Question 12 Marks
The cylinder of radius $12 cm$ have filled the $20 cm$ with water. One piece of iron drop in the stands of water goes up $6.75 cm$. Find the radius of sphere piece.
AnswerRadius of cylinder = 12 cm
Height of cylinder = 6.75 cm
Volume of water $= \pi r^2h = \pi \times 12 \times 12 \times 6.75 cm^3$
Let the radius of iron sphere piece = R cm.
∵ Volume of sphere = volume of water
$\frac{4}{3} \pi R^3=\pi \times 12 \times 12 \times 6.75$
$R^3=\frac{\pi \times 12 \times 12 \times 6.75 \times 3}{4 \pi}$
$R^3 = 729$
$R =\sqrt[3]{729}=9 cm$
Hence radius of sphere piece = 9 cm.
View full question & answer→Question 22 Marks
Water flows at the rate of $10 m$ per minute through a cylindrical pipe $5 mm$ of diameter. How much time would it take to fill a conical vessel whose diameter at he surface is $40 cm$ and depth is $24 cm$?
AnswerVolume that flows in $1 min = [ \pi \times (0.25)2 \times 1000 ] cm^3$
Volume of the conical vessel $=\frac{1}{3} \pi \times(20)^2 \times 24 cm ^3$
Required time $=\frac{\frac{1}{3} \pi \times(20)^2 \times 24}{\pi \times(0.25)^2 \times 1000}$
= 51 min 12 sec. View full question & answer→Question 32 Marks
A metal container in the form of a cylinder is surmounted by a hemisphere of the same radius. The internal height of the cylinder is $7\ m$ and the internal radius is $3.\ m.$ Calculate: the internal volume of the container in $m^3$.
Answer
Radius of the cylinder $= 3.5\ m$
Height $= 7\ m$
Volume of the container $=\pi r ^2 h +\frac{2}{3} \pi r ^3$
$=\left(\frac{22}{7} \times 3.5 \times 3.5 \times 7\right)+\left(\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5\right) $
$ =\frac{539}{2}+\frac{539}{6}$
$ =\frac{1617+539}{6}$
$ =\frac{2156}{6}$
$= 359.33\ m^3$ View full question & answer→Question 42 Marks
The ratio of the base area and the curved surface of a conical tent is 40: 41. If the height is 18 m, Find the air capacity of the tent in terms of n.
AnswerGiven: $\frac{\text { base area }}{\text { curved surface }}=\frac{40}{41}$
$\Rightarrow \frac{\pi r^2}{\pi r \sqrt{h^2+r^2}}=\frac{40}{41}$ ...( Where h is the height and r is the radius of conical tent)
$\Rightarrow \frac{r}{\sqrt{18^2+r^2}}=\frac{40}{41}$ ....( ∵ h = 18 m)
⇒ r = 80 m
∴ Air capacity = $\frac{1}{3} \pi(80)^2 \times 18$
Air capacity = 38,400 π cu m.
View full question & answer→Question 52 Marks
The radius and height of a cylinder are in the ratio of $5: 7$ and its volume is 550 cm . Find its radius. (Take $\pi=22 / 7$ )
AnswerLet the radius of the base and height of the cylinder be $5x \ cm$ and $7x \ cm$ respectively.
Then,
Volume $= 550 cm^3$
$\Rightarrow \frac{22}{7} \times(5 x)^2 \times 7 x=550 \ldots$ ( Use,$ r = 5x, h = 7x$ and volume $= \pi r^2h )$
$\Rightarrow \frac{22}{7} \times 25 x^2 \times 7 x=550$
$\Rightarrow 22 x 25x^3 = 550$
$\Rightarrow 550x^3 = 550$
$\Rightarrow x^3 = 1$
$\Rightarrow x = 1$
Hence, radius of the cyclinder $= 5x cm = ( 5 x 1 ) cm = 5 cm.$
View full question & answer→Question 62 Marks
The diameter of a garden roller is $1.4\ m$ and it is $2\ m$ long. How much area will it cover in $5$ revolutions? (Take $\pi = 22/7$)
AnswerClearly, Area covered = Curved surface area x No. of revolutions
Here, r $=\frac{1.4}{2}$ m = 0.7 m and h = 2 m
∴ Curved surface area = 2πrh
$=2 \times \frac{22}{7} \times 0.7 \times 2 m ^2$
$=8.8 m ^2$
Here, Area covered = $8.8 m^2 x 5 = 44 m^2$.
View full question & answer→Question 72 Marks
Water is being pumped out through a circular pipe whose external diameter is 7 cm. If the flow of water is 72 cm per second how many litres of water are being pumped out in one hour.
AnswerVolume of water that will be pumped out in 1 second
$=\pi\left(\frac{7}{2}\right)^2(72)$ cu cm
∴ Volume of water that will be pumped out in one hour
$=\pi\left(\frac{7}{2}\right)^2(72)(3600)$ cu cm
= 99,79,200 cu cm
= 9979.2 litres.
View full question & answer→Question 82 Marks
A glass cylinder with a diameter $20 \ cm$ water to a height of $9 \ cm$ . A metal cube of $8 \ cm$ edge is immersed in it completely. Calculate the height by which water will rise in the cylinder. (Take $\pi=3.142$ )
AnswerSuppose the water rises by $h cm.$
Clearly water in the cylinder forms a cylinder of height h cm and radius $10 cm.$
$\therefore$ The volume of the water displaced = Volume of the cube of edge $8 cm.$
$\Rightarrow \pi r^2h = 8^3$
$\Rightarrow 3.142 x 10^2 x h = 8 x 8 x 8$
⇒$\frac{8 \times 8 \times 8}{3.142 \times 10 \times 10}$ $cm = h$
$\Rightarrow h = 1.6 cm.$
View full question & answer→Question 92 Marks
A well $28.8 \ m$ deep and of diameter $2 \ m$ is dug up. The soil dug out is spread all around the well to make a platform $1 \ m$ high considering the fact losse soil settled to a height in the ratio $6 : 5$ find the width of the platform.
AnswerVolume of Soil drug out $= \pi (1)^2(28.8) cu m.$
$\therefore$ Volume of soil that would be settled on the platform is $\frac{5 \times \pi \times 28.8}{6}$ cu. m.
Let the width of the platform be r cm. Then,
$\pi (1 + r^2) - (1)^2 (1) = \frac{5 \times \pi \times 28.8}{6}$
$\Rightarrow (1 + r)^2 - 1 = \frac{5 \times 28.8}{6}= 24$
$\Rightarrow (1 + r)^2= 25$
$\Rightarrow 1 + r = 5$
$\Rightarrow r = 5 - 1$
$\Rightarrow r = 4 cm.$
View full question & answer→Question 102 Marks
A vessel is in he form of an inverted cone. Its height is $11\ cm.,$ and the radius of its top which is open is $2.5\ cm.$ It is filled with water up to the rim. When lead shots, each of which is a sphere of radius $0.25\ cm.,$ are dropped $2$ into the vessel,$\frac{2}{5}$th of the water flows out. Find the number of lead shots dropped into the vessel.
AnswerVolume of $n$ lead shots $=$ volume of water displaced.
$n \times \frac{4}{3} \pi r^3=\frac{2}{5} \times \frac{1}{3} \pi R^2 H$
$\therefore n=\frac{\frac{2}{5} \times \frac{1}{3} \pi R^2 H}{\frac{4}{3} \pi r^3}$
$\therefore n=\frac{2 R^2 H}{5 \times 4 r^3}$
$\therefore n=\frac{2 \times 2.5^2 \times 11}{5 \times 4 \times(0.25)^3}$
$\therefore n = 440$ shots.
View full question & answer→Question 112 Marks
A road roller is cylindrical in shape, its circular end has a diameter of $1.4 m$ and its width is $4 m$. It is used to level a play ground measuring $70 m \times 40 m$. Find the minimum number of complete revolutions that the roller must take in order to cover the entire ground once.
AnswerCurved surface area of the road roller = 2πrh
$=2 \times \frac{22}{7} \times 0.7 \times 4=17.6 cm ^2$
Area of the play ground $= 70 \times 40 = 2800 m^2$
∴ Number of revolutions to cover the entire ground = $\frac{2800}{17.6}=159 \frac{1}{11}$
∴ Number of complete revolutions = 160.
View full question & answer→Question 122 Marks
A glass cylinder with a diameter $20 \ cm$ has water to a height of $9 \ cm$. A metal cube of 8 cm edge is immersed in it completely. Calculate the height by which the water will up in the cylinder. Answer correct of the nearest mm. (Take $\pi = 3.142)$
AnswerLet the height by which the water ups be $= h cm$
Volume of the increase in water $= 3.142 \times 10 \times 10 \times h cm^3$
Volume of the cube $= 8 \times 8 \times 8 cm^3$
Both the above volumes are equal
$\therefore 3.142 \times 10 \times 10 \times h = 8 \times 8 \times 8$
$h =\frac{8 \times 8 \times 8}{3.142 \times 10 \times 10}=1.6 cm$
The height $h = 16 mm.$
View full question & answer→Question 132 Marks
Find the weight of a lead pipe $35 \ cm$ long. The external diameter of the pipe is $2.4 \ cm$ and the thickness of the pipe is $2mm,$ given $1 \ cm^3$ of lead weighs 10 gm.
AnswerExternal radius $'R' = 1.2 cm$
Internal radius $'r' = 1.0 cm ...$(Since internal radius = external radius - thickness)
Height $'h' = 35 cm$
$\therefore$ Volume of the pipe $= \pi h[ R^2- r^2]$
$=\frac{22}{7} \times 35 \times\left[(1.2)^2-(1.0)^2\right]$
$= 48.4 cm^3$
$\therefore $ Weight of lead pie $= 10 \times 48.4 = 484 gm.$
View full question & answer→Question 142 Marks
The total surface area of a hollow metal cylinder, open at both ends of external radius $8 \ cm$ and height 10 cm is $338\pi cm^2.$ Taking r to be inner radius, write down an equation in r and use it to state the thickness of the metal in the cylinder.
AnswerSurface area of the hollow cylinder $= 338\pi cm^2$
$2\pi (8)10 + 2\pi (r)10 + 2[ \pi (8)^2 - \pi (r)^2] = 338 \pi$
$\Rightarrow 160 + 20r + 2(64 - r^2) = 338$
$\Rightarrow - 2r^2 + 20r - 50 = 0$
$\Rightarrow r^2 - 10r + 25 = 0$
$\Rightarrow (r - 5)^2 = 0$
$\Rightarrow r = 5$
$\therefore$ Thickness of the metal in the cylinder $= 8 - r = 8 - 5 = 3 cm.$
View full question & answer→Question 152 Marks
A hemispherical bowl of diameter $7.2\ cm$ is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius $4.8\ cm.$ Find the height of the cone.
AnswerA volume of hemispherical bowl $=\frac{2}{3} \pi r^3=\frac{2}{3} \pi(3.6)^3 cm ^3$
Volume of cone $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi \times(4.8)^2 \times h= cm ^3$
But Volume of bowl $=$ Volumw of cone
$=\frac{2}{3} \pi \times(3.6)^3=\frac{1}{3} \pi \times(4.8)^2 \times h$
$h=\frac{2 \times 3.6 \times 3.6 \times 3.6}{4.8 \times 4.8}=4.05\ cm$
View full question & answer→Question 162 Marks
A sphere cut out from a side of 7 cm cubes. Find the volume of this sphere?
Answer∵ Diameter of sphere equal to sides of cube.
∴ Radius of sphere = $=\frac{7}{2} cm$
Volume of sphere $=\frac{4}{3} \pi r^3$
Volume of sphere $=\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$
Volume of sphere $=\frac{539}{3}=179.66 cm ^3$.
View full question & answer→Question 172 Marks
If the perimeter of a semi-circular protractor is 36 cm. Find its diameter.
AnswerLet r cm be the radius of the protractor, then
Perimeter = πr + 2r
According to the question,
⇒ 36 = πr + 2r
⇒ 36 = r $\left(\frac{22}{7}+2\right)$
⇒ r = 7 cm.
∴ The diameter of the protractor is ( 2 x 7) cm = 14 cm.
View full question & answer→Question 182 Marks
A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
AnswerLet, the height of cone $= h$ and height of hemisphere $= H$
$\therefore$ Volume of cone $=$ Volume of hemisphere
$\frac{1}{3} \pi r^2 h=\frac{2}{3} \pi r^3$
$\frac{1}{3} \pi r^2 h=\frac{2}{3} \pi r^2 H \quad \ldots(\because H = r )$
$\frac{h}{H}=\frac{2 \pi r^2 \times 3}{3 \pi r^2}$
$\frac{h}{H}=\frac{2}{1}$
$h: H = 2: 1.$
View full question & answer→Question 192 Marks
A spherical cannon ball, $28\ cm$ in diameter is melted and recast into a right circular conical mould, the base of which is $35\ cm$ in diameter. Find the height of the cone, correct to one place of decimal.
AnswerLet h be the height of conee. Then
$\frac{1}{3} \pi\left(\frac{35}{2}\right)^2 h=\frac{4}{3} \pi(14)^3$ ....(Because the volume of conical mould is the same as that of the spherical cannon ball.)
$\Rightarrow h =\frac{4 \times 14 \times 14 \times 14 \times 2 \times 2}{35 \times 35}$
$ \Rightarrow h =\frac{896}{25}$
$ \Rightarrow h =35.84\ cm .$
View full question & answer→Question 202 Marks
There is a ratio $1: 4$ between the surface area of two spheres, find the ratio between their radius.
AnswerLet radius of spheres are $r_1$ and $r_2$.
So, surface area of spheres is $4\pi r_1{}^2$ and $4\pi r_2{}^2$.
Ratio of surface area = $\frac{4 \pi r_1^2}{4 \pi r_2^2}=\frac{1}{4}$
$=\frac{r_1^2}{r_2^2}=\frac{1}{4}$
$=\left(\frac{r_1}{r_2}\right)^2=\frac{1}{4}$
$=\frac{r_1}{r_2}=\sqrt{\frac{1}{4}}$
$=\frac{r_1}{r_2}=\frac{1}{2}$
Hence, the ratio between their radius $= 1: 2.$
View full question & answer→Question 212 Marks
There is surface area and volume of a sphere equal, find the radius of sphere.
AnswerLet the radius of sphere $= r$
$\because$ Volume of sphere = Surface area of sphere
$\frac{4}{3} \pi r^3=4 \pi r^2$
$\Rightarrow r^3=\frac{4 \pi r^2 \times 3}{4} \pi$
$\Rightarrow r^3 = 3r^2$
$\Rightarrow r = 3\ cm$
Hence, the radius of sphere $= 3\ cm.$
View full question & answer→Question 222 Marks
The volume of a sphere is $905 1/7 cm^3$, find its diameter.
AnswerVolume of a sphere $=905 \frac{1}{7}$
$\frac{4}{3} \pi r^3=\frac{6336}{7}$
$r^3=\frac{6336 \times 7 \times 3}{4 \times 22 \times 7}$
$r^3=216$
$r=\sqrt[3]{216}$
$r=6 cm$
$\therefore$ Diameter of a sphere $= 2r = 2 \times 6 = 12\ cm.$
View full question & answer→Question 232 Marks
Find the volume and surface area of a sphere of diameter $21 cm$.
AnswerDiameter of sphere = 21 cm
∴ Radius of sphere = $\frac{21}{2} cm$
∴ Surface area of sphere = $4\pi r^2$
$=4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$
$= 1386 cm^2$
Volume of sphere = $\frac{4}{3} \pi r^3$
$=4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$
$= 4851 cm^3$.
View full question & answer→Question 242 Marks
The radius and height of cone are in the ratio $3 : 4$. If its volume is $301.44 cm^3$. What is its radius? What is its slant height? (Take $\pi = 3.14$)
AnswerLet the radius of cone be 3x cm and the height 4x cm, then
Volume of cone = $\frac{1}{3} \pi r^2 h$
$\Rightarrow \frac{1}{3} \pi(3 x)^2(4 x)=301.44$
$\Rightarrow x^3 = 8$
$\Rightarrow x = 2$
Thus, radius of cone is 6 cm and height 8 cm.
Now,
slant height of cone = $=\sqrt{(6)^2+(8)^2}=10 cm$.
View full question & answer→Question 252 Marks
The diameter of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surface areas.
AnswerLet radius of each cone = r
Ratio between their slant heights = 5:4
Let slant height of the first cone = 5x
and slant height of second cone = 4x
Therefore, curved surface area of the first cone $=\pi r \times(5 x)=5 \pi r x$
curved surface area of the second cone $=\pi r l=\pi r \times(4 x)=4 \pi r x$
Hence, ratio between them $=5 \pi r x: 4 \pi r x=5: 4$
View full question & answer→Question 262 Marks
Find the area of a circle whose circumference is 22 cm.
AnswerLet r be the radius of the circle, then circumference = 2πr
⇒ 2πr = 22 cm
$\Rightarrow r =\frac{22}{2 \pi}=\frac{22 \times 7}{2 \times 22}=\frac{7}{2} cm$
$\therefore$ Area of circle $=\pi r^2$
$\therefore$ Area of circle $=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} cm ^2=\frac{77}{2} cm ^2$
$\therefore$ Area of circle $=38.5 cm ^2$.
View full question & answer→