Question 12 Marks
The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius $20\ cm$ and height 10m. How much concreate mixture would be required to build $14$ such pillars?
AnswerRadius (r) of pillar = 20cm $=\frac{20}{100}\text{m}$
Height (h) of pillar $= 10m$
$\therefore$ Volume of 1 pillar $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times\frac{20}{100}\times\frac{20}{100}\times10\Big)\text{m}^3$
$=\frac{44}{35}\text{m}^3$
$\Rightarrow Volume of concreate mixture in $14$
pillars $=14\times\frac{44}{35}=17.6\text{m}^3$
View full question & answer→Question 22 Marks
Find the length of the longest pole that can be put in a room of dimension $(10\ m \times 10\ m \times 5\ m).$
AnswerLength of the longest pole = length of the diagonal of the room
$=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}\text{m}$
$=\sqrt{10^2+10^2+5^2}\text{m}$
$=\sqrt{100+100+25}$
$=\sqrt{225}=15\text{m}$
View full question & answer→Question 32 Marks
The outer diameter of a spherical shell is $12\ cm$ and its inner diameter is $8\ cm$. Find the volume of metal contained in the shell. Also, find its outer surface area. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerOuter radius of the spherical shell $= 6cm$
Inner radius of the spherical shell $= 4\ cm$
Volume of metal contained in the shell $=\frac{4}{3}\times\frac{22}{7}(6^3-4^3)$
$=\frac{88}{21}\times(216-64)$
$=\frac{88}{21}\times152$
$=636.95\text{cm}^3$
$\therefore$ Outer surface area $=4\times\frac{22}{7}\times6\times6$
$=452.57\text{cm}^2$
View full question & answer→Question 42 Marks
The surface area of sphere is $(576\pi)\text{cm}^2.$ Find its volume.$\big(\text{Take}\ \pi=\frac{22}{7}\big).$
AnswerSurface area of the sphere $=(576\pi)\text{cm}^2$ Suppose that $r \ cm$ is the radius of the sphere. Then $4\pi\text{r}^2=576\pi$
$\Rightarrow\text{r}^2=\frac{576}{4}=144$
$\Rightarrow\text{r}=12\text{cm}$
$\therefore$ Volume of the sphere $=\frac{4}{3}\times\pi\times12\times12\times12\text{cm}^3$
$=2304\text{cm}^3$
View full question & answer→Question 52 Marks
A solid metallic cuboid of dimensions $(9\ m \times 8\ m \times 2\ m)$ is melted and recast into solid cubes of edge $2\ m$. Find the number of cubes so formed.
AnswerVolume of the solid metallic cuboid $=9 \mathrm{~m} \times 8 \mathrm{~m} \times 2 \mathrm{~m}=144 \mathrm{~m} 3$
Volume of each solid cube $=(\text { Edge })^3=(2)^3=8 \mathrm{~m}^3 \therefore$
Number of cubes formed $=\frac{\text { Volume of the solid metallic cuboid }}{\text { Volume of each solid cube }}=\frac{144}{8}=18$ Thus, the number of cubes so formed is $18.$
View full question & answer→Question 62 Marks
Find the total surface area of a cone, if its slant height is $21\ m$ and diameter of its base is $24\ m$. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerRadius of a cone, $r = 12\ cm$
Slant height of a cone, $l = 21\ cm$
Totalsurface area of a cone $=\pi\text{r}(\text{l}+\text{r})$
$=\Big[\frac{22}{7}\times12(21+12)\Big]\text{m}^2$
$=\Big(\frac{22}{7}\times12\times33\Big)\text{m}^2$
$=1244.57\text{m}^2$
View full question & answer→Question 72 Marks
A godown measures $40\ m \times 25\ m \times 15\ m$. Find the maximum number of wooden crates, each measuring $1.5\ m \times 1.25\ m \times 0.5\ m$, that can be stored in the godown.
AnswerVolume of the godown $=40 \mathrm{~m} \times 25 \mathrm{~m} \times 15 \mathrm{~m}=15000 \mathrm{~m}^3$
Volume of each wooden create $=1.5 \mathrm{~m} \times 1.25 \mathrm{~m} \times 0.5 \mathrm{~m}=$$0.9375 \mathrm{~m}^3$
$\therefore$ Maximum number of wooden creates that can be stored in the godown $=\frac{\text { Volume of the godown }}{\text { Volume of each wooden crate }}$
$=\frac{15000}{0.9375}=16000$
View full question & answer→Question 82 Marks
A cylindeical tub of radius $12\ cm$ contains water to a depth of $20\ cm$. A sphrical iron ball is dropped into the tub and thus the leval of water is raised by $6.75\ cm$. What is the radius of the ball.
AnswerSuppose that the radius of the ball is $r \ cm$.
Radius of the cylindrical tub $= 12\ cm$
Depth of the tub $= 20\ cm$
Now, volume of the ball = volume of water raised in the cylinder $\Rightarrow\frac{4}{3}\pi\text{r}^3=\pi\times12^2\times6.75$
$\Rightarrow\text{r}^3=\frac{144\times6.75\times3}{4}$
$=36\times6.5\times3=729$
$\Rightarrow\text{r}=9\text{cm}$
$\therefore$ The radius of the ball is $9\ cm.$
View full question & answer→Question 92 Marks
Two cones have their heights in the ratio $1 : 3$ and the radii of their bases in the ratio $3 : 1$. Show that their volumes are in the ratio $3 : 1.$
AnswerLet their heights be h and $3h$ And, their radii be $3r$ and $r.$
Then, $\text{V}_1=\frac{1}{3} \pi(3\text{r})^2\times\text{h}$ And, $\text{V}_2=\frac{1}{3} \pi\text{r}^2\times3\text{h}$
$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\frac{\frac{1}{3}\pi(3\text{r})^2\times\text{h}}{\frac{1}{3}\pi\text{r}^2\times3\text{h}}=\frac{3}{1}$
$\therefore\text{V}_1:\text{V}_2=3:1$
View full question & answer→Question 102 Marks
The volume of a cuboid is $1536 \mathrm{~m}^3$. Its length is $16 \ m$ , and its breadth and height are in the ratio $3: 2$. Find the breadth and height of the cuboid.
AnswerLength of the cuboid $=16 \mathrm{~m}$ Suposs that the breadth and height of the cuboid are $3 x \mathrm{~m}$ and $2 \times m ,$ respectively.
Then $1536=16 \times 3 \mathrm{x} \times 2 \mathrm{x} $
$\Rightarrow 1536=16 \times 6 \mathrm{x}^2 $
$\Rightarrow \mathrm{x}^2=\frac{1536}{96}=16 $
$\Rightarrow \mathrm{x}=\sqrt{16}=4$
$\therefore$ The breadth and height of the cuboid are $12 \ m$ and $8 \ m$, respectively.
View full question & answer→Question 112 Marks
A cuboidal water tank is $6 \ m$ long, $5 \ m$ wide and $4.5 \ m$ deep. How many litres of water can it hold? (Given, $1 \mathrm{~m}^3=1000$ litres.)
AnswerVolume of water in the tank $=$ Length $\times$ Breadth $\times$ Height $=6 \times 5 \times 4.5=135 \mathrm{~m}^3$
$\therefore$ Volume of water in litres $=135 \times 1000=135000 \mathrm{~L}\left(1 \mathrm{~m}^3=1000 \mathrm{~L}\right)$
Thus, the water tank can hold 135000 L of water.
View full question & answer→Question 122 Marks
The surface areas of two spheres are in the $1 : 4$. Find the ratio of their volumes.
AnswerSuppose that the radii of the spheres are $r$ and $R$.
We have: $\frac{4\pi\text{r}^2}{4\pi\text{R}^2}=\frac{1}{4}$
$\Rightarrow\frac{\text{r}}{\text{R}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
Now, ratio of the volumes $=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{R}^3}=\Big(\frac{\text{r}}{\text{R}}\Big)^3$
$=\Big(\frac{1}{2}\Big)^3=\frac{1}{8}$
$\therefore$ The ratio of the volumes of the sphere is $1 : 8.$
View full question & answer→Question 132 Marks
A hemispherical bowl of internal radius $9\ cm$ contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3cm and height $4\ cm$ How many bottles required to empty the bowl?
AnswerInternal radius of the hemispherical bowl $= 9\ cm$
Radius of a cylindrical shaped bottle $= 1.5\ cm$
Height of a bottle $= 4\ cm$
Number of bottles required to empty the bowl $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{hemispherical}\ \text{bowl}}{\text{Volume}\ \text{of}\ \text{a}\ \text{cylindrical}\ \text{shaped}\ \text{bottle}}$
$=\frac{\frac{2}{3}\pi\times9^3}{\pi\times1.5^2\times4}$
$=\frac{2\times9\times9\times9}{3\times1.5\times1.5\times4}$
$=54$
$\therefore 54$ bottles are required to empty the bowl.
View full question & answer→Question 142 Marks
The volume of a cube is $512 \mathrm{~cm}^3$. Find its surface area.
AnswerSuppose that the side of the given cube is $x \mathrm{~cm}$.Volume of the cube $=512 \mathrm{~cm}^3$
Then $512=x^3$
$\Rightarrow x=\sqrt[3]{512}=8$
i.e., the side of the cube is $8 \ cm .$
$\therefore$ Surface area of the cube $=6 \mathrm{x}^2 \mathrm{~cm}^2=6 \times 8^2 \mathrm{~cm}^2=384 \mathrm{~cm}^2$
View full question & answer→Question 152 Marks
A matchbox measures $4\ cm \times 2.5\ cm \times 1.5\ cm$. What is the volume of a packet containing $12$ such matchboxes?
AnswerVolume of each matchbox $=4 \times 2.5 \times 1.5=15 \mathrm{~cm}^3$
$\therefore$ Volume of $12$ matchbox $=12 \times 15=180 \mathrm{~cm}^3$
Thus, the volume of a packet containing $12$ such matchboxes is $180 \mathrm{~cm}^3$.
View full question & answer→Question 162 Marks
A patient in a hospital is given soup in a cylindrical bowl of diameter $7\ cm$. If the bowl is filled with soup to a height of $4\ cm, $how much soup the hospital has to prepare daily to serve $250$ patients?
AnswerRadius $(r)$ of cylindrical bowl $=\left(\frac{7}{2}\right) \mathrm{cm}=3.5 \mathrm{~cm}$ Height $(h)$ up to which the bowl is filled with soup $=4 \mathrm{~cm}$
Volume of soup in 1bowl $=\mathrm{pr}^2 \mathrm{~h}=\left(\frac{22}{7} \times(3.5)^2 \times 4\right) \mathrm{cm}^3=154 \mathrm{~cm}^3$
Hence, volume of soup in 250 bowls $=(250 \times$ 154) $\mathrm{cm}^3=38500 \mathrm{~cm}^3=38.5$ litres
Thus, the hospital will have to prepare $38.5$ litres of soup daily to serve 250 patients.
View full question & answer→Question 172 Marks
In a shower, $5\ cm$ of rain falls. Find the volume of water that falls on $2$ hectares of ground.
Answer$\text { Volume of the water that falls on the ground }=\text { area of ground } \times \text { depth }$
$=20000 \times 0.05 \mathrm{~m}^3$
$=1000 \mathrm{~m}^3$
View full question & answer→Question 182 Marks
Find the curved surface area of a cone with base radius $5.25\ cm$ and slant height 10cm. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$
AnswerRadius of a cone, $r = 5.25\ cm$
Slant height of a cone, $l = 10\ cm$
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times5.25\times10\Big)\text{cm}^2$
$=165\text{cm}^2$
View full question & answer→Question 192 Marks
A right circular cone is $3.6\ cm$ high and the radius of its base is $1.6\ cm$. It is melted and recast into a right circular cone having base radius $1.2\ cm$. Find its height.
AnswerHere, height of cone $= 3.6\ cm$ and radius $= 1.6\ cm$
After melting, its radius $= 1.2\ cm$
Volume of original cone = Volume of cone after melting $\therefore\frac{1}{3}\pi\times1.6\times1.6\times3.6$
$=\frac{1}{3}\pi\times1.2\times1.2\times\text{h}$
$\Rightarrow\text{h}=\frac{\frac{1}{3}\pi\times1.6\times1.6\times3.6}{\frac{1}{3}\pi\times1.2\times1.2}=6.4\text{cm}$
$\therefore$ height of new cone $= 6.4\ cm.$
View full question & answer→Question 202 Marks
Find the volume, the lateral surface area, the totle surface area and the diagonal of a cube, each of whose edges measures $9\ m.$ $\big(\text{Take}\sqrt{3}=1.73\big)$
AnswerHere, $a=9 \mathrm{~m}$
Volume of the cube $=a^3=9^3 \mathrm{~m}^3=729 \mathrm{~m}^3$
Lateral surface area of the cube $=4 a^2=4 \times 9^2 \mathrm{~m}^2=4 \times$
$81 \mathrm{~m}^2=324 \mathrm{~m}^2$
Total surface area of the cube $=6 \mathrm{a}^2 6 \times 9^2 \mathrm{~m}^2=6 \times 81 \mathrm{~m}^2=486 \mathrm{~m}^2$
$\therefore$ Diagonal of the cube $=\sqrt{3} \mathrm{a}=\sqrt{3} \times 9=15.57 \mathrm{~m}$
View full question & answer→Question 212 Marks
How many persons can be accommodated in a dining hall of dimension $(20m \times 16m \times 4.5m)$, assuming that each person requires $5$ cubic metres of air?
AnswerVolume of the dining hall $=(20 \times 16 \times 4.5) \mathrm{m}^3=1440 \mathrm{~m}^3$
Volume of air required by each person $=5 \mathrm{~m}^3 \therefore$
Capacity of the dining hall $=\frac{\text { Volume of dining hall }}{\text { Volume of air required by each each person }}=\frac{1440}{5}=288$ persons
View full question & answer→Question 222 Marks
A conical pit of diameter $3.5\ m is 12\ m$ deep. What is its capacity in kilolitres? $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$ Hint: $1m^3 = 1$ kilolitre
AnswerRadius of a conical pit, $\text{r}=\frac{3.5}{2}\text{m}$ Depth of a conical pit, $h = 12m.$
Volume of the conical pit $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times\frac{3.5}{2}\times\frac{3.5}{2}\times12\Big)\text{m}^3$
$=38.5\text{m}^3$
$=38.5\ \text{kiloletre}$
View full question & answer→Question 232 Marks
The radii of two spheres are in the ratio $1 : 2.$ Find the ratio of their surface areas.
AnswerSuppose that the radii are $r$ and $2r.$
Now, ratio of the surface areas $=\frac{4\pi\text{r}^2}{4\pi(2\text{r})^2}=\frac{\text{r}^2}{4\text{r}^2}=\frac{1}{4}$ $=1:4$
$\therefore$ The ratio of their surface areas is $1 : 4.$
View full question & answer→Question 242 Marks
Three cubes of metal with edges $3\ cm, 4\ cm$ and $5\ cm$ respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.
AnswerThree cubes of metal with edges $3 \mathrm{~cm}, 4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ are melted to form a single cube.
$\therefore$ Volume of the new cube $=$ sum of the volumes the old cubes $=\left(3^3+4^3+5^3\right) \mathrm{cm}^3=(27+64+125) \mathrm{cm}^3=216 \mathrm{~cm}^3$
Suppose the edge of the new cube $=\mathrm{x} \mathrm{~cm}$
Then we have: Then $216=\mathrm{x}^3 $
$\Rightarrow \mathrm{x}=\sqrt[3]{216}=6$
$\therefore$ Lateral surface area of the new cube $=4 \mathrm{x}^2 \mathrm{~cm}^2=4 \times$
$6^2 \mathrm{~cm}^2=144 \mathrm{~cm}^2$
View full question & answer→