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$r^{2} h$

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

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

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

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