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