S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 C

 Exercise 15 C

Question 1

Ans: (i) Given, 

Radius = 3cm

Height = 4cm

Slant height (l) =$\sqrt{r^{2}+h^{2}}$

$=\sqrt{(3)^{2}+(4)^{2}}$

$=\sqrt{3+16}$

$=\sqrt{25}$

$=5 \mathrm{~cm}$

Curved surface = $\pi r l$

$=\pi \times 3 \times 5$

$=15 \pi \mathrm{cm}^{2}$

Area of base = $\pi r^{2}$

$\pi \times 3 \times 3$

$=9 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=15 \pi+9 \pi$

$=24 \pi \mathrm{cm}^{2}$

Volume  =  $=\frac{1}{3} \pi r^{2} h$

$-\frac{1}{3} \pi \times 3 \times 3 \times 4 $

$=12 \pi \mathrm{cm}^{3}$

(ii) Given, 

Radius = 20cm

Slant height = 25 cm

$l^{2}=r^{2}+1^{2}$

$\left(25^{2}-r^{2}=h^{2}\right.$

$\sqrt{625}-(20)^{2}=1^{2}$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

Curved surface = πrl

$=\pi \times 20 \times 25$

$=500 \pi \mathrm{cm}^{2}$

area of base =  π$r^{2}$

$=\pi \times 200 \times 20$

$=400 \pi \mathrm{cm}^{2} .$

Total surface area = $\pi r l+\pi r^{2}$

$\begin{aligned}=& 500 \pi+400 \pi \\=& 900 \pi \mathrm{cm}^{2} \end{aligned}$

Volume $=\frac{1}{3} \pi{r}^{2} h$

$=2000 \pi \mathrm{cm}^{3}$

(iii) Given, 

Height = 18cm

Slant height = 30 cm

$\therefore \quad l^{2}=h^{2}+r^{2} .$

$(30)^{2}=(18)^{2}+r^{2}$

$900-324=r^{2} .$

$\sqrt{576}=r$

$24=r$

Curved surface = πrl 

$=\pi \times 24 \times 30$

$=720 π c m^{2}$

Area of base = $\pi r^{2}$

$=\pi \times 24 \times 24$

$=576 \pi \mathrm{m}^{2}$

Total surface area= $\pi r l+\pi r^{2}$

729π + 576π

$=1296 \mathrm π{cm}^{2}$

Volume = $\frac{1}{3} \pi r^{2} h$

$=576 \times 6 \pi .$

$=3456 \pi \mathrm{cm}^{3}$

(iv) Given, 

Radius = 27cm

Height = 36cm

$\begin{aligned} \because \quad l &=\sqrt{h^{2}+r^{2}} \\ &=\sqrt{(36)^{2}+(27)^{2}} \\ &=\sqrt{1296+729} \\ &=\sqrt{2025} \\ &=45 \mathrm{~cm}\end{aligned}$

Curved surface =  $\pi$ rl

= $\pi$ $\times 27 \times 45$

$=1215 \pi \mathrm{cm}^{2} .$

Area of base =  $\pi r^{2}$ 

$=\pi \times 27 \times 27$

$=729 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=1215 \pi+729 \pi$

$=1944 \pi \mathrm{cm}^{2}$

Volume = $\frac{1}{3}$ $\pi r l+\pi r^{2}h$

$=8748 \pi \mathrm{cm}^{3}$

(v) Given, 

Radius = 5cm,

Curved surface = 65π $\left(cm^{2}\right)$

  

∴ Curved surface =πrl 

65π = π $5\times l$

$\frac{65} π { π  \times 5}=l$

$l=13 \mathrm{~cm}$

$\because \quad l^{2}=h^{2}+r  ^{2}$

$(13)^{2}=h^{2}+(5)^{2}$

$16 y-25=h^{2}$

$144=h^{2}$

$\sqrt{144}=h$

$12=h$

$h=12 \mathrm{~cm} .$

Area of base = $πr^{2}$

$=\pi \times 5 \times 5$

$=25 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=65 \pi+25 \pi$

$=90 \pi \mathrm{m}^{2} .$

Volume $=\frac{1}{3}\pi r^{2} h$

$=\frac{1}{3} \pi \times(5)^{2} \times 12$

$=25 \times 4 \pi$

$=100 \pi \mathrm{cm}^{3}$

(vi) Given,

Radius = 35cm

Total surface area  = $13860 \mathrm{~cm}^{2}$

Total surface area = πrl+π $r^{2}$

$13860:=\pi r(l+r) .$

$4410 \pi=\pi \times 35(l+35) .$

$4410 \pi=35 \pi(l+75)$

$126=l+35$

$126-35=l$

$91=l$

$\ell=91 \mathrm{~cm} .$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\(91)^{2} &=h^{2}+(35)^{2} \end{aligned}$

$8281-1225=h^{2}$

$\sqrt{8281-1225}=h$

$\sqrt{7056}=h$

$84=h$

$h=84 \mathrm{~cm}$

Curved surface = $\pi rl$

$=\pi \times 35 \times 91$

$=3185 \pi \mathrm{cm}^{2}$

Area of base $=\pi r^{2}$

$=\pi \times 35 \times 35$

$=1225 \pi c m^{2}$

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi \times 35 \times 35 \times 84$

$=34300 \pi \mathrm{cm}^{3}$

Question 2

Ans: (i) Given, 

Height = 8m, 

Area of base = $156 \mathrm{~m}^{2}$

$\begin{aligned} \therefore \text { Area of base } &=\pi r^{2} \\ 156 &=\frac{22}{7} \times \r^{2} \\ \frac{156 \times 7}{22} &=r^{2} \end{aligned}$

$\frac{546}{11}=r^{2}$

$r^{2}=\frac{546}{11} \mathrm{~m}$

ஃ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{546}{11} \times 8$

$=2 \times 26 \times 8$

$=416 \mathrm{~m}^{3}$

(ii) Given,

Slant height (l)= 17cm,

Radius (r)= 8cm

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-r^{2}=h^{2} .$

$(17)^{2}-(8)^{2}=h^{2}$

$\sqrt{289-64}=h .$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 15$

$=176 \times 40$

$=\frac{7.040}{7}$

$=1005.71 \mathrm{~cm}^{3}$

(iii) Given, 

Height = 8cm,

Slant length = 10cm,

$\therefore \quad l^{2} = h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2}$

$(10)^{2}-(8)^{2}=r^{2}$

$100-64=r^{2}$

$\sqrt{36}=r$

$6=r$

$r=6 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 8$

$=\frac{44 \times 48}{7}$

=301.71$cm^{3}$

(iv) Given, 

Height=5cm 

Perimeter of base = 8cm

Perimeter of base= $2 \pi r$

$8=2 \times \frac{22}{7} \times r$

$\frac{{8} \times 7}{2 \times 22}=6$

$\frac{14}{11}=r$

$r=\frac{14}{11} \mathrm{~cm}$

∴ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{14}{11} \times \frac{14}{11} \times 5$

$=\frac{4 \times 70}{33}$

$=\frac{280}{33}$

$=8.48 . \mathrm{cm}^{3}$

Question 3

Ans: (i) Given, 

Height = 8m,

Slant height = 10m

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2} .$

$(10)^{2}-(8)^{2}=r^{2}$

$\sqrt{100-64}=r$

$\sqrt{36}=r$

$6=r$

r=6 m

∴ Curved surface area = πrl

$\begin{aligned} &=\frac{22}{7} \times 6 \times 10 \\=& \frac{1320}{7} \end{aligned}$

=199.6 $m^{2}$

(ii) Given, 

Perimeter of base = 88cm,

slant height = 2dm = $2 \times 10cm=20cm

∴ Perimeter of base $=2 \pi r$

88= $=2 \times \frac{22}{7} \times r$

$\frac{88 \times 7}{2 \times 22}=r$

r=14cm

Curved surface area $=\pi r l$

$\begin{aligned} & \frac{22}{7} \times 14 \times 20 \\=& 44 \times 20 \\=& 880 \mathrm{~cm}^{2} \end{aligned}$

(iii) Given,

Area of base= $154 \mathrm{~cm}^{2}$

Height 24cm

Area of base= $\pi r^{2}$

$154=\frac{22}{7} \times r^{2}$

$\frac{154 \times 22}{7}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~cm}$

$\therefore \ell=\sqrt{h^{2}+r^{2}}$

l=$\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=25 \mathrm{~cm}$

So 

Curved surface area = π rl

$=22 \times 25$

$=550 \mathrm{~cm}^{2}$

Question 4

Ans: Given,

Radius = 5cm

Volume = 50π $\mathrm{Cm}^{3}$

Volume= $\frac{1}{3}π r^{2} h$

$50 \pi=\frac{1}{3} \pi \times 5 \times 5 \times \mathrm{h}$

$50 \pi=\frac{\pi}{3} \times 25 \mathrm{~h}$

$\frac{50 \times \pi \times 3}{\pi \times 25}=h$

$\frac{150}{25}=h$

$6=h$

$h=6 \mathrm{~cm}$

Hence, the height of the cone is 6cm

Question 5

Ans: Given,

Radius = 11.3cm

Curved surface area = $710(cm)^{2}$

Curved surface area $=\pi rl$

$710=\frac{355}{113} \times \frac{11.3}{10} \times l$

$710=\frac{355}{10} l$

$\frac{710 \times 10}{355}=l$

$\frac{7100}{355}=l$

$20=l$

$l=20 \mathrm{~cm}$

Hence , slant height is 20cm

Question 6

Ans:  Let the radius be r and height be h

Volume =$\frac{1}{3} \pi r^{2} h$

According to question 

Radius =$\frac{r}{2}$

and height= h

Volume = $\frac{1}{3} \pi\left(\frac{8}{2}\right)^{2} h$

$=\frac{1}{3} \pi \frac{r^{2}}{4} h=$ $\frac{\pi^{2} h}{12}$

∴ Ratio = $\frac{\pi r^{2} h}{12}=\frac{1}{3} \pi r^{2} h$

$=\frac{\frac{\pi r^{2} h}{12}}{\frac{1 \pi x^{2} h}{3}}$

$=\frac{1}{4}=1: 4 .$

Question 7

Ans: Given 

Curved surface area = $264 m^{2}$

Slant height = 12m

Curved surface area = πrl

$264=\frac{22}{7} \times r \times 12$

$\frac{264 \times 7}{22 \times 12}=r$

$7=r$

$r=7 \mathrm{~m}$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l^{2}-r^{2} &=h^{2} \end{aligned}$

Question 8

Ans: According to question, 

Height (h) = $2 \times$ diameter 

Diameter = $\frac{\text { height }}{2}$

$\therefore \quad \operatorname{Radius}=\frac{h}{2} \times \frac{1}{2}=\frac{h}{4}$.

Volume $=36 \pi \mathrm{cm}^{3}$

$\frac{1}{3} \pi r^{2} h=36 \pi$

$\frac{1}{3} \times \pi \times \frac{h}{4} \times \frac{h}{4} \times \pi=36 \pi .$

$\frac{h^{3}}{48} \pi=36 \pi$

$h^{3}=\sqrt{728}$

$h=\sqrt[3]{1728}=\sqrt{12 \times 12 \times 12}=12 . \mathrm{cm}$

Question 9

Ans: Given,

Radius and height of cone are in ratio = 3:4

Let radius be 3x 

and height be 4x 

Volume = 301.44 $(cm)^{3}$

$\because \quad$ Volume =\frac{1}{3} \pi r^{2} h$

$301.44=\frac{1}{3} \times 3.14$  $\times(3x)^{2}$  $\times$(4x)$

$301.44=3.14 \times 12 x^{3}$

$\frac{301.44}{3.14 \times 12}=x^{3}$

$\frac{30144}{3768}=x^{3}$

$8 =x^{3}$

$x^{3}=8$

$x=\sqrt[3]{8}$

$x=\sqrt{2 \times 2 \times 2}$

$x=2$

$\begin{aligned} \therefore \text { Radius } &=3 \times \\ &=3 \times 2 \\ &=6 \mathrm{~cm} . \end{aligned}$

and Hight $=4 x$

$=4 \times 2$

=8

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l &=\sqrt{(8)^{2}+(6)^{2}} \\ l &=\sqrt{64+36} \\ l &=\sqrt{100} \\ l &=10 \mathrm{~cm} \end{aligned}$

Question 10

Ans: Given ,

The ratio of radius and slant height of cone = 4:7

Curved surface area $=792 \mathrm{~cm}^{2}$

Let radius be 4x and 

slant height be 7x

Curved surface area = πrl

$792=\frac{22}{7} \times 4 x \times 7 x$

$792=22 \times 4 x^{2}$

$792=88 x^{2}$

$\frac{792}{88}=x^{2}$

$9=x^{2}$

$x^{2}=9$

$x=\sqrt{9}$

$x=3$

Radius =4x

$=4 \times 3$

$=12 \mathrm{~cm}$

Question 11

Ans: Given, 

The ratio of radii of two cones = 3:5

Let r1 be 3x and r2 be 5x

Volume of first cone = $\frac{1}{3} \pi r_{1}^{2} h$

$=\frac{1}{3}π  (3x)^{2}\times h$

= $3 \pi x^{2} h .$

Volume of second cone =  $\frac{1}{3} \pi r_{2}^{2} h$

$\begin{aligned} &=\frac{1}{3} \pi \times (5 x)^{2} x h \\=& \frac{1}{3} \pi 25 x^{2} h \\=& \frac{25 \pi x^{2} h}{3} \end{aligned}$

ration of their volumes 

$=3 \pi x^{2}h \frac{25 \pi x^{2} 1}{3}$

$=\frac{3 \pi x^{2} h}{\frac{25 \pi x^{2} h}{3}}$

$=\frac{9}{25}$

$=9: 25$

Question 12

Ans: Given, 

Circumference = 44m

Height = 10m

Circumference = 2πr

$44=2 \times \frac{22}{7} \times r$

$\frac{44 \times 7}{44}=r$

$7=r$,

$\therefore l^{2}=h^{2}+r^{2} .$

$l=\sqrt{(10)^{2}+(7)^{2}}$

$l=\sqrt{100+49}$

$l=\sqrt{149 \mathrm{~m} .}$

∴ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times \sqrt{149}$

$\begin{aligned} &=22 \sqrt{149}\\=& 22 \times 12.2 . \\=& 268.4 \mathrm{~m}^{2} . \end{aligned}$

Width of canvas(d)=  2cm= $\frac{2}{100} \mathrm{~m}$

$\therefore$ Area $=l \times b$

$268.4=l \times \frac{2}{100}$

$\frac{26840}{2}=l$

=13429=l

Hence, the length of canvas is 13420m

Question 13

Ans: Given, 

Radius =7m

Height = 24m

$\because$ Slant height $(l)=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{2.5 \times 25}$

$l=25 \mathrm{~m}$

Curved surface area =  πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

Width of canvas (b) = 5m

Area = L $\times$ B

$550=L \times 3$

$\frac{550}{5}=L$

$110=L$

$L=110 \mathrm{~m}$

$\therefore$ Length $=110 \mathrm{~m}$.

Question 14

Ans: Given,

Volume =  $1232 \mathrm{~m}^{3} .$

Area of base = $154 \mathrm{~m}^{2}$

Area of base = $\pi r^{2}$

$154=$ $\frac{22}{7} \times r^{2}$

$\frac{154 \times 7}{22}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~m}$

Volume $=\frac{1}{3} \pi r^{2} h$

$1232=$ $\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h$

$\frac{1232 \times 3}{22 \times 7}=h $

$8 \times 3=h$

$24=h$

$h=24 m$

$l^{2}=h^{2}+x^{2}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{25 \times 25}$

$l=25 \mathrm{~m}$

ஃ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

∴ The area of canvas $=550 \mathrm{~m}^{2}$

Question 15

Ans: Given, 

In the tent, accommodate is available for 11 persons and each person must have $4r^{2}$ of space on the ground 

So , aera of base of the tent = $11 \times{4}$

$44 m^{2}$

Air is required for each person to breadth = $20m^{3}$

Volume of air = $20 \times 11$

$=220 \mathrm{~m}^{3}$

Area of base = π$r^{2}$

$44=\frac{22}{7} \times r^{2}$

$\frac{44 \times 7}{22}=r^{2}$

$r^{2}=14 \mathrm{~m} .$

Volume = $=\frac{1}{3} \pi{r}^{2} h$

$220=\frac{1}{3}\times $\frac{22}{7}\times 14 \times h$

$220=\frac{44h}{3}$

$\frac{220 \times 3}{44}=h$

$15=h$

h=15 m

Hence , the height of the cone = 15m

Question 16

Ans: False , because the volume of a cone is one third $\left(\frac{1}{3}\right)$ of the volume of cylinder of the same radius and height 

Question 17

Ans: Given,

Height of cylinder = 9cm

and radius= $\frac{40}{2} cm=20 c m$

Height of cone = 108cm

∴ Volume of cylinder = $\pi r^{2} h$

$=\frac{22}{7} \times 20 \times 20 \times 9$

$=\frac{440 \times 180}{7}$

$=\frac{79200}{7} \mathrm{~cm}^{2} .$

ஃ Volume of cone $\frac{79200}{7}$ (Given volume of cylinder is equal to volume of cone)

Volume of cone = $=\frac{1}{3} \pi r^{2}$h

$\frac{79200}{7}=\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 108$

$\frac{79200 \times 7 \times 3}{7 \times 22 \times 108}=r^{2}$

$\frac{79200 \times 21}{7 \times 22 \times 108}=r^{2}$

$\frac{79200}{792}=r^{2}$

$100=r^{2}$

$r^{2}=100$

$r=\sqrt{100}$

$r=10$

Hence, the radius of cone is 10cm

Question 18

Ans: Given, 

Radius, of cone and cylinder (r) = 7m,

Height of the cylinder (h1)=8m

And height of the conical of (h2)= 4m

(IMAGE TO BE ADDED)

∴ l =$\sqrt{h2^{2}+r^{2}}$

$=\sqrt{(4)^{2}+(7)^{2}}$ $=\sqrt{16+49}$ $=8.06M$

∴ Area of canvas = Curved surface area of cylinder + Curved surface area of cone 

= 2πrh_{1}$ + πrl

$=2 \times \frac{22}{7} \times 7 \times 8+\frac{22}{7} \times 7 \times 8.06 .$

$=352+177.32 .$

$=529.32 . \mathrm{m}^{2}$

Question 19

Ans: Given, 

Height of cylinder = 3m 

Radius of cylinder = $\frac{105}{2} \mathrm{~m}$

and slant height = 53m

Total area of the canvas = Curved surface area of cylinder +curved surface area of cone 

= 2πrh+πrl
$=22 \times \frac{22}{7} \times $3+\frac{2 x}{7}\times\frac{105}{2} \times 53$

$=22 \times 45+11 \times 795 .$

$=990+8745$

$=9735 \mathrm{~m}^{2}$

Question 20

Ans: Given, 

Edge of a cube = 9cm

Diameter of cone = 9cm (because equal to edge of cube)

Then radius = $\frac{9}{2} cm$

and height of cone = 9cm (because equal to edge of cube)

Volume of largest cone = $\frac{1}{3}πr^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 9$

$=\frac{99 \times 27}{14}$

$=\frac{2673}{14}$

$=190.93 . \mathrm{cm}^{3}$

Question 21

Ans:  Given,

Height of cylinder(h1) = 3m, 

Total height of tent = 13.4m

Height of conical part (h2) = 13.5-3

=10.5m

(IMAGE TO BE ADDED)

Radius = 14 m

$l=\sqrt{h_{2}^{2}+r^{2}}$

$l=\sqrt{(10.5)^{2}+(14)^{2}}$

$l=\sqrt{110.25+196}$

$l=\sqrt{306.25}$

$l=17.5 \mathrm{~m}$

Question 22

Ans:  (IMAGE TO BE ADDED)

Given, 

Height of the cylinder (h1)= 32cm

and radius (r1) = 18cm

Height of the conical= 24 cm

Volume of sand in it = π$r_{1}^{2}$ h

$=\frac{22}{7} \times 18 \times 18 \times 32$

$=\frac{396 \times 576}{7}$

$=\frac{228096 \mathrm{cm}^{3}}{7}$

(IMAGE TO BE ADDED)

So volume of conical heap of the sand = $\frac{228096 \mathrm{~cm}^{3}}{7}$

Volume of conical heap = $\frac{1}{3} \pi r^{2} h$

$\frac{228096}{7}=$ $\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24$

$\frac{22.8096}{7}=$  $\frac{176}{7} r^{2}$

$\frac{228096 \times 7}{7 \times 176}$ $=r^{2}$

$\frac{228096}{176}=r^{2}$

$1296=r^{2} .$

$r^{2}=1296$

$r=\sqrt{1296}$

$r=36 \mathrm{~cm}$

(i) Radius of cone = 36cm

(ii) $l=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(36)^{2}}$

$l=\sqrt{576+1296}$

$l=\sqrt{1872}$

l=43.3cm

Height the slant height of heap is 43.3cm

Question 23

Ans:  (IMAGE TO BE ADDED)

Given, 

Height cylinder = 8cm

Radius of cylinder = 6cm

Volume of cylinder = $\pi r^{2} h$

$=31416 \times 6 \times 6 \times 8$

$=18.8496 \times 48$

$=904.7808 \mathrm{~cm}^{3}$

Radius of cone = 6cm 

Height of cone = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2}h$

$=62832 \times 48$

$=301.5936 \mathrm{~cm}^{3}$

 

∴ Volume of remaining solid = Volume of cylinder - Volume of cone 

$=904.7808-301.5936$

$=603.1872 \mathrm{~cm}^{3}$

Question 24

Ans: (IMAGE TO BE ADDED)

Given,

Radius , of cylinder = 3cm 

and height of cylinder = 5cm 

Volume of cylinder = $\pi r^{2} h$

$=\frac{21}{7} \times 3 \times 3 \times 5$

$=\frac{990}{7} \mathrm{~cm}^{2}$

Radius of conical portion = $\frac{3}{2} c m$

and Height of conical portion = $\frac{8}{9} c h$

Volume of conical portion = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{8}{9}$

$=\frac{44}{21} \mathrm{~cm}^{3}$

Metal in the remaining part = Volume of cylinder - Volume of conical portion 

$=\frac{990}{7}-\frac{44}{21}$

$=\frac{2970-44}{21}$

$=\frac{2926}{21}$

According to the question 

$\frac{2926}{21}: \frac{44}{21}$

$\frac{\frac{2926}{21}}{\frac{4 4}{21}}$

$=\frac{2926 \times 21}{21 \times 44}$

$\frac{1463}{22}=$

$=133: 2$

Question 25

Ans:   (IMAGE TO BE ADDED)

Given, 

Radius of cylinder = $\frac{7}{2} \mathrm{~cm}$

Its height = 8cm

Volume of cylinder = $\pi r^{2} h$

$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 8$

$=22 \times 14$

$=308 \mathrm{~cm}^{3} .$

Radius of cone $=\frac{7}{4} \mathrm{~cm}$

Its height = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7}\times \frac{7}{4} \times \frac{7}{4} \times 8$

$=\frac{77}{3} \mathrm{~cm}^{3}$

Volume of water required to fill the vessel= Volume of cylinder -Volume of cone 

$=308-\frac{77}{3}$

$=\frac{924-77}{3 .}$

$=\frac{847}{3}$

$=28 \frac{1}{3} \mathrm{~cm}^{3}$

According to question,

Height of cone = $1 \frac{3}{4}=\frac{7}{7} \mathrm{Cm} x$

its radius = 2cm

Volume of cone =  $\frac{1}{3} \pi r^{2} h$

=$\frac{22}{3} \mathrm{~cm}^{3}$

Change in volume of cones 

$=\frac{77}{3}-\frac{22}{3}$

$=\frac{77-22}{3}$

$=\frac{55}{3} \mathrm{~cm}^{3}$

Let the drop in water level be h cm

Volume = π$r^{2} h$

$\frac{55}{3}=$ $\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h $

$\frac{55 \times 2}{3 \times 11 \times 7}=h$

$\frac{10}{21}=h $

$h=\frac{10}{21} \mathrm{~cm}$

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 B

  Exercise 15 B

Question 1 

Ans: (i) Given,

r= 7cm, 'h=8cm

∴ Volume = $\pi r^{2} h$

$=\frac{22}{7} \times 7 \times 7+8$

$=22 \times 56$

$=1232 \mathrm{~cm}^{3}$

(ii) Given,

$\begin{aligned}&r=7 \mathrm{~cm} \\&h=12 \mathrm{~cm} .\end{aligned}$

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

$\begin{aligned} &=\frac{22}{7} \times 7 \times 7 \times 12 \\ &=22 \times 84 \\ &=1848 \mathrm{~cm}^{3} \end{aligned}$

(iii) Given,

$r=14 \mathrm{~cm}$

$\mathrm{h}=16 \mathrm{~cm} .$

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

$-\frac{22}{7} \times 14 \times 14 \times 6$

$=44 \times 224$

$=9856 \mathrm{~cm}^{3}$

(iv) Given, 

r= 21cm,

h=40cm

Volume = $\pi r^{2} h$,

$=\frac{22}{7} \times 21 \times 21 \times 40$

$=66 \times 840$

$=\quad 55440 \mathrm{~cm}^{3}$

Question 2

Ans: (a) Given, 

Volume, = $44 \mathrm{Cm}^{3}$

Height $=3.5 \mathrm{~cm}$

ஃ Volume = $\pi r^{2}h$

$44=\frac{22}{7}\times r^{2}\times \frac{3.5}{10}$

$ 44=11 r^{2}$

$ \frac{44}{11}=r^{2}$

$4=r^{2}$

$r^{2}=4 .$

$r=\sqrt{4} .$

$r=2 \mathrm{cm} $

∴ Diameter = 2r 

$=2 \times 2$

$=4 c \mathrm{~m}$

(b) Given, 

Volume , = $385 \mathrm{~cm}^{3}$

Height $=1 \mathrm{dm}=10 \mathrm{~cm}$

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

$385=\frac{22}{7} \times r^{2} \times 10$

$\frac{49}{4}=r^{2}$

$r^{2}=\frac{49}{4}$

$r=\sqrt{\frac{49}{4}}$

$r=\frac{7}{2}$

∴ Diameter = 2r = $2 \times \frac{7}{2}=7 \mathrm{~cm}$

 

Question 3

Ans: (a) Given,

Volume = $66 \mathrm{~cm}^{3}$

Radius = 2cm

∴ Volume = π$(r^{2}$h

$\frac{66 \times 7}{22 \times 4}=h$

$\frac{21}{4}=h$

$h=\frac{21}{4}$

$h=5.25 \mathrm{~cm}$

(b) Given,

Volume = 4litres= 4000 $\mathrm{Cm}^{3}$

Radius= 5cm

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

$4000=\frac{22}{7} \times 5 \times 5 \times h$

$\frac{4000 \times 7}{22 \times 25}=h$

$\frac{80 \times 7}{11}$

$\frac{560}{11}=h$

$h=\frac{510}{11} \mathrm{~cm}$

Question 4

Ans: Given,

Height(h) = 7m

Radius(r) = $\frac{20}{2}=100 \mathrm{~cm}$ $=\frac{10}{100}$ $=\frac{1}{10}m$

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

$=\frac{11}{50}$

According to question,

Total weight $=\frac{11}{50} \times 225$

$=\frac{99}{2}$

$=49.5 \mathrm{~kg}$

Question 5

Ans: Giver,

Internal radius. $(r)=3 \mathrm{~cm}$

Thickness of pipe= $1 \mathrm{~cm}$

$\therefore$ Outer radius $(R)=3+1$=4cm

Length = 6cm

$\begin{aligned} \therefore \quad \text { Volume } &=\pi R^{2} h-\pi r^{2} h . \\ &=\pi h\left(R^{2}-r^{2}\right) . \\ &=\frac{22}{7} \times 6\left((4)^{2}-(3)^{2}\right) \end{aligned}$

$=\frac{132}{7}(16-9)$

$=\frac{132}{7} \times 7 $

$=132 \mathrm{cm}^{3}$

Question 6

Ans: Given,

Sum of radius, of the base and the height of a cylinder (h+r) = 37cm,

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

∴ Total surface area = $2πr h+2πr^{2}$

$1628=2 \pi r(h+r)$

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

r=7

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

So, $h+r=37$

$h+7=37$.

$h=377$

$h=30 \mathrm{~cm}$.

$\begin{aligned} \text { Volume } &=\pi r^{2} h \\ &=\frac{22}{7} \times 7 \times 7 \times 30 \\ &=22 \times 210 \\ &=4620 \mathrm{~cm}^{3} \end{aligned}$

Question 7

Ans: Given

Capacity of a cylindrical tank = $6160 \mathrm{m}^{3}$

Radius $(r)  \frac{28}{2}=14 \mathrm{~m}$

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

$6160=\frac{22}{7} \times 14 \times 14 \times h$

$\frac{6160}{22 \times 281}=h $

h= 10m

Area of curved surface of taken inner sides = $2 \pi r h$

$=2 \times \frac{22}{7} \times 14 \times 10$

$=44 \times 20$

$=880 \mathrm{m}^{2}$

Question 8

Ans: Given,

Curved surface area of cylinder = $4400 \mathrm{~cm}^{2}$

Circumference $=110 \mathrm{~cm} $

Circumference $=2 \pi r$

$110=2 \times \frac{22}{7} \times r$

$\frac{110 \times 7}{2 \times 22}=r$

$\frac{35}{2}$=r

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

(i) Curved Surface area $=2$ πrh

4400= $=\frac{2}\times \frac{22}{7} \times \frac{35}{2} \times h$

$\frac{4400}{22 \times 5}=h$

$40=h$

$h=40$ cm

hence the height of cylinder is 40cm

(ii) Volume = $\pi r^{2} h$

$=110 \times 350$

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

Question 9

Ans: (i) Given ,

Height of the wall = 20 meter 

Radius = $\frac{2}{2}=1$

$\because$ Volume =  πr^{2}$

$\begin{aligned} &=\frac{22}{7} \times 1 \times 1 \times 20 \\=& \frac{440}{7} \\=& 62 \frac{6}{7} \mathrm{~m}^{3} . \end{aligned}$

(ii) Curved surface area = $2 π r h$

$\begin{aligned} &=2 \times \frac{22}{7} \times 1 \times 20 \\=& \frac{44 \times 20}{7} \\=& \frac{880}{7} \end{aligned}$

Rate of plastering the inner surface = Rs 5  per m².

Total cost = 880\7 x 5

$=\frac{4400}{7}$

$=7628.57 $

Question 10

Ans: Giver

Radius of cylinder $=\frac{20}{2}=10 \mathrm{~cm}

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

(i) Curved surface area $=2 \pi r$

$1000=2 \times \frac{22}{7} \times 10 \times h$

$\frac{175}{11}=h$

$h=\frac{175}{11}$

$h=15.9 \mathrm{~cm} .$

(ii) $\begin{aligned} \text { Volume } &=\pi r^{2}h \\ &=3.14 \times 10 \times 10 \times 15-9 . \\ &=31.4 \times 159 . \\ &=4992.6 \mathrm{~cm}^{3}\end{aligned}$

Question 11

Ans: Given, 

Radius =  $\frac{35}{2} \mathrm{~cm}$

height $=1.2 \mathrm{~m}=1.2 \times 100 \mathrm{~cm}=120 \mathrm{~cm}$.

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

$=110 \times 120$

$=13200 \mathrm{~cm}^{2} $

(ii) Capacity = $\pi r^{2} h$

 $=55 \times 2100$

$=115500 \mathrm{~cm} .$

=115.5 liters 

Question 12

Ans: Given, 

Radius of cylindrical glass = $\frac{8}{2}=4 \mathrm{~cm}$

and its height = 15cm

Radius, of cylindrical vessel = $\frac{30}{2}=15 \mathrm{~cm}$

and its height = 80 cm

Volume of cylindrical glass= $=\pi r^{2} \mathrm{~h}$

$\pi \times 4 \times 4 \times 15$

$=240 \pi$

Volume of cylindrical vessel $=\pi r^{2} h$

$=\pi \times 15 \times 15 \times 80$

$=\pi \times 225 \times 80$

$=18000 \pi$.

$\therefore$ Number of glasses $=\frac{\text { Volume of vessel }}{\text { Volume of glass}}

$=\frac{18000 \pi}{240 \pi}$ 

$=75$ glasses

Question 13

Ans: Given, 

R= 22m

r= 20 

h=$\frac{7}{100} m$

$\pi R^{2} h-\pi{r}^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times \frac{7}{100}(R+r)(R-r)$

$=\frac{22}{7}\times{7}{100}(22) \times 18=18.48 \mathrm{~mm}$

Question 14

Ans: Given

Radius of iron cylindrical block = $=\frac{0.5m}{20}$ =$\frac{1}{4} \mathrm{~m}=$0.25m= $0.25\times $100cm  =25cm

Length = 3.5m= 3.5 $\times 100$cm = 350 cm

Volume of block =  $\pi r^{2} b$

$=\frac{22}{7} \times 25 \times 25 \times 350$

$=550 \times 1250$

$=687500 \mathrm{~cm}^{3}$

So , volume of base = $687500 \mathrm{~cm}^{3}$

Area of square base $=25 \times 25$

$=625 \mathrm{~cm}^{2}$

Height of bar= $\frac {Volume of bar}{Area of square base}$

$=\frac{687500}{625}$

$=1100 \mathrm{~cm} $

$=\frac{1100}{100}$

$=11 \mathrm{~m}$

Question 15

Ans: Given, 

Length of swimming pole (l) = 70m

Breadth (b)= 44m

and depth (h) =  3m

Volume = lbh

$=70 \times 40 \times 3$

$=924012^{3}$

Radius of pipe= $\frac{14}{2} c m=7 cm=\frac{7}{160} \mathrm{~m}$

Volume = $πr^{2}h

9240= $\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times h$ (Volume = 9240)

$\frac{9240 \times 100 \times 100}{22 \times 7}=h$

$\frac{92400000}{154}=h .$

$600000=h .$

$\therefore h=600000$

Let be the distance 

$\therefore \quad$ Distance $=$ speed \times $ Time.

$600000=2 \times$ Time. (speed = 2m)

$\frac{600000}{2}=$ Time.

$83 \frac{1}{3}$ hours

Question 16

Ans: Given, 

External radius of a hollow cylinder = $\frac{12}{2}=6 \mathrm{~cm}$

Internal radius of a hollow cylinder = 6_ 0.25

= 5.75cm

Length (h)= 15cm

Volume of hollow cylinder = $\pi R^{2} h-\pi r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 15\left((6)^{2}-\left(5.70^{2}\right)\right.$

$=\frac{330}{7}(36-33.0625) .$

$=\frac{330}{7} \times 2.9375$

$=\frac{969.375}{7 }$

Radius of solid cylinder $\frac{2}{2}=1 \mathrm{~cm}$ (given).

$\begin{aligned} \therefore \text { Volume } &=\pi r^{2} h \\ \frac{969-375}{7} &=\frac{22}{7} \times 1 \times 1 \times h \end{aligned}$

$\frac{969.375}{22}=h$ 

$\therefore h=\frac{969.375}{22}$

$h=44.0625 \mathrm{~cm} $

Question 17

Ans: Given, 

Internal radius of tube= $\frac{11.2}{20} \mathrm{~cm}=5.6 \mathrm{~cm}$

$\operatorname{Length}(h)=21 \mathrm{~cm}$.

Thickness $=0.4 \mathrm{~cm}$.

∴ Outer radius 5.6+0.4

=6cm

Volume of Metal = $\pi R^{2} h - r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$\left.=\frac{22}{7}\times 21 \times(6)^{2}-(5-6)^{2}\right)$

$=66 \times(36-31.36)$

$=66 \times 464$

$=306.24 \mathrm{cm}^{3}$

$=306.2 \mathrm{~cm}^{3}$

Question 18

Ans: Given, 

Volume of water = 1000 lit = $1000 \times 1000 \mathrm{~cm}^{3}=1000000 \mathrm{~cm}^{3}$

Radius of pip = 0.6cm

∴ Volume = $\pi r^{2} h$

$1000000=\frac{22}{7} \times 0-6 \times 0.6 \times h .$

$\frac{1000000 \times 7}{22 \times 0-6 \times 0.6}=h .$

$\frac{700000000}{7.92}=h .$

$883838.38=h .$

$h=883838-38 \mathrm{~cm}$

Let height (h) be the distance, 

Distance = speed $\times$ times 

$883838.38=8 \times$ time.

$\frac{88383838}{800}=$ time

$110479.7975=$ Time

Time = 110479.7975sec

Time = $\frac{110479.7975}{60 \times 60}$ hours 

$=\frac{11047967975}{36000060}$ hours

$=30.69$ hours 

Question 19

Ans: Given,

Radius =$\frac{28}{2} \mathrm{~cm}=14 \mathrm{~cm}$

Height = 72cm

$\therefore$ Volume $\pi r 2 h$

$=\frac{22}{7} \times 14 \times 14 \times 72$

$=44 \times 1008$

$=44352 \mathrm{~cm}^{3}$

Length f tank = 66cm

Breadth of tank = 28cm

Volume = $=l \times b \times h$

$\begin{aligned} 44352 &=66 \times 28 \times h . \\ 44352 &=1848 h . \\ \frac{44352}{1848} &=h . \\ 24 &=h \\ \therefore h &=24 \mathrm{~cm} . \end{aligned}$

Hence, the height of the water level in the tank is 24 cm

Question 20

Ans: Given,

Radius of cylindrical vessel = $\frac{14}{2} c m=7 cm$ 

Height of water = $8 \frac{9}{14}(cm )=\frac{121}{14}$

Volume of water= $\pi r^{2} h$

$=1331 \mathrm{~cm}^{3}$

Volume of cube = $=1331 \mathrm{~cm}^{3}$

Volume of cube= $a^{3}$

$1331=a^{3}$

$\sqrt[3]{1331}=a$

$\sqrt{11 \times 11 \times 11}=a$

$a=11$

Hence , the length of the edge is 11cm

Question 21

Ans: Let the radius of the cylinder = r

And height = h

Volume = $\pi r^{2} h$

If the radius is halved 

$\therefore \quad r=\frac{r}{2}$

Height = h

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

$=\pi \frac{r^{2}}{4} h$

According to question 

$=\frac{\pi r^{2}}{4} h=\pi r^{2} h$

$=\frac{1}{4}$= 1:4

Question 22

Ans: Given,

Length of sheet = 22cm

and breadth = 12cm

If it is folded breadth wire, 

∴ Circumference = 12cm

and height = 22cm

Circumference = 2πr

$12=2 \times \frac{22}{7} \times r$

$\frac{21}{11}=r$

$y=\frac{21}{11} \mathrm{~cm}$

∴ Volume =πz $=r^{2} h$

$=\frac{22}{7} \times \frac{21}{11} \times \frac{21}{11} \times 22$

$=\frac{462 \times 42}{77}$

$=\frac{19404}{72}$

$=269.5 \mathrm{~cm}^{3}$

IF it is folded length wire 

$\therefore$ circumference  $=22 \mathrm{~cm}$. 

and Height: $12 \mathrm{~cm}$.

So circumference =$2 \pi r$

$22=2 \times \frac{22}{7} \times r$

$\frac{22 \times 7}{2 \times 22}=r$

$\frac{7}{2}=r$

$r=\frac{7}{2} cm$

Volume =$\left.\pi r^{2}\right)h$

$\frac{27}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$

$=22 \times 21$

$=482cm^{3}$

According to question, 

$=462-269-5$

$=192.5 \mathrm{~cm}^{3}$

Question 23

Ans: Given, 

Depth of a wall = 20m

Radius = $\frac{7}{2} \mathrm{~m}$

Volume of earth = $\pi r^{2} h$

$=22 \times 35$

$=770 \mathrm{m}^{3}$

Length of platform= 22m,

Breadth = 14m

Volume of platform = $770 \mathrm{~m} 3$

Volume = lbh 

770= $22 \times 14 \times h$

$\frac{770}{22 \times 14}=h$

$\frac{770}{308}=h$

$2-5=h$

$h=2.5 \mathrm{~m}$

Hence , the height of the platform is 2.5m

Question 24

Ans: Given,

Height of cylindrical barrel of pen = 7cm

Radius = $\frac{5}{2} \cdot mm =$ $\frac{5}{2} \times \frac{1}{10} \mathrm{~cm}$=1\4cm

Volume of ink in it =  $\pi r^{2} h$

$-\frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 7$

$=\frac{11}{8} \mathrm{~cm}^{3}$

Volume of link in bottle =  $\frac{1}{5}l=\frac{1}{5} \times 1000 \mathrm{~cm}=200 \mathrm{cm}^{3}$

$\therefore$ Total number of barrels $=200 \div \frac{11}{8}$

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

$=\frac{1600}{11}$

Word written in one barrel = 310 words 

Total number of words = $\frac{1600}{11} \times 320$

$=\frac{496000}{11 .}$

$=45080.90 .$

$=45090$ words 

Question 25

 

Ans: Given , 

Length of a rectangular box= 40cm,

Breadth = 30cm and 

Height = 25cm

Volume = lbh 

$=40 \times 30 \times 25$

$=1200 \times 25$

$=30000 \mathrm{~cm}^{3}$

 Radius of cylindrical tin= 17.5cm

Volume = $30000 \mathrm{~cm}^{3}$

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

$30000=3-14 \times 17.5 \times 17.5 \times h $

$30000=3-14 \times 306.25 \dot{x h}$

$30000=961.625h $

$\frac{30000000}{961625}=h$

$31 \cdot 2$

$h=31-2 \mathrm{~cm}$

Hence the height of the cylindrical tin is 31.2cm

Question 26

 

Ans: Given, 

Diameter of a circular tank = 17.5m

So, the radius = $\frac{175}{20}=\frac{35}{4}=$ 8.75m

Outer radius = 8.75+4

= 12.75m

Height= 2m 

Volume of the embankment = $\pi R^{2} h-\pi r^{2} h$

$\begin{aligned} & \pi h\left(R^{2}-r^{2}\right) \\=& \frac{22}{7} \times 2\left((12-75)^{2}-(8.75)^{2}\right) \\=& \frac{44}{7}(162.5625-76.5625) \\=& \frac{44}{7} \times 86 . \\=& \frac{3784}{7} \\ 540.57 \mathrm{~m}^{3} . \end{aligned}$

 

∴ Volume of earth of the tank = $540.57 \mathrm{~m}^{3}$

$\begin{aligned} \therefore \text { Volyme } &=\pi r^{2} h \\ 540.57 &=\frac{22}{7} \times 8.75 \times 8.75 \times h . \end{aligned}$

$540.57=\frac{1684.375 \mathrm{~h}}{7}$

$2.25=h$

$h=2.25 \mathrm{~m} .$

Hence the depth of the circular tank is 2.25m

Question 27

 

Ans: Given, 

Speed of water = $7 m=700 \mathrm{~cm}$

Internal radius= $\frac{2}{2} c m=1 c m$

Radius of tank = 40cn

Time =$\frac{1}{2}$ hour $=$ $\frac{1}{2} \times 60 \times 60=\frac{3600}{2} \mathrm{sec}=$

=1800sec 

∴ Distance  (h) = Times into speed 

$=1800 \times 700$

$=1260000 \mathrm{~cm}$

Volume $=\pi r^{2} \mathrm{~h}$. 

=$\frac{22}{7}1 \times1 \times \times 1260000$

$=3960000 \mathrm{~cm}^{3}$

∴ Volume of water in the tank =$3960000 \mathrm{cm}^{3} $

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

$3960000=\frac{22}{7} \times 40 \times 40 \times h$.

$\frac{17325}{22}=h$

$787.5=h$

$h=787.5 \mathrm{cm} $

Question 28

 

Ans: Given, 

Radius of pipe = $\frac{7}{2} cm=\frac{7}{200}$ m

Speed = 36 km\hr = $36 \times \frac{1000}{60}$ = 600m

Radius of tank = $35 \mathrm{~cm}=\frac{35}{100} \mathrm{~m}$

Height = 1m

∴ Volume of tank =π$r^{2}h$

$\frac{22}{7} \times \frac{35}{100} \times \frac{35}{100} \times 1$

$=\frac{77}{200} \mathrm{~m}^{3}$

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

$\frac{77}{200}=\frac{22}{7} \times \frac{7}{200} \times \frac{7}{200} \times h .$

$\frac{2200}{221}=h$

$100=h .$

$h=100$

Let height (h) be the distance 

Distance = speed into Time

100= $600 \times Time$

Time $=\frac{100}{600}$

Time $=0.167 \mathrm{~min} $

Question 29

 

Ans: Given, 

Radius of coin =$\frac{1.5}{20} \mathrm{~cm}=\frac{3}{4} \mathrm{~cm}$

and its thickness = 0.2 cm

Volume of one coin = $\pi r^{2} h$

$=\frac{22}{7} \times \frac{3}{4} \times \frac{3}{4} \times 0.2 .$

$=\frac{39.6}{112}$

$=0.35 \mathrm{~cm}^{3}$

Radius of cylinder = $\frac{4.5}{20}=\frac{9}{4} \mathrm{~cm}$

Height = 10cm

Volume of cylinder = $\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \\=& \frac{17820}{112} \\=& 159.11 \mathrm{~cm}^{3} . \end{aligned}$

Volume of cylinder = volume of one coin X Number of coin

159.11 = $0.35 \times$ Number of coin.

$\frac{159 \cdot 11}{0.35}$ =Number of coin.

$454.6=$ Number of coin

Number of coin $=454.6$

Question 30

 

Ans: Given

Weight of  $1cm^{3}=21 g$

Length of pipe = $1 \mathrm{~m}=100 \mathrm{~cm}$

Internal radius = $\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm}$

External radius = 1.5+1= 2.5cm

∴ Volume = Volume of outer - Volume of inner surface 

$π R^{2} h$-$π r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 100\left((2.5)^{2}-(1.5)^{2}\right)$

$=\frac{2200}{7} \times(6.25-2.25) .$

$=\frac{2200}{7} \times 4 $

$=\frac{8800}{7} \mathrm{~cm}^{3} $

∴ Total Weight of metal = $\frac{8800}{7} \times {21}$

$=26400 \mathrm{~g}$

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15 A

 Exercise 15 A

Question 1

Ans: (i) Given

$\begin{aligned}&h=12 \mathrm{~cm}, \\&r=7 \mathrm{~cm} .\end{aligned}$

$\therefore$ Curved surface arca $=2 \pi r h$'

$\begin{aligned} &=2 \times \frac{22}{7} \times 7 \times 12 \\ &=44 \times 12 \\ &=528 \mathrm{~cm}^{2}\end{aligned}$

(ii) Given,

$\begin{aligned}&h=10 \mathrm{~cm}, \\&r=7 \mathrm{~cm} .\end{aligned}$

Curved Surface area = $2 \pi r h$

$=440 \mathrm{~cm}^{2}$

Total surface area $=2 πrh (h+r)$

$\begin{aligned}&=2 \times \frac{22}{7} \times 7(10+7) \\&=44 \times 17 \\&=748 \mathrm{~cm}^{2}\end{aligned}$

(iii) Given

$\begin{aligned}&h=5 \mathrm{~cm}, \\&r=21 \mathrm{~cm}\end{aligned}$

∴ Curved Surface area = 2πr h

$=2 \times \frac{22}{7} \times 21 \times 5$

$=44 \times 15$

$=660 \mathrm{~cm}^{2}$

Total Surface area = $2 \pi r(h+r)$

$=2 \times \frac{22}{7} \times 21(5+21)$

$=44 \times 3 \times 26$

$=132 \times 26$

$=3432 \mathrm{~cm}^{2}$

(iv) Given

$\begin{aligned}&h=20 \mathrm{~cm} \\&r=14 \mathrm{~cm}\end{aligned}$

$\therefore$ Curved Surface area $=2πrh 

 $\begin{aligned} &=2 \times \frac{22}{7} \times 14 \times 20 \\=& 44 \times 40 \end{aligned}$

$=1760 \mathrm{~cm}^{2}$

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

$=2 \times \frac{22}{7} \times 14(20+14)$

$=44 \times 2 \times 34$

$=88 \times 34$

$=2992 \mathrm{~cm}^{2}$

(v) Given h= 16m

r= 10.5m

 Curved surface arca=2πrh 

$=44 \times 24$

$=105.5 m^{2}$

Total surface area $=2πr (h+r)$

$=66 \times 26.5$

$=1749 \mathrm{~m}^{2}$

Question 2

Ans: Given,

Radius (r) = $\frac{7}{2} m$

Depth (h) = 4m

$\therefore$ Total area of the-wet surface.

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

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

$=\frac{22}{7} \times \frac{7}{2}\left(2 \times 4+\frac{7}{2}\right)$

$\begin{aligned} &=11\left(8+\frac{7}{2}\right) \\ &=11\left(\frac{16+7}{2}\right) \\ &=11 \times \frac{23}{2} \\ &=\frac{253}{2} \\ &=126.5 \mathrm{~m}^{2} . \end{aligned}$

Question 3

Ans: Given, 

External Radius, (r) $\frac{14}{2}=7 \mathrm{~cm} .$

Thickness = 2cm

So, inner radius (r) = 7-2=5cm

Height (h) = 20cm

Total surface area = $2 \pi R h+2 \pi r h+2 \pi R^{2}-2 \pi r^{2} .$

$\begin{aligned} &=2 \pi\left(R h+r h+R^{2}-r^{2}\right) . \\ &=2 \times \frac{22}{7}\left(7 \times 20+5 \times 20+(7)^{2}-(5)^{2}\right) \\ &=\frac{44}{7}(140+100+49-25) \\ &=\frac{44}{7}(240+24) . \\ &=\frac{44}{7} \times 264 . \\ &=\frac{11616}{7} \\=& 1659.43 \mathrm{~cm}^{2} \end{aligned}$

Question 4

Ans: Given, 

Curved Surface Area = $110 \mathrm{~cm}^{2}$

Heights $(h)=5 \mathrm{~cm}$.

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

$110=2 \times \frac{22}{7} \times r \times 5 .$

$110=\frac{44 \times 5}{7} \times r$

$\frac{110\times 7}{44\times 5}=r$

$\frac{7}{2}=r$

$3 \cdot 5=r $

Hence the radius of the cylinder is 3.5cm

Question 5

Ans: Given, 

Curved Surface area = $13 \cdot 2 \mathrm{~cm}^{2}$

Radius $=6 \mathrm{~cm}$.

$\therefore \quad$ Curved surface arca= $2 \pi r h$

$13.2=2 \times \frac{22}{7} \times 6 \times \mathrm{h}$

$\frac{13.2}{10}=\frac{44 \times 6}{7} \times h$

$\frac{132 \times 7}{44 \times 6 \times 10}=h$

$\frac{132 \times 7}{44 \times 60}$=h .

$\begin{aligned} \frac{7}{20} &=h \\ 0.35 &=h \\ \therefore h &=0.35 \mathrm{~cm}\end{aligned}$

Hence the height of the cylinder is 0.35cm

Question 6

Ans: Given,

Radius of garden roller (r) =$\frac{75}{2}cm$

width $(\mathrm{h})=105 \mathrm{~cm}$

 

Curved surface area = $2 \pi \mathrm{rh}$

$=2 \times \frac{22}{7} \times \frac{75}{2} \times 105$

$=22 \times 15 \times 75$

$=330 \times 75$

$=24750 \mathrm{~cm}^{2} .$

Area covered in 14 revolutions 

$=24750 \times 14$

$=346500 \mathrm{~cm}^{2} .$

Question 7

Ans: Given, 

Radius of each cylindrical pillar = $=\frac{50}{2} \mathrm{~m}=25 \mathrm{~m}=\frac{25}{100} \mathrm{~m}$

Height = 4m

$\therefore$ Curved surface area $=2πrh

$=6.28 \mathrm{~m}^{2}$

Area of 10 cylinder = $=6.28 \times 10$

$=62.8 \mathrm{~m}^{2}$

Cost of painting = 50 paise per $m^{2}$

Total cost = $=\frac{628}{10} \times \frac{50}{100}$

$=\frac{314}{10}$

$=₹ 31.4 $

Question 8

Ans: Given, 

Radius of a roller = $\frac{84}{2}=42 \mathrm{~cm}$

Height $(h)=120 \mathrm{~cm}$

Curved surface area = $2 \pi r h$

$=2 \times \frac{22}{7} \times 42 \times 120 .$

$=44 \times 420$

$=31680 \mathrm{~cm}^{2} .$

$=\frac{31680}{100 \times 100}\mathrm{m}^{2}$

$=\frac{3168}{1000} \mathrm{~m}^{2}$

Area of leveling playground by 500 revolution 

$=\frac{3168}{1000} \times 500$

$=\frac{15840}{10}$

$=1584 \mathrm{~m}^{2}$

Cost of leveling the ground at the rate of 30 paisa per meter square

$=1584 \times \frac{30}{100}$

$=\frac{4752}{10}$

$=₹ 475.2 $

Question 9

Ans: Given, 

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

Diameter of wire =5mm = 1cm

$=0.5 \mathrm{~cm}$

$2 \pi r h=1000 \mathrm{~cm}^{2}$

$\therefore$ Number of coils $=\frac{h}{\text { Diameter wire }}$

$=\frac{h}{\frac{0.5}{10}}$

$=\frac{10h}{5}$=2h

So total length of wire = $2 \pi R \times 2 h$

$=2 \times 2 \pi \mathrm{Rh} .$

$=2 \times 1000 .$

$=2000 \mathrm{~cm}$

$=\frac{2000}{100} \mathrm{m} $

=20m

Question 10

Ans: Given,

Height = 20cm

Exterior radius = $\frac{25}{2} \mathrm{~cm}=12.5 \mathrm{~cm}$

Thickness of pipe = 1cm

$\begin{aligned} \therefore \quad \text { Interior radius }(x) &=12.5-1 \\ &=11.5 \mathrm{~cm} \end{aligned}$

∴ Total surface area = $2 \pi R h+2 \pi r h+2 r R^{2}-2 \pi r^{2}$

$\begin{aligned} &=2 \pi \cdot\left(R h+r h+R^{2}-r^{2}\right) \\ &=2 \times \frac{22}{7}\left(12-5 \times 20+11.5 \times 20+(12-5)^{2}-(11-5)^{2}\right) \\ &=\frac{44}{7}(250+230+156.25-132-25) . \\ &=\frac{44}{7}(480+24) \\ &=\frac{44}{7} \times 504 \\ &=\frac{22176}{7} \\ &=3168 \mathrm{~cm}^{2}\end{aligned}$

Question 11

Ans: Given,

No. of circular plates = 50 

Radius of each plates = 7cm

Thickness = 0.5cm

∴ Height thickness of 50 plates = $0.5 \times 50$

$25 \mathrm{~cm}$

∴ Curved surface area = 2πrh

$=2 \times \frac{22}{7} \times 7 \times 25$

$=44 \times 25$

$=1100 \mathrm{~cm}^{2} .$

Area of top and bottom = $2 \pi r^{2}$ 

$=2 \times \frac{22}{7} \times-7-x 7$

$=44 \times 7$

$=308 \mathrm{~cm}^{2} .$

∴ Total surface area = Curved surface area + Area of top and bottom

$=1100+308$

$=1408 \mathrm{~cm}^{2} .$

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15C

 Exercise 15C

Question 1

Ans: (i) Given, 

Radius = 3cm

Height = 4cm

Slant height (l) =$\sqrt{r^{2}+h^{2}}$

$=\sqrt{(3)^{2}+(4)^{2}}$

$=\sqrt{3+16}$

$=\sqrt{25}$

$=5 \mathrm{~cm}$

Curved surface = $\pi r l$

$=\pi \times 3 \times 5$

$=15 \pi \mathrm{cm}^{2}$

Area of base = $\pi r^{2}$

$\pi \times 3 \times 3$

$=9 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=15 \pi+9 \pi$

$=24 \pi \mathrm{cm}^{2}$

Volume  =  $=\frac{1}{3} \pi r^{2} h$

$-\frac{1}{3} \pi \times 3 \times 3 \times 4 $

$=12 \pi \mathrm{cm}^{3}$

(ii) Given, 

Radius = 20cm

Slant height = 25 cm

$l^{2}=r^{2}+1^{2}$

$\left(25^{2}-r^{2}=h^{2}\right.$

$\sqrt{625}-(20)^{2}=1^{2}$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

Curved surface = πrl

$=\pi \times 20 \times 25$

$=500 \pi \mathrm{cm}^{2}$

area of base =  π$r^{2}$

$=\pi \times 200 \times 20$

$=400 \pi \mathrm{cm}^{2} .$

Total surface area = $\pi r l+\pi r^{2}$

$\begin{aligned}=& 500 \pi+400 \pi \\=& 900 \pi \mathrm{cm}^{2} \end{aligned}$

Volume $=\frac{1}{3} \pi{r}^{2} h$

$=2000 \pi \mathrm{cm}^{3}$

(iii) Given, 

Height = 18cm

Slant height = 30 cm

$\therefore \quad l^{2}=h^{2}+r^{2} .$

$(30)^{2}=(18)^{2}+r^{2}$

$900-324=r^{2} .$

$\sqrt{576}=r$

$24=r$

Curved surface = πrl 

$=\pi \times 24 \times 30$

$=720 π c m^{2}$

Area of base = $\pi r^{2}$

$=\pi \times 24 \times 24$

$=576 \pi \mathrm{m}^{2}$

Total surface area= $\pi r l+\pi r^{2}$

729π + 576π

$=1296 \mathrm π{cm}^{2}$

Volume = $\frac{1}{3} \pi r^{2} h$

$=576 \times 6 \pi .$

$=3456 \pi \mathrm{cm}^{3}$

(iv) Given, 

Radius = 27cm

Height = 36cm

$\begin{aligned} \because \quad l &=\sqrt{h^{2}+r^{2}} \\ &=\sqrt{(36)^{2}+(27)^{2}} \\ &=\sqrt{1296+729} \\ &=\sqrt{2025} \\ &=45 \mathrm{~cm}\end{aligned}$

Curved surface =  $\pi$ rl

= $\pi$ $\times 27 \times 45$

$=1215 \pi \mathrm{cm}^{2} .$

Area of base =  $\pi r^{2}$ 

$=\pi \times 27 \times 27$

$=729 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=1215 \pi+729 \pi$

$=1944 \pi \mathrm{cm}^{2}$

Volume = $\frac{1}{3}$ $\pi r l+\pi r^{2}h$

$=8748 \pi \mathrm{cm}^{3}$

(v) Given, 

Radius = 5cm,

Curved surface = 65π $\left(cm^{2}\right)$

  

∴ Curved surface =πrl 

65π = π $5\times l$

$\frac{65} π { π  \times 5}=l$

$l=13 \mathrm{~cm}$

$\because \quad l^{2}=h^{2}+r  ^{2}$

$(13)^{2}=h^{2}+(5)^{2}$

$16 y-25=h^{2}$

$144=h^{2}$

$\sqrt{144}=h$

$12=h$

$h=12 \mathrm{~cm} .$

Area of base = $πr^{2}$

$=\pi \times 5 \times 5$

$=25 \pi \mathrm{cm}^{2}$

Total surface area = $\pi r l+\pi r^{2}$

$=65 \pi+25 \pi$

$=90 \pi \mathrm{m}^{2} .$

Volume $=\frac{1}{3}\pi r^{2} h$

$=\frac{1}{3} \pi \times(5)^{2} \times 12$

$=25 \times 4 \pi$

$=100 \pi \mathrm{cm}^{3}$

(vi) Given,

Radius = 35cm

Total surface area  = $13860 \mathrm{~cm}^{2}$

Total surface area = πrl+π $r^{2}$

$13860:=\pi r(l+r) .$

$4410 \pi=\pi \times 35(l+35) .$

$4410 \pi=35 \pi(l+75)$

$126=l+35$

$126-35=l$

$91=l$

$\ell=91 \mathrm{~cm} .$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\(91)^{2} &=h^{2}+(35)^{2} \end{aligned}$

$8281-1225=h^{2}$

$\sqrt{8281-1225}=h$

$\sqrt{7056}=h$

$84=h$

$h=84 \mathrm{~cm}$

Curved surface = $\pi rl$

$=\pi \times 35 \times 91$

$=3185 \pi \mathrm{cm}^{2}$

Area of base $=\pi r^{2}$

$=\pi \times 35 \times 35$

$=1225 \pi c m^{2}$

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi \times 35 \times 35 \times 84$

$=34300 \pi \mathrm{cm}^{3}$

Question 2

Ans: (i) Given, 

Height = 8m, 

Area of base = $156 \mathrm{~m}^{2}$

$\begin{aligned} \therefore \text { Area of base } &=\pi r^{2} \\ 156 &=\frac{22}{7} \times \r^{2} \\ \frac{156 \times 7}{22} &=r^{2} \end{aligned}$

$\frac{546}{11}=r^{2}$

$r^{2}=\frac{546}{11} \mathrm{~m}$

ஃ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{546}{11} \times 8$

$=2 \times 26 \times 8$

$=416 \mathrm{~m}^{3}$

(ii) Given,

Slant height (l)= 17cm,

Radius (r)= 8cm

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-r^{2}=h^{2} .$

$(17)^{2}-(8)^{2}=h^{2}$

$\sqrt{289-64}=h .$

$\sqrt{225}=h$

$15=h$

$h=15 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 15$

$=176 \times 40$

$=\frac{7.040}{7}$

$=1005.71 \mathrm{~cm}^{3}$

(iii) Given, 

Height = 8cm,

Slant length = 10cm,

$\therefore \quad l^{2} = h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2}$

$(10)^{2}-(8)^{2}=r^{2}$

$100-64=r^{2}$

$\sqrt{36}=r$

$6=r$

$r=6 \mathrm{~cm}$

$\therefore$ Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 8$

$=\frac{44 \times 48}{7}$

=301.71$cm^{3}$

(iv) Given, 

Height=5cm 

Perimeter of base = 8cm

Perimeter of base= $2 \pi r$

$8=2 \times \frac{22}{7} \times r$

$\frac{{8} \times 7}{2 \times 22}=6$

$\frac{14}{11}=r$

$r=\frac{14}{11} \mathrm{~cm}$

∴ Volume = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{14}{11} \times \frac{14}{11} \times 5$

$=\frac{4 \times 70}{33}$

$=\frac{280}{33}$

$=8.48 . \mathrm{cm}^{3}$

Question 3

Ans: (i) Given, 

Height = 8m,

Slant height = 10m

$\therefore l^{2}=h^{2}+r^{2}$

$l^{2}-h^{2}=r^{2} .$

$(10)^{2}-(8)^{2}=r^{2}$

$\sqrt{100-64}=r$

$\sqrt{36}=r$

$6=r$

r=6 m

∴ Curved surface area = πrl

$\begin{aligned} &=\frac{22}{7} \times 6 \times 10 \\=& \frac{1320}{7} \end{aligned}$

=199.6 $m^{2}$

(ii) Given, 

Perimeter of base = 88cm,

slant height = 2dm = $2 \times 10cm=20cm

∴ Perimeter of base $=2 \pi r$

88= $=2 \times \frac{22}{7} \times r$

$\frac{88 \times 7}{2 \times 22}=r$

r=14cm

Curved surface area $=\pi r l$

$\begin{aligned} & \frac{22}{7} \times 14 \times 20 \\=& 44 \times 20 \\=& 880 \mathrm{~cm}^{2} \end{aligned}$

(iii) Given,

Area of base= $154 \mathrm{~cm}^{2}$

Height 24cm

Area of base= $\pi r^{2}$

$154=\frac{22}{7} \times r^{2}$

$\frac{154 \times 22}{7}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~cm}$

$\therefore \ell=\sqrt{h^{2}+r^{2}}$

l=$\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=25 \mathrm{~cm}$

So 

Curved surface area = π rl

$=22 \times 25$

$=550 \mathrm{~cm}^{2}$

Question 4

Ans: Given,

Radius = 5cm

Volume = 50π $\mathrm{Cm}^{3}$

Volume= $\frac{1}{3}π r^{2} h$

$50 \pi=\frac{1}{3} \pi \times 5 \times 5 \times \mathrm{h}$

$50 \pi=\frac{\pi}{3} \times 25 \mathrm{~h}$

$\frac{50 \times \pi \times 3}{\pi \times 25}=h$

$\frac{150}{25}=h$

$6=h$

$h=6 \mathrm{~cm}$

Hence, the height of the cone is 6cm

Question 5

Ans: Given,

Radius = 11.3cm

Curved surface area = $710(cm)^{2}$

Curved surface area $=\pi rl$

$710=\frac{355}{113} \times \frac{11.3}{10} \times l$

$710=\frac{355}{10} l$

$\frac{710 \times 10}{355}=l$

$\frac{7100}{355}=l$

$20=l$

$l=20 \mathrm{~cm}$

Hence , slant height is 20cm

Question 6

Ans:  Let the radius be r and height be h

Volume =$\frac{1}{3} \pi r^{2} h$

According to question 

Radius =$\frac{r}{2}$

and height= h

Volume = $\frac{1}{3} \pi\left(\frac{8}{2}\right)^{2} h$

$=\frac{1}{3} \pi \frac{r^{2}}{4} h=$ $\frac{\pi^{2} h}{12}$

∴ Ratio = $\frac{\pi r^{2} h}{12}=\frac{1}{3} \pi r^{2} h$

$=\frac{\frac{\pi r^{2} h}{12}}{\frac{1 \pi x^{2} h}{3}}$

$=\frac{1}{4}=1: 4 .$

Question 7

Ans: Given 

Curved surface area = $264 m^{2}$

Slant height = 12m

Curved surface area = πrl

$264=\frac{22}{7} \times r \times 12$

$\frac{264 \times 7}{22 \times 12}=r$

$7=r$

$r=7 \mathrm{~m}$

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l^{2}-r^{2} &=h^{2} \end{aligned}$

Question 8

Ans: According to question, 

Height (h) = $2 \times$ diameter 

Diameter = $\frac{\text { height }}{2}$

$\therefore \quad \operatorname{Radius}=\frac{h}{2} \times \frac{1}{2}=\frac{h}{4}$.

Volume $=36 \pi \mathrm{cm}^{3}$

$\frac{1}{3} \pi r^{2} h=36 \pi$

$\frac{1}{3} \times \pi \times \frac{h}{4} \times \frac{h}{4} \times \pi=36 \pi .$

$\frac{h^{3}}{48} \pi=36 \pi$

$h^{3}=\sqrt{728}$

$h=\sqrt[3]{1728}=\sqrt{12 \times 12 \times 12}=12 . \mathrm{cm}$

Question 9

Ans: Given,

Radius and height of cone are in ratio = 3:4

Let radius be 3x 

and height be 4x 

Volume = 301.44 $(cm)^{3}$

$\because \quad$ Volume =\frac{1}{3} \pi r^{2} h$

$301.44=\frac{1}{3} \times 3.14$  $\times(3x)^{2}$  $\times$(4x)$

$301.44=3.14 \times 12 x^{3}$

$\frac{301.44}{3.14 \times 12}=x^{3}$

$\frac{30144}{3768}=x^{3}$

$8 =x^{3}$

$x^{3}=8$

$x=\sqrt[3]{8}$

$x=\sqrt{2 \times 2 \times 2}$

$x=2$

$\begin{aligned} \therefore \text { Radius } &=3 \times \\ &=3 \times 2 \\ &=6 \mathrm{~cm} . \end{aligned}$

and Hight $=4 x$

$=4 \times 2$

=8

$\begin{aligned} \therefore l^{2} &=h^{2}+r^{2} \\ l &=\sqrt{(8)^{2}+(6)^{2}} \\ l &=\sqrt{64+36} \\ l &=\sqrt{100} \\ l &=10 \mathrm{~cm} \end{aligned}$

Question 10

Ans: Given ,

The ratio of radius and slant height of cone = 4:7

Curved surface area $=792 \mathrm{~cm}^{2}$

Let radius be 4x and 

slant height be 7x

Curved surface area = πrl

$792=\frac{22}{7} \times 4 x \times 7 x$

$792=22 \times 4 x^{2}$

$792=88 x^{2}$

$\frac{792}{88}=x^{2}$

$9=x^{2}$

$x^{2}=9$

$x=\sqrt{9}$

$x=3$

Radius =4x

$=4 \times 3$

$=12 \mathrm{~cm}$

Question 11

Ans: Given, 

The ratio of radii of two cones = 3:5

Let r1 be 3x and r2 be 5x

Volume of first cone = $\frac{1}{3} \pi r_{1}^{2} h$

$=\frac{1}{3}π  (3x)^{2}\times h$

= $3 \pi x^{2} h .$

Volume of second cone =  $\frac{1}{3} \pi r_{2}^{2} h$

$\begin{aligned} &=\frac{1}{3} \pi \times (5 x)^{2} x h \\=& \frac{1}{3} \pi 25 x^{2} h \\=& \frac{25 \pi x^{2} h}{3} \end{aligned}$

ration of their volumes 

$=3 \pi x^{2}h \frac{25 \pi x^{2} 1}{3}$

$=\frac{3 \pi x^{2} h}{\frac{25 \pi x^{2} h}{3}}$

$=\frac{9}{25}$

$=9: 25$

Question 12

Ans: Given, 

Circumference = 44m

Height = 10m

Circumference = 2πr

$44=2 \times \frac{22}{7} \times r$

$\frac{44 \times 7}{44}=r$

$7=r$,

$\therefore l^{2}=h^{2}+r^{2} .$

$l=\sqrt{(10)^{2}+(7)^{2}}$

$l=\sqrt{100+49}$

$l=\sqrt{149 \mathrm{~m} .}$

∴ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times \sqrt{149}$

$\begin{aligned} &=22 \sqrt{149}\\=& 22 \times 12.2 . \\=& 268.4 \mathrm{~m}^{2} . \end{aligned}$

Width of canvas(d)=  2cm= $\frac{2}{100} \mathrm{~m}$

$\therefore$ Area $=l \times b$

$268.4=l \times \frac{2}{100}$

$\frac{26840}{2}=l$

=13429=l

Hence, the length of canvas is 13420m

Question 13

Ans: Given, 

Radius =7m

Height = 24m

$\because$ Slant height $(l)=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{2.5 \times 25}$

$l=25 \mathrm{~m}$

Curved surface area =  πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

Width of canvas (b) = 5m

Area = L $\times$ B

$550=L \times 3$

$\frac{550}{5}=L$

$110=L$

$L=110 \mathrm{~m}$

$\therefore$ Length $=110 \mathrm{~m}$.

Question 14

Ans: Given,

Volume =  $1232 \mathrm{~m}^{3} .$

Area of base = $154 \mathrm{~m}^{2}$

Area of base = $\pi r^{2}$

$154=$ $\frac{22}{7} \times r^{2}$

$\frac{154 \times 7}{22}=r^{2}$

$49=r^{2}$

$\sqrt{49}=r$

$7=r$

$r=7 \mathrm{~m}$

Volume $=\frac{1}{3} \pi r^{2} h$

$1232=$ $\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times h$

$\frac{1232 \times 3}{22 \times 7}=h $

$8 \times 3=h$

$24=h$

$h=24 m$

$l^{2}=h^{2}+x^{2}$

$l=\sqrt{(24)^{2}+(7)^{2}}$

$l=\sqrt{576+49}$

$l=\sqrt{625}$

$l=\sqrt{25 \times 25}$

$l=25 \mathrm{~m}$

ஃ Curved surface area = πrl

$=\frac{22}{7} \times 7 \times 25$

$=22 \times 25$

$=550 \mathrm{~m}^{2}$

∴ The area of canvas $=550 \mathrm{~m}^{2}$

Question 15

Ans: Given, 

In the tent, accommodate is available for 11 persons and each person must have $4r^{2}$ of space on the ground 

So , aera of base of the tent = $11 \times{4}$

$44 m^{2}$

Air is required for each person to breadth = $20m^{3}$

Volume of air = $20 \times 11$

$=220 \mathrm{~m}^{3}$

Area of base = π$r^{2}$

$44=\frac{22}{7} \times r^{2}$

$\frac{44 \times 7}{22}=r^{2}$

$r^{2}=14 \mathrm{~m} .$

Volume = $=\frac{1}{3} \pi{r}^{2} h$

$220=\frac{1}{3}\times $\frac{22}{7}\times 14 \times h$

$220=\frac{44h}{3}$

$\frac{220 \times 3}{44}=h$

$15=h$

h=15 m

Hence , the height of the cone = 15m

Question 16

Ans: False , because the volume of a cone is one third $\left(\frac{1}{3}\right)$ of the volume of cylinder of the same radius and height 

Question 17

Ans: Given,

Height of cylinder = 9cm

and radius= $\frac{40}{2} cm=20 c m$

Height of cone = 108cm

∴ Volume of cylinder = $\pi r^{2} h$

$=\frac{22}{7} \times 20 \times 20 \times 9$

$=\frac{440 \times 180}{7}$

$=\frac{79200}{7} \mathrm{~cm}^{2} .$

ஃ Volume of cone $\frac{79200}{7}$ (Given volume of cylinder is equal to volume of cone)

Volume of cone = $=\frac{1}{3} \pi r^{2}$h

$\frac{79200}{7}=\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 108$

$\frac{79200 \times 7 \times 3}{7 \times 22 \times 108}=r^{2}$

$\frac{79200 \times 21}{7 \times 22 \times 108}=r^{2}$

$\frac{79200}{792}=r^{2}$

$100=r^{2}$

$r^{2}=100$

$r=\sqrt{100}$

$r=10$

Hence, the radius of cone is 10cm

Question 18

Ans: Given, 

Radius, of cone and cylinder (r) = 7m,

Height of the cylinder (h1)=8m

And height of the conical of (h2)= 4m

(IMAGE TO BE ADDED)

∴ l =$\sqrt{h2^{2}+r^{2}}$

$=\sqrt{(4)^{2}+(7)^{2}}$ $=\sqrt{16+49}$ $=8.06M$

∴ Area of canvas = Curved surface area of cylinder + Curved surface area of cone 

= 2πrh_{1}$ + πrl

$=2 \times \frac{22}{7} \times 7 \times 8+\frac{22}{7} \times 7 \times 8.06 .$

$=352+177.32 .$

$=529.32 . \mathrm{m}^{2}$

Question 19

Ans: Given, 

Height of cylinder = 3m 

Radius of cylinder = $\frac{105}{2} \mathrm{~m}$

and slant height = 53m

Total area of the canvas = Curved surface area of cylinder +curved surface area of cone 

= 2πrh+πrl
$=22 \times \frac{22}{7} \times $3+\frac{2 x}{7}\times\frac{105}{2} \times 53$

$=22 \times 45+11 \times 795 .$

$=990+8745$

$=9735 \mathrm{~m}^{2}$

Question 20

Ans: Given, 

Edge of a cube = 9cm

Diameter of cone = 9cm (because equal to edge of cube)

Then radius = $\frac{9}{2} cm$

and height of cone = 9cm (because equal to edge of cube)

Volume of largest cone = $\frac{1}{3}πr^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 9$

$=\frac{99 \times 27}{14}$

$=\frac{2673}{14}$

$=190.93 . \mathrm{cm}^{3}$

Question 21

Ans:  Given,

Height of cylinder(h1) = 3m, 

Total height of tent = 13.4m

Height of conical part (h2) = 13.5-3

=10.5m

(IMAGE TO BE ADDED)

Radius = 14 m

$l=\sqrt{h_{2}^{2}+r^{2}}$

$l=\sqrt{(10.5)^{2}+(14)^{2}}$

$l=\sqrt{110.25+196}$

$l=\sqrt{306.25}$

$l=17.5 \mathrm{~m}$

Question 22

Ans:  (IMAGE TO BE ADDED)

Given, 

Height of the cylinder (h1)= 32cm

and radius (r1) = 18cm

Height of the conical= 24 cm

Volume of sand in it = π$r_{1}^{2}$ h

$=\frac{22}{7} \times 18 \times 18 \times 32$

$=\frac{396 \times 576}{7}$

$=\frac{228096 \mathrm{cm}^{3}}{7}$

(IMAGE TO BE ADDED)

So volume of conical heap of the sand = $\frac{228096 \mathrm{~cm}^{3}}{7}$

Volume of conical heap = $\frac{1}{3} \pi r^{2} h$

$\frac{228096}{7}=$ $\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24$

$\frac{22.8096}{7}=$  $\frac{176}{7} r^{2}$

$\frac{228096 \times 7}{7 \times 176}$ $=r^{2}$

$\frac{228096}{176}=r^{2}$

$1296=r^{2} .$

$r^{2}=1296$

$r=\sqrt{1296}$

$r=36 \mathrm{~cm}$

(i) Radius of cone = 36cm

(ii) $l=\sqrt{h^{2}+r^{2}}$

$l=\sqrt{(24)^{2}+(36)^{2}}$

$l=\sqrt{576+1296}$

$l=\sqrt{1872}$

l=43.3cm

Height the slant height of heap is 43.3cm

Question 23

Ans:  (IMAGE TO BE ADDED)

Given, 

Height cylinder = 8cm

Radius of cylinder = 6cm

Volume of cylinder = $\pi r^{2} h$

$=31416 \times 6 \times 6 \times 8$

$=18.8496 \times 48$

$=904.7808 \mathrm{~cm}^{3}$

Radius of cone = 6cm 

Height of cone = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2}h$

$=62832 \times 48$

$=301.5936 \mathrm{~cm}^{3}$

 

∴ Volume of remaining solid = Volume of cylinder - Volume of cone 

$=904.7808-301.5936$

$=603.1872 \mathrm{~cm}^{3}$

Question 24

Ans: (IMAGE TO BE ADDED)

Given,

Radius , of cylinder = 3cm 

and height of cylinder = 5cm 

Volume of cylinder = $\pi r^{2} h$

$=\frac{21}{7} \times 3 \times 3 \times 5$

$=\frac{990}{7} \mathrm{~cm}^{2}$

Radius of conical portion = $\frac{3}{2} c m$

and Height of conical portion = $\frac{8}{9} c h$

Volume of conical portion = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{8}{9}$

$=\frac{44}{21} \mathrm{~cm}^{3}$

Metal in the remaining part = Volume of cylinder - Volume of conical portion 

$=\frac{990}{7}-\frac{44}{21}$

$=\frac{2970-44}{21}$

$=\frac{2926}{21}$

According to the question 

$\frac{2926}{21}: \frac{44}{21}$

$\frac{\frac{2926}{21}}{\frac{4 4}{21}}$

$=\frac{2926 \times 21}{21 \times 44}$

$\frac{1463}{22}=$

$=133: 2$

Question 25

Ans:   (IMAGE TO BE ADDED)

Given, 

Radius of cylinder = $\frac{7}{2} \mathrm{~cm}$

Its height = 8cm

Volume of cylinder = $\pi r^{2} h$

$\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 8$

$=22 \times 14$

$=308 \mathrm{~cm}^{3} .$

Radius of cone $=\frac{7}{4} \mathrm{~cm}$

Its height = 8cm

Volume of cone = $\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7}\times \frac{7}{4} \times \frac{7}{4} \times 8$

$=\frac{77}{3} \mathrm{~cm}^{3}$

Volume of water required to fill the vessel= Volume of cylinder -Volume of cone 

$=308-\frac{77}{3}$

$=\frac{924-77}{3 .}$

$=\frac{847}{3}$

$=28 \frac{1}{3} \mathrm{~cm}^{3}$

According to question,

Height of cone = $1 \frac{3}{4}=\frac{7}{7} \mathrm{Cm} x$

its radius = 2cm

Volume of cone =  $\frac{1}{3} \pi r^{2} h$

=$\frac{22}{3} \mathrm{~cm}^{3}$

Change in volume of cones 

$=\frac{77}{3}-\frac{22}{3}$

$=\frac{77-22}{3}$

$=\frac{55}{3} \mathrm{~cm}^{3}$

Let the drop in water level be h cm

Volume = π$r^{2} h$

$\frac{55}{3}=$ $\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h $

$\frac{55 \times 2}{3 \times 11 \times 7}=h$

$\frac{10}{21}=h $

$h=\frac{10}{21} \mathrm{~cm}$

S Chand Class 10 CHAPTER 15 Three Dimensional Solids Exercise 15B

  Exercise 15B

Question 1 

Ans: (i) Given,

r= 7cm, 'h=8cm

∴ Volume = $\pi r^{2} h$

$=\frac{22}{7} \times 7 \times 7+8$

$=22 \times 56$

$=1232 \mathrm{~cm}^{3}$

(ii) Given,

$\begin{aligned}&r=7 \mathrm{~cm} \\&h=12 \mathrm{~cm} .\end{aligned}$

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

$\begin{aligned} &=\frac{22}{7} \times 7 \times 7 \times 12 \\ &=22 \times 84 \\ &=1848 \mathrm{~cm}^{3} \end{aligned}$

(iii) Given,

$r=14 \mathrm{~cm}$

$\mathrm{h}=16 \mathrm{~cm} .$

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

$-\frac{22}{7} \times 14 \times 14 \times 6$

$=44 \times 224$

$=9856 \mathrm{~cm}^{3}$

(iv) Given, 

r= 21cm,

h=40cm

Volume = $\pi r^{2} h$,

$=\frac{22}{7} \times 21 \times 21 \times 40$

$=66 \times 840$

$=\quad 55440 \mathrm{~cm}^{3}$

Question 2

Ans: (a) Given, 

Volume, = $44 \mathrm{Cm}^{3}$

Height $=3.5 \mathrm{~cm}$

ஃ Volume = $\pi r^{2}h$

$44=\frac{22}{7}\times r^{2}\times \frac{3.5}{10}$

$ 44=11 r^{2}$

$ \frac{44}{11}=r^{2}$

$4=r^{2}$

$r^{2}=4 .$

$r=\sqrt{4} .$

$r=2 \mathrm{cm} $

∴ Diameter = 2r 

$=2 \times 2$

$=4 c \mathrm{~m}$

(b) Given, 

Volume , = $385 \mathrm{~cm}^{3}$

Height $=1 \mathrm{dm}=10 \mathrm{~cm}$

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

$385=\frac{22}{7} \times r^{2} \times 10$

$\frac{49}{4}=r^{2}$

$r^{2}=\frac{49}{4}$

$r=\sqrt{\frac{49}{4}}$

$r=\frac{7}{2}$

∴ Diameter = 2r = $2 \times \frac{7}{2}=7 \mathrm{~cm}$

 

Question 3

Ans: (a) Given,

Volume = $66 \mathrm{~cm}^{3}$

Radius = 2cm

∴ Volume = π$(r^{2}$h

$\frac{66 \times 7}{22 \times 4}=h$

$\frac{21}{4}=h$

$h=\frac{21}{4}$

$h=5.25 \mathrm{~cm}$

(b) Given,

Volume = 4litres= 4000 $\mathrm{Cm}^{3}$

Radius= 5cm

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

$4000=\frac{22}{7} \times 5 \times 5 \times h$

$\frac{4000 \times 7}{22 \times 25}=h$

$\frac{80 \times 7}{11}$

$\frac{560}{11}=h$

$h=\frac{510}{11} \mathrm{~cm}$

Question 4

Ans: Given,

Height(h) = 7m

Radius(r) = $\frac{20}{2}=100 \mathrm{~cm}$ $=\frac{10}{100}$ $=\frac{1}{10}m$

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

$=\frac{11}{50}$

According to question,

Total weight $=\frac{11}{50} \times 225$

$=\frac{99}{2}$

$=49.5 \mathrm{~kg}$

Question 5

Ans: Giver,

Internal radius. $(r)=3 \mathrm{~cm}$

Thickness of pipe= $1 \mathrm{~cm}$

$\therefore$ Outer radius $(R)=3+1$=4cm

Length = 6cm

$\begin{aligned} \therefore \quad \text { Volume } &=\pi R^{2} h-\pi r^{2} h . \\ &=\pi h\left(R^{2}-r^{2}\right) . \\ &=\frac{22}{7} \times 6\left((4)^{2}-(3)^{2}\right) \end{aligned}$

$=\frac{132}{7}(16-9)$

$=\frac{132}{7} \times 7 $

$=132 \mathrm{cm}^{3}$

Question 6

Ans: Given,

Sum of radius, of the base and the height of a cylinder (h+r) = 37cm,

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

∴ Total surface area = $2πr h+2πr^{2}$

$1628=2 \pi r(h+r)$

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

r=7

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

So, $h+r=37$

$h+7=37$.

$h=377$

$h=30 \mathrm{~cm}$.

$\begin{aligned} \text { Volume } &=\pi r^{2} h \\ &=\frac{22}{7} \times 7 \times 7 \times 30 \\ &=22 \times 210 \\ &=4620 \mathrm{~cm}^{3} \end{aligned}$

Question 7

Ans: Given

Capacity of a cylindrical tank = $6160 \mathrm{m}^{3}$

Radius $(r)  \frac{28}{2}=14 \mathrm{~m}$

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

$6160=\frac{22}{7} \times 14 \times 14 \times h$

$\frac{6160}{22 \times 281}=h $

h= 10m

Area of curved surface of taken inner sides = $2 \pi r h$

$=2 \times \frac{22}{7} \times 14 \times 10$

$=44 \times 20$

$=880 \mathrm{m}^{2}$

Question 8

Ans: Given,

Curved surface area of cylinder = $4400 \mathrm{~cm}^{2}$

Circumference $=110 \mathrm{~cm} $

Circumference $=2 \pi r$

$110=2 \times \frac{22}{7} \times r$

$\frac{110 \times 7}{2 \times 22}=r$

$\frac{35}{2}$=r

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

(i) Curved Surface area $=2$ πrh

4400= $=\frac{2}\times \frac{22}{7} \times \frac{35}{2} \times h$

$\frac{4400}{22 \times 5}=h$

$40=h$

$h=40$ cm

hence the height of cylinder is 40cm

(ii) Volume = $\pi r^{2} h$

$=110 \times 350$

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

Question 9

Ans: (i) Given ,

Height of the wall = 20 meter 

Radius = $\frac{2}{2}=1$

$\because$ Volume =  πr^{2}$

$\begin{aligned} &=\frac{22}{7} \times 1 \times 1 \times 20 \\=& \frac{440}{7} \\=& 62 \frac{6}{7} \mathrm{~m}^{3} . \end{aligned}$

(ii) Curved surface area = $2 π r h$

$\begin{aligned} &=2 \times \frac{22}{7} \times 1 \times 20 \\=& \frac{44 \times 20}{7} \\=& \frac{880}{7} \end{aligned}$

Rate of plastering the inner surface = Rs 5  per m².

Total cost = 880\7 x 5

$=\frac{4400}{7}$

$=7628.57 $

Question 10

Ans: Giver

Radius of cylinder $=\frac{20}{2}=10 \mathrm{~cm}

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

(i) Curved surface area $=2 \pi r$

$1000=2 \times \frac{22}{7} \times 10 \times h$

$\frac{175}{11}=h$

$h=\frac{175}{11}$

$h=15.9 \mathrm{~cm} .$

(ii) $\begin{aligned} \text { Volume } &=\pi r^{2}h \\ &=3.14 \times 10 \times 10 \times 15-9 . \\ &=31.4 \times 159 . \\ &=4992.6 \mathrm{~cm}^{3}\end{aligned}$

Question 11

Ans: Given, 

Radius =  $\frac{35}{2} \mathrm{~cm}$

height $=1.2 \mathrm{~m}=1.2 \times 100 \mathrm{~cm}=120 \mathrm{~cm}$.

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

$=110 \times 120$

$=13200 \mathrm{~cm}^{2} $

(ii) Capacity = $\pi r^{2} h$

 $=55 \times 2100$

$=115500 \mathrm{~cm} .$

=115.5 liters 

Question 12

Ans: Given, 

Radius of cylindrical glass = $\frac{8}{2}=4 \mathrm{~cm}$

and its height = 15cm

Radius, of cylindrical vessel = $\frac{30}{2}=15 \mathrm{~cm}$

and its height = 80 cm

Volume of cylindrical glass= $=\pi r^{2} \mathrm{~h}$

$\pi \times 4 \times 4 \times 15$

$=240 \pi$

Volume of cylindrical vessel $=\pi r^{2} h$

$=\pi \times 15 \times 15 \times 80$

$=\pi \times 225 \times 80$

$=18000 \pi$.

$\therefore$ Number of glasses $=\frac{\text { Volume of vessel }}{\text { Volume of glass}}

$=\frac{18000 \pi}{240 \pi}$ 

$=75$ glasses

Question 13

Ans: Given, 

R= 22m

r= 20 

h=$\frac{7}{100} m$

$\pi R^{2} h-\pi{r}^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times \frac{7}{100}(R+r)(R-r)$

$=\frac{22}{7}\times{7}{100}(22) \times 18=18.48 \mathrm{~mm}$

Question 14

Ans: Given

Radius of iron cylindrical block = $=\frac{0.5m}{20}$ =$\frac{1}{4} \mathrm{~m}=$0.25m= $0.25\times $100cm  =25cm

Length = 3.5m= 3.5 $\times 100$cm = 350 cm

Volume of block =  $\pi r^{2} b$

$=\frac{22}{7} \times 25 \times 25 \times 350$

$=550 \times 1250$

$=687500 \mathrm{~cm}^{3}$

So , volume of base = $687500 \mathrm{~cm}^{3}$

Area of square base $=25 \times 25$

$=625 \mathrm{~cm}^{2}$

Height of bar= $\frac {Volume of bar}{Area of square base}$

$=\frac{687500}{625}$

$=1100 \mathrm{~cm} $

$=\frac{1100}{100}$

$=11 \mathrm{~m}$

Question 15

Ans: Given, 

Length of swimming pole (l) = 70m

Breadth (b)= 44m

and depth (h) =  3m

Volume = lbh

$=70 \times 40 \times 3$

$=924012^{3}$

Radius of pipe= $\frac{14}{2} c m=7 cm=\frac{7}{160} \mathrm{~m}$

Volume = $πr^{2}h

9240= $\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times h$ (Volume = 9240)

$\frac{9240 \times 100 \times 100}{22 \times 7}=h$

$\frac{92400000}{154}=h .$

$600000=h .$

$\therefore h=600000$

Let be the distance 

$\therefore \quad$ Distance $=$ speed \times $ Time.

$600000=2 \times$ Time. (speed = 2m)

$\frac{600000}{2}=$ Time.

$83 \frac{1}{3}$ hours

Question 16

Ans: Given, 

External radius of a hollow cylinder = $\frac{12}{2}=6 \mathrm{~cm}$

Internal radius of a hollow cylinder = 6_ 0.25

= 5.75cm

Length (h)= 15cm

Volume of hollow cylinder = $\pi R^{2} h-\pi r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 15\left((6)^{2}-\left(5.70^{2}\right)\right.$

$=\frac{330}{7}(36-33.0625) .$

$=\frac{330}{7} \times 2.9375$

$=\frac{969.375}{7 }$

Radius of solid cylinder $\frac{2}{2}=1 \mathrm{~cm}$ (given).

$\begin{aligned} \therefore \text { Volume } &=\pi r^{2} h \\ \frac{969-375}{7} &=\frac{22}{7} \times 1 \times 1 \times h \end{aligned}$

$\frac{969.375}{22}=h$ 

$\therefore h=\frac{969.375}{22}$

$h=44.0625 \mathrm{~cm} $

Question 17

Ans: Given, 

Internal radius of tube= $\frac{11.2}{20} \mathrm{~cm}=5.6 \mathrm{~cm}$

$\operatorname{Length}(h)=21 \mathrm{~cm}$.

Thickness $=0.4 \mathrm{~cm}$.

∴ Outer radius 5.6+0.4

=6cm

Volume of Metal = $\pi R^{2} h - r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$\left.=\frac{22}{7}\times 21 \times(6)^{2}-(5-6)^{2}\right)$

$=66 \times(36-31.36)$

$=66 \times 464$

$=306.24 \mathrm{cm}^{3}$

$=306.2 \mathrm{~cm}^{3}$

Question 18

Ans: Given, 

Volume of water = 1000 lit = $1000 \times 1000 \mathrm{~cm}^{3}=1000000 \mathrm{~cm}^{3}$

Radius of pip = 0.6cm

∴ Volume = $\pi r^{2} h$

$1000000=\frac{22}{7} \times 0-6 \times 0.6 \times h .$

$\frac{1000000 \times 7}{22 \times 0-6 \times 0.6}=h .$

$\frac{700000000}{7.92}=h .$

$883838.38=h .$

$h=883838-38 \mathrm{~cm}$

Let height (h) be the distance, 

Distance = speed $\times$ times 

$883838.38=8 \times$ time.

$\frac{88383838}{800}=$ time

$110479.7975=$ Time

Time = 110479.7975sec

Time = $\frac{110479.7975}{60 \times 60}$ hours 

$=\frac{11047967975}{36000060}$ hours

$=30.69$ hours 

Question 19

Ans: Given,

Radius =$\frac{28}{2} \mathrm{~cm}=14 \mathrm{~cm}$

Height = 72cm

$\therefore$ Volume $\pi r 2 h$

$=\frac{22}{7} \times 14 \times 14 \times 72$

$=44 \times 1008$

$=44352 \mathrm{~cm}^{3}$

Length f tank = 66cm

Breadth of tank = 28cm

Volume = $=l \times b \times h$

$\begin{aligned} 44352 &=66 \times 28 \times h . \\ 44352 &=1848 h . \\ \frac{44352}{1848} &=h . \\ 24 &=h \\ \therefore h &=24 \mathrm{~cm} . \end{aligned}$

Hence, the height of the water level in the tank is 24 cm

Question 20

Ans: Given,

Radius of cylindrical vessel = $\frac{14}{2} c m=7 cm$ 

Height of water = $8 \frac{9}{14}(cm )=\frac{121}{14}$

Volume of water= $\pi r^{2} h$

$=1331 \mathrm{~cm}^{3}$

Volume of cube = $=1331 \mathrm{~cm}^{3}$

Volume of cube= $a^{3}$

$1331=a^{3}$

$\sqrt[3]{1331}=a$

$\sqrt{11 \times 11 \times 11}=a$

$a=11$

Hence , the length of the edge is 11cm

Question 21

Ans: Let the radius of the cylinder = r

And height = h

Volume = $\pi r^{2} h$

If the radius is halved 

$\therefore \quad r=\frac{r}{2}$

Height = h

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

$=\pi \frac{r^{2}}{4} h$

According to question 

$=\frac{\pi r^{2}}{4} h=\pi r^{2} h$

$=\frac{1}{4}$= 1:4

Question 22

Ans: Given,

Length of sheet = 22cm

and breadth = 12cm

If it is folded breadth wire, 

∴ Circumference = 12cm

and height = 22cm

Circumference = 2πr

$12=2 \times \frac{22}{7} \times r$

$\frac{21}{11}=r$

$y=\frac{21}{11} \mathrm{~cm}$

∴ Volume =πz $=r^{2} h$

$=\frac{22}{7} \times \frac{21}{11} \times \frac{21}{11} \times 22$

$=\frac{462 \times 42}{77}$

$=\frac{19404}{72}$

$=269.5 \mathrm{~cm}^{3}$

IF it is folded length wire 

$\therefore$ circumference  $=22 \mathrm{~cm}$. 

and Height: $12 \mathrm{~cm}$.

So circumference =$2 \pi r$

$22=2 \times \frac{22}{7} \times r$

$\frac{22 \times 7}{2 \times 22}=r$

$\frac{7}{2}=r$

$r=\frac{7}{2} cm$

Volume =$\left.\pi r^{2}\right)h$

$\frac{27}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$

$=22 \times 21$

$=482cm^{3}$

According to question, 

$=462-269-5$

$=192.5 \mathrm{~cm}^{3}$

Question 23

Ans: Given, 

Depth of a wall = 20m

Radius = $\frac{7}{2} \mathrm{~m}$

Volume of earth = $\pi r^{2} h$

$=22 \times 35$

$=770 \mathrm{m}^{3}$

Length of platform= 22m,

Breadth = 14m

Volume of platform = $770 \mathrm{~m} 3$

Volume = lbh 

770= $22 \times 14 \times h$

$\frac{770}{22 \times 14}=h$

$\frac{770}{308}=h$

$2-5=h$

$h=2.5 \mathrm{~m}$

Hence , the height of the platform is 2.5m

Question 24

Ans: Given,

Height of cylindrical barrel of pen = 7cm

Radius = $\frac{5}{2} \cdot mm =$ $\frac{5}{2} \times \frac{1}{10} \mathrm{~cm}$=1\4cm

Volume of ink in it =  $\pi r^{2} h$

$-\frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 7$

$=\frac{11}{8} \mathrm{~cm}^{3}$

Volume of link in bottle =  $\frac{1}{5}l=\frac{1}{5} \times 1000 \mathrm{~cm}=200 \mathrm{cm}^{3}$

$\therefore$ Total number of barrels $=200 \div \frac{11}{8}$

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

$=\frac{1600}{11}$

Word written in one barrel = 310 words 

Total number of words = $\frac{1600}{11} \times 320$

$=\frac{496000}{11 .}$

$=45080.90 .$

$=45090$ words 

Question 25

 

Ans: Given , 

Length of a rectangular box= 40cm,

Breadth = 30cm and 

Height = 25cm

Volume = lbh 

$=40 \times 30 \times 25$

$=1200 \times 25$

$=30000 \mathrm{~cm}^{3}$

 Radius of cylindrical tin= 17.5cm

Volume = $30000 \mathrm{~cm}^{3}$

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

$30000=3-14 \times 17.5 \times 17.5 \times h $

$30000=3-14 \times 306.25 \dot{x h}$

$30000=961.625h $

$\frac{30000000}{961625}=h$

$31 \cdot 2$

$h=31-2 \mathrm{~cm}$

Hence the height of the cylindrical tin is 31.2cm

Question 26

 

Ans: Given, 

Diameter of a circular tank = 17.5m

So, the radius = $\frac{175}{20}=\frac{35}{4}=$ 8.75m

Outer radius = 8.75+4

= 12.75m

Height= 2m 

Volume of the embankment = $\pi R^{2} h-\pi r^{2} h$

$\begin{aligned} & \pi h\left(R^{2}-r^{2}\right) \\=& \frac{22}{7} \times 2\left((12-75)^{2}-(8.75)^{2}\right) \\=& \frac{44}{7}(162.5625-76.5625) \\=& \frac{44}{7} \times 86 . \\=& \frac{3784}{7} \\ 540.57 \mathrm{~m}^{3} . \end{aligned}$

 

∴ Volume of earth of the tank = $540.57 \mathrm{~m}^{3}$

$\begin{aligned} \therefore \text { Volyme } &=\pi r^{2} h \\ 540.57 &=\frac{22}{7} \times 8.75 \times 8.75 \times h . \end{aligned}$

$540.57=\frac{1684.375 \mathrm{~h}}{7}$

$2.25=h$

$h=2.25 \mathrm{~m} .$

Hence the depth of the circular tank is 2.25m

Question 27

 

Ans: Given, 

Speed of water = $7 m=700 \mathrm{~cm}$

Internal radius= $\frac{2}{2} c m=1 c m$

Radius of tank = 40cn

Time =$\frac{1}{2}$ hour $=$ $\frac{1}{2} \times 60 \times 60=\frac{3600}{2} \mathrm{sec}=$

=1800sec 

∴ Distance  (h) = Times into speed 

$=1800 \times 700$

$=1260000 \mathrm{~cm}$

Volume $=\pi r^{2} \mathrm{~h}$. 

=$\frac{22}{7}1 \times1 \times \times 1260000$

$=3960000 \mathrm{~cm}^{3}$

∴ Volume of water in the tank =$3960000 \mathrm{cm}^{3} $

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

$3960000=\frac{22}{7} \times 40 \times 40 \times h$.

$\frac{17325}{22}=h$

$787.5=h$

$h=787.5 \mathrm{cm} $

Question 28

 

Ans: Given, 

Radius of pipe = $\frac{7}{2} cm=\frac{7}{200}$ m

Speed = 36 km\hr = $36 \times \frac{1000}{60}$ = 600m

Radius of tank = $35 \mathrm{~cm}=\frac{35}{100} \mathrm{~m}$

Height = 1m

∴ Volume of tank =π$r^{2}h$

$\frac{22}{7} \times \frac{35}{100} \times \frac{35}{100} \times 1$

$=\frac{77}{200} \mathrm{~m}^{3}$

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

$\frac{77}{200}=\frac{22}{7} \times \frac{7}{200} \times \frac{7}{200} \times h .$

$\frac{2200}{221}=h$

$100=h .$

$h=100$

Let height (h) be the distance 

Distance = speed into Time

100= $600 \times Time$

Time $=\frac{100}{600}$

Time $=0.167 \mathrm{~min} $

Question 29

 

Ans: Given, 

Radius of coin =$\frac{1.5}{20} \mathrm{~cm}=\frac{3}{4} \mathrm{~cm}$

and its thickness = 0.2 cm

Volume of one coin = $\pi r^{2} h$

$=\frac{22}{7} \times \frac{3}{4} \times \frac{3}{4} \times 0.2 .$

$=\frac{39.6}{112}$

$=0.35 \mathrm{~cm}^{3}$

Radius of cylinder = $\frac{4.5}{20}=\frac{9}{4} \mathrm{~cm}$

Height = 10cm

Volume of cylinder = $\pi r^{2} h$

$\begin{aligned} &=\frac{22}{7} \times \frac{9}{4} \times \frac{9}{4} \times 10 \\=& \frac{17820}{112} \\=& 159.11 \mathrm{~cm}^{3} . \end{aligned}$

Volume of cylinder = volume of one coin X Number of coin

159.11 = $0.35 \times$ Number of coin.

$\frac{159 \cdot 11}{0.35}$ =Number of coin.

$454.6=$ Number of coin

Number of coin $=454.6$

Question 30

 

Ans: Given

Weight of  $1cm^{3}=21 g$

Length of pipe = $1 \mathrm{~m}=100 \mathrm{~cm}$

Internal radius = $\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm}$

External radius = 1.5+1= 2.5cm

∴ Volume = Volume of outer - Volume of inner surface 

$π R^{2} h$-$π r^{2} h$

$=\pi h\left(R^{2}-r^{2}\right)$

$=\frac{22}{7} \times 100\left((2.5)^{2}-(1.5)^{2}\right)$

$=\frac{2200}{7} \times(6.25-2.25) .$

$=\frac{2200}{7} \times 4 $

$=\frac{8800}{7} \mathrm{~cm}^{3} $

∴ Total Weight of metal = $\frac{8800}{7} \times {21}$

=26400 g