Question 12 Marks
A cylindrical bucket $28\ cm$ in diameter and $72\ cm$ high is full of water. The water is emptied into a rectangular tank $66\ cm$ long and $28\ cm$ wide. Find the height of the water level in the tank.
AnswerLet $h$ be the height of rectangular tank = volume of cylindrical bucket
$66\times28\times\text{h}=\pi\Big(\frac{28}{2}\Big)^2\times72$
$66\times28\times\text{h}=\frac{22}{7}\times14\times14\times72$
$\text{h}=\frac{22\times2\times14\times72}{66\times28}$
$\text{h}=24\text{cm}$
Hence, the height of rectangular tank is $24\ cm$.
View full question & answer→Question 22 Marks
A rectangular tank $15m$ long and $11m$ broad is required to receive entire liquid contents from a fully cylindrical tank of internal diameter $21m$ and length $5m$. Find the least height of the tank that will serve the purpose.
AnswerSuppose height of the rectangular tank is equal to $h$.
Length of the tank $= 15m$
Breadth of the tank $= 11m$
Further,
length of cylindrical tank $= 5m$
Radius of cylindrical tank $=\frac{21}{2}\text{m}$
To find out the least height of the tank, equate the volumes of two tanks.
$15\times11\times\text{h}\times\pi\Big(\frac{21}{2}\Big)^2\times5$
$\Rightarrow\text{h}=\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{5}{15}\times\frac{1}{11}$
$\Rightarrow\text{h}=\frac{21}{2}$
$\Rightarrow\text{h}=10.5$.
Hence, the least height of the tank is equal to $10.5$.
View full question & answer→Question 32 Marks
A spherical ball of iron has been melted and made into smaller balls. If the radius of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made?
AnswerLet the radius of larger ball = $r$
Then volume $=\frac{4}{3}\pi\text{r}^3$
Now radius of one of the smaller balls $=\frac{\text{r}}{4}$
$\therefore$ Volume of each smaller balls $=\frac{4}{3}\pi\Big(\frac{\text{r}}{4}\Big)^3$
$=\frac{4}{3}\pi\frac{\text{r}^3}{64}=\frac{1}{64}\Big(\frac{4}{3}\pi\text{r}^3\Big)$
$\therefore$ Number of balls can be made
$=\frac{4}{3}\pi\text{r}^3\div\frac{1}{64}\Big(\frac{4}{3}\pi\text{r}^3\Big)$
$=1\div\frac{1}{64}=1\times\frac{64}{1}=64$
View full question & answer→Question 42 Marks
$25$ circular plates, each of radius $10.5\ cm$ and thickness $1.6\ cm$, are placed one above the other to form a solid circular cylinder. Find the curved surface area and
the volume of the cylinder so formed.
AnswerWe have $25$ circular plates, each with radius $= 10.5\ cm$ and thickness $= 1.6\ cm.$
These plates are stacked on top of one another.
So, the total height of the arrangement becomes $= 1.6 \times 25 = 40\ cm$
Volume of this arrangement $=\pi\text{r}^2\text{h}=\pi(10.5)^2\times40=13860\text{cm} ^2$
Curved surface area $=2\pi\text{rh}=2\pi\times10.5\times40=2640\text{cm}^2$
Hence, Volume $= 13860\ cm^3$ and $C.S.A$ $= 2640\ cm^2$
View full question & answer→Question 52 Marks
A solid metallic sphere of radius $5.6\ cm$ is melted and solid cones each of radius $2.8\ cm$ and height $3.2\ cm$ are made. Find the number of such cones formed.
AnswerLet the number of such cones formed be $n$
Now, volume of solid metallic sphere = volume of n solid cones
$\Rightarrow\frac{4}{3}\times\frac{22}{7}\times(5.6)^3=\text{n}\times\frac{1}{3}\times\frac{22}{7}\times(2.8)^2\times3.2$
$\Rightarrow4\times(5.6)^3=\text{n}\times(2.8)^2\times3.2$
$\Rightarrow\text{n}=28$
View full question & answer→Question 62 Marks
A rectangular vessel of dimensions $20\ cm \times 16\ cm \times 11\ cm$ is full of water. This water is poured into a conical vessel. The top of the conical vessel has its radius $10\ cm$. If the conical vessel is filled completely, determine its height.
AnswerVolume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\pi\text{r}^2\text{h}=20\times16\times11$
Height of cone (h) $=\frac{3\times20\times16\times11\times7}{22\times100}$
$=33.6\text{cm}$
View full question & answer→Question 72 Marks
In the given figure, from a cuboidal solid metalic block, of dimensions $15\ cm \times 10\ cm \times 5\ cm$, a cylindrical hole of diameter $7\ cm$ is drilled out. Find the surface area of the remaining block. $\Big(\text{use}\ \pi=\frac{22}{7}\Big)$
AnswerSurface area of the remaining block
= Total Surface area of cubic block + Curved Surface area of cylinder − $2$(Area of circular base)
$=2(\text{lb}+\text{bh}+\text{lh})+2\pi\text{rh}-2\pi\text{r}^2$
$=2(15\times10\times+10\times5+15\times5)\\+2\times\frac{22}{7}\times\frac{7}{2}\times5-2\times\frac{22}{7}\times\Big(\frac{7}{2}\Big)^2$
$=2\times275+110-77$
$=583\text{cm}^2$
View full question & answer→Question 82 Marks
The largest sphere is to be curved out of a right circular cylinder of radius $7\ cm$ and height $14\ cm$. Find the volume of the sphere.
AnswerRadius of cylinder $(r) = 7\ cm$
and height $(h) = 14\ cm$
The diameter of the largest sphere curved out of the given cylinder = diameter of the cylinder
$= 2 \times 7 = 14\ cm$
$= 2 \times 7 = 14\ cm$
$\therefore$ Radius $= 7\ cm$
$\therefore$ Volume $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times7\times7\times7\text{cm}^3$
$=1437.33=1437\text{cm}^3$ View full question & answer→Question 92 Marks
A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14\ cm$ and the total height of the vessel is $13\ cm$. Find the inner surface area of the vessel.
AnswerWe have to find the inner surface area of a vessel which is in the form of a hemisphere mounted by a hollow cylinder.
Radius of hemisphere and cylinder $(r) = 7\ cm$
Total height of vessel $(r + h) = 13\ cm$
So, the inner surface area of a vessel,
$=2\pi\text{r}(\text{r}+\text{h})$
$=2\times\frac{22}{7}\times7\times13\text{ cm}^2$
$=572\text{cm}^2$
View full question & answer→Question 102 Marks
An wooden toy is made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is $10\ cm$, and its base is of radius $3.5\ cm$, find the volume of wood in the toy.$\Big(\text{use}\ \pi=\frac{22}{7}\Big)$
AnswerVolume of wood in the toy = Volume of cylinder - $2$(Volume of hemisphere)
$=\pi\text{r}^2\text{h}-2\times\frac{2}{3}\pi\text{r}^3$
$=\frac{22}{7}\times(3.5)^2\times10-2\times\frac{2}{3}\times(3.5)^3$
$=385-179.67$
$=205.33\text{cm}^3$
View full question & answer→Question 112 Marks
The radii of the ends of a bucket of height $24\ cm$ are $15\ cm$ and $5\ cm$. Find its capacity. $(\text{Take}\ \pi=\frac{22}{7})$
AnswerHeight of a bucket $= 24cm$
$R = 15\ cm$
$r = 5\ cm$
Therefore,
Capacity of the bucket
$=\frac{\pi\text{h}}{3}[\text{h}^2+\text{Rr}+\text{r}^2]$
$=\frac{22}{7}\times\frac{24}{3}\times\Big[(15)^2+15\times5+(5)^2\Big]$
$=8171.42\text{cm}^3$
View full question & answer→Question 122 Marks
The largest possible sphere is carved out of a wooden solid cube of side $7\ cm$. Find the volume of the wood left.$(\text{use}\ \pi=\frac{22}{7})$
AnswerThe radius of the largest possible sphere is carved out of a wooden solid cube is equal to the half of the side of the cube.
Radius of sphere $=\frac{7}{2}=3.5$
Volume of the wood left = Volume of cube - Volume of sphere
$=(\text{side})^3-\frac{4}{3}\pi\text{r}^3$
$=7^3-\frac{4}{3}\times\frac{22}{7}\times(3.5)^3$
$=343-179.67$
$=163.33\text{cm}^3$
View full question & answer→Question 132 Marks
The largest cone is curved out from one face of solid cube of side $21\ cm$. Find the volume of the remaining solid.
AnswerThe radius of the largest possible cone is carved out of a solid cube is equal to the half of the side of the cube.
Also, the height of the cone is equal to the side of the cube.
Radius of the cone $=\frac{21}{2}=10.5\text{cm}$
Volume of the remaining solid = Volume of cube - Volume of cone
$=(\text{Side})^3-\frac{1}{3}\pi\text{r}^2\text{h}$
$=(21)^3-\frac{1}{3}\times\frac{22}{7}\times(10.5)^2\times21$
$=9261-2425.5$
$=6835.5\text{cm}^3$
View full question & answer→Question 142 Marks
Find the number of coins, $1.5\ cm$ is diameter and $0.2\ cm$ thick, to be melted to form a right circular cylinder of height $10\ cm$ and diameter $4.5\ cm$.
AnswerVolume of one coin $=\pi\times(1.5)^2\times\frac{0.2}{4}$
Volume of cylinder $=\pi\times(4.5)^2\times\frac{10}{4}$
So number of coins to be melted
$=\frac{\text{volume of cylinder}}{\text{volume of each coin}}$
$=\Big(\frac{4.5}{1.5}\Big)^2\times\frac{10}{0.2}$
$=9\times50$
$=450$
View full question & answer→Question 152 Marks
A bucket, made of metal sheet, is in the form of a cone whose height is $35\ cm$ and radii of circular ends are $30\ cm$ and $12\ cm$. How many litres of milk it contains if it is full to the brim? If the milk is sold at $Rs\ 40$ per litre, find the amount received by the person.
AnswerRadii of the bucket in the form of frustum of cone $= 30\ cm$
and 12cm Depth of the bucket $= 35\ cm$
capacity of the bucket $=\frac{1}{3}\pi\text{h}(\text{r}_1^2+\text{r}^2_2+\text{r}_1\text{r}_2)$
$=\frac{1}{3}\times\frac{22}{7}\times35\big[(30)^2+(12)^2+(30\times12)\big]$
$=\frac{110}{3}[900+144+360]$
$=\frac{110}{3}\times1404=51480\text{cm}^2=51.48\ \text{liters}$
Price of milk per liter $= ₹ 40$
Total amount recieved by milkman $= ₹ 41 \times 51.48 = 2059.20$
View full question & answer→Question 162 Marks
$150$ spherical marbles, each of diameter $1.4\ cm$ are dropped in a cylindrical vessel of diameter $7\ cm$ containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.
AnswerLet the rise in the level of water in the vessel be $h\ cm$.
Now, volume of $150$ spherical maeble = volume of water displace in vessel
$\Rightarrow150\times\frac{4}{3}\times\frac{22}{7}\times(\frac{1.4}{2})^3=\frac{22}{7}\times(\frac{7}{2})^2\times\text{h}$
$\Rightarrow200\times(0.7)^3=(\frac{7}{2})^2\times\text{h}$
$\Rightarrow\text{h}=5.6\text{cm}$
View full question & answer→Question 172 Marks
Find the depth of a cylindrical tank of radius $28m$, if its capacity is equal to that of a rectangular tank of size $28m \times 16m \times 11m.$
AnswerLet $x$ be the depth of cylindrical tank.
The radius of tank $r = 28m$.
Since,
The volume of cylindrical tank = volume of rectangular tank
$\pi\text{r}^2\text{x}=28\times16\times11$
$\Rightarrow\frac{22}{7}\times28\times28\times\text{x}=28\times16\times11$
$\text{x}=\frac{28\times16\times11\times7}{22\times28\times28}$
$\text{x}=\frac{16}{8}$
$\text{x}=2\text{m}$
Thus, the depth of cylindrical tank $= 2m$.
View full question & answer→Question 182 Marks
The radii of the ends of a frustum of a right circular cone are $5m$ and $8m$ and its lateral height is $5m$. Find the lateral surface and volume of the frustum.
AnswerLateral surface area of frustum
$=\pi(\text{r}+\text{R})\ \text{l}$
$=\pi(5+8)\times5$
$=204.28\text{m}^2$
Height of cone
$\text{h}=\sqrt{5^2=(\text{R}-\text{r}^2)}$
$=\sqrt{5^2-(8-5)^2}$
$=\sqrt{13}=4\text{cm}$
Volume $=\frac{\pi\times4}{3}(8^2+5^2+40)=540.56\text{cm}^3$
View full question & answer→