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 $=\pi r^{2} h$
$4000=\frac{22}{7} \times 5 \times 5 \times h$
$\frac{4000 \times 7}{22 \times 25}=h$
$\frac{80 \times 7}{11}$
$\frac{560}{11}=h$
$h=\frac{510}{11} \mathrm{~cm}$
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 $
$=132 \mathrm{cm}^{3}$
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, $h+r=37$
$h+7=37$.
$h=377$
$h=30 \mathrm{~cm}$.
$\begin{aligned} \text { Volume } &=\pi r^{2} h \\ &=\frac{22}{7} \times 7 \times 7 \times 30 \\ &=22 \times 210 \\ &=4620 \mathrm{~cm}^{3} \end{aligned}$
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 = $2 \pi r h$
$=2 \times \frac{22}{7} \times 14 \times 10$
$=44 \times 20$
$=880 \mathrm{m}^{2}$
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= $=\frac{2}\times \frac{22}{7} \times \frac{35}{2} \times h$
$\frac{4400}{22 \times 5}=h$
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) $\begin{aligned} \text { Volume } &=\pi r^{2}h \\ &=3.14 \times 10 \times 10 \times 15-9 . \\ &=31.4 \times 159 . \\ &=4992.6 \mathrm{~cm}^{3}\end{aligned}$
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
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
R= 22m
r= 20
h=$\frac{7}{100} m$
$\pi R^{2} h-\pi{r}^{2} h$
$=\pi h\left(R^{2}-r^{2}\right)$
$=\frac{22}{7} \times \frac{7}{100}(R+r)(R-r)$
$=\frac{22}{7}\times{7}{100}(22) \times 18=18.48 \mathrm{~mm}$
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} $
$=\frac{1100}{100}$
$=11 \mathrm{~m}$
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
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} $
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 = $\pi R^{2} h - r^{2} h$
$=\pi h\left(R^{2}-r^{2}\right)$
$\left.=\frac{22}{7}\times 21 \times(6)^{2}-(5-6)^{2}\right)$
$=66 \times(36-31.36)$
$=66 \times 464$
$=306.24 \mathrm{cm}^{3}$
$=306.2 \mathrm{~cm}^{3}$
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
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
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
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,
$=462-269-5$
$=192.5 \mathrm{~cm}^{3}$
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
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
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
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
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} $
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 $=\pi r^{2} h .$
$\frac{77}{200}=\frac{22}{7} \times \frac{7}{200} \times \frac{7}{200} \times h .$
$\frac{2200}{221}=h$
Let height (h) be the distance
Distance = speed into Time
100= $600 \times Time$
Time $=\frac{100}{600}$
Time $=0.167 \mathrm{~min} $
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 $=454.6$
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
