\right]$

$=66(6+5.6)(6-5.6) \mathrm{cm}^{3}$

$=66 \times 11.6 \times 0.4 \mathrm{~cm}^{3}=306.24 \mathrm{~cm}^{3}$

Question 19

The given figure shows a metal pipe 77 cm long. The inner diameter of a cross-section is 4 cm and the outer one is 4.4 cm. Find its

(i) inner curved surface area

(ii) outer curved surface area

(iii) total surface area.

Sol :

In the given figure,

Length of metal pipe (h) = 77 cm

Inner diameter = 4 cm

$\therefore$ Radius $(r)=\frac{4}{2}=2 \mathrm{~cm}$

and outer radius $(\mathrm{R})=\frac{4.4}{2}=2.2 \mathrm{~cm}$

(i) Inner curved area $=2 \pi r h$

$=2 \times \frac{22}{7} \times 2 \times 77 \mathrm{~cm}^{2}=968 \mathrm{~cm}^{3}$

(ii) Outer curved area $=2 \pi \mathrm{R} h$

$=2 \times \frac{22}{7} \times 2.2 \times 77 \mathrm{~cm}^{3}=1064.8 \mathrm{~cm}^{2}$

(iii) Total surface area = Inner area + Outer area + Area of two rings

$=968+1064.8+2 \times \frac{22}{7}\left(2.2^{2}-2^{2}\right)$

$=2032.8+\frac{44}{7}(4.84-4)$

$=2032.8+\frac{44}{7} \times 0.84$

$=2032.8+5.28=2038.08 \mathrm{~cm}^{2}$

Question 20

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Sol :

Diameter of the pencil = 7 mm

Radius $(R)=\frac{7}{2} m m=\frac{7}{20}$ cm

Diameter of graphite (lead) = 1 mm

$\therefore$ Radius $(r)=\frac{1}{2}=\frac{1}{20} \mathrm{~cm}$

Length of pencil $(h)=14 \mathrm{~cm}$

$\therefore$ Volume of graphite $=\pi r^{2} h$

$=\frac{22}{7} \times \frac{1}{20} \times \frac{1}{20} \times 14=\frac{11}{100} \mathrm{~cm}^{3}$

$=0.11 \mathrm{~cm}^{3}$

and volume of wood $=\pi \mathrm{R}^{2} h-\pi r^{2} h$

$=\pi h\left[\mathrm{R}^{2}-r^{2}\right]=\frac{22}{7} \times 14\left[\left(\frac{7}{20}\right)^{2}-\left(\frac{1}{20}\right)^{2}\right]$

$=44\left[\frac{49}{400}-\frac{1}{400}\right]=44 \times \frac{48}{400} \mathrm{~cm}^{3}$

$=\frac{528}{100}=5.28 \mathrm{~cm}^{3}$

Question 21

A soft drink is available in two packs

(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and

(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Sol :

(i) Base of the tin of rectangular base = 5 cm × 4 cm

Height = 15 cm

Volume = lbh = 5 × 4 × 15 = 300 cm³

(ii) Base diameter of cylindrical plastic cylinder = 7 cm

$\therefore$ Radius $(r)=\frac{7}{2} \mathrm{~cm}$

and height $(h)=10 \mathrm{~cm}$

$\therefore$ Volume $=\pi r^{2} h$

$=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 10=385 \mathrm{~cm}^{3}$

It is clear that cylindrical container has greater capacity and $385-300-85 \mathrm{~cm}^{3}$ more.

Question 22

A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm³ of iron weighs 8 g.

Sol :

Length of cylindrical roller (h) = 2 m = 200 cm

Diameter = 35 cm

Inner radius $=\frac{35}{2} \mathrm{~cm}$

Thickness = 7 cm

$\therefore$ Outer radius $=\frac{35}{2}+7=\frac{49}{2} \mathrm{~cm}$

and inner height =200 cm

$\therefore$ Volume of the iron in roller

$=\pi h\left[\mathrm{R}^{2}-r^{2}\right]=\frac{22}{7} \times 200\left[\left(\frac{49}{2}\right)^{2}-\left(\frac{35}{2}\right)^{2}\right]$

$=\frac{4400}{7}\left[\frac{84}{2} \times \frac{14}{2}\right]=\frac{4400}{7} \times 42 \times 7 \mathrm{~cm}^{3}$

$=184800 \mathrm{~cm}^{3}$

Weight of $1 \mathrm{~cm}^{3}=8 \mathrm{gm}$

$\therefore$ Total weight $=184800 \times 8 \mathrm{gm}$

$=1478400 \mathrm{gm}=1478.4 \mathrm{~kg}$