Question 13 Marks
If the heights of two right circular cones are in the ratio $1 : 2$ and the perimeters of their bases are in the ratio $3 : 4,$ what is the ratio of their volumes$?$
AnswerGiven that
ratio of height of right circular cones
$h_1: h_2=1: 2$
ratio of base of perimeter
$2\pi\text{r}_1:2\pi\text{r}_2=3:4$
$\Rightarrow\text{r}_1:\text{r}_2=3:4$
therefore,
the ratio of volume of their cones
$=\text{v}_1:\text{v}_2=\frac{1}{3}\pi\text{r}^2_1\text{h}_1:\frac{1}{3}\pi\text{r}^2_2\text{h}_2$
$=\frac{\frac{1}{3}\pi\text{r}^2_1\text{h}_1}{\frac{1}{3}\pi\text{r}^2_2\text{h}_2}$
$=\frac{\text{r}^2_1\text{h}_1}{\text{r}^2_2}=\Big(\frac{3}{4}^2\Big)\times\frac{1}{2}$
$=\frac{9}{32}$
$\Rightarrow\text{V}_1:\text{V}_2=9:32$
Hence, the ratio of their volumes are $9:32$
View full question & answer→Question 23 Marks
A well of diameter $3m$ is dug $14m$ deep. The earth taken out of it has been spread evenly all around it to a width of 4m to form an embankment. Find the height of the embankment.
AnswerThe inner radius of the well is $\frac{3}{2}\text{m}$ and the height is $14m$. Therefore, the volume of the Earth taken out of it is
$\text{V}_1=\pi\times\Big(\frac{3}{2}\Big)^2\times14\text{m}^3$
The inner and outer radii of the embankment are $\frac{3}{2}\text{m}$ and $4+\frac{3}{2}=\frac{11}{2}$ respectively. Let the height of the embankment be h. Therefore, the volume of the embankment is
$\text{V}_2=\pi\times\Big\{\Big(\frac{11}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2\Big\}\times\text{h m}^3$
Since, the volume of the well is same as the volume of the embankment; we have
$V_1= V_2$
$\Rightarrow\pi\times\Big(\frac{3}{2}\Big)^2\times14=\pi\times\Big\{\Big(\frac{11}{2}\Big)-\Big(\frac{3}{2}\Big)^2\Big\}\times\text{h}$
$\Rightarrow\text{h}=\frac{9\times14}{112}$
$\Rightarrow\text{h}=\frac{9}{8}\text{m} $
Hence, the height of the embankment is $\text{h}=\frac{9}{8}\text{m}$
View full question & answer→Question 33 Marks
A cylindrical tank full of water is emptied by a pipe at the rate of $225$ litres per minute. How much time will it take to empty half the tank, if the diameter of its base is $3m$ and its height is $3.5m?$
AnswerDiameter of cylindrical tank $= 3m$
height $(h) = 3.5m = \frac{7}{2}\text{m}$
$\therefore$ Radius $(r) = \frac{3}{2}\text{m}$
Volume of water filled in it
$=\pi\text{r}^2\text{h}=\frac{22}{7}\times\frac{3}{2}\times\frac{3}{2}\times\frac{7}{2}\text{m}^3$
$=\frac{99}{4}\text{m}^3$
Volume of water in half the tank
$=\frac{99}{4\times2}\times1000=\frac{9900}{8}\text{l}$
water flow at the rate of $225$ l per min.
$\therefore$ Total time taken to empting the bank
$=\frac{99000}{8\times225}$
$= 55\ \text{minutes}$
View full question & answer→Question 43 Marks
What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?
AnswerGiven that:
Eact solid has the same diameter and height
Therefore,
$r_1=r_2=r_3=r=h_1=h_2$
The volume of cylinder $\text{V}_1=\pi\text{r}_1^2\text{h}_1$
The volume of cone $\text{V}_2=\pi\text{r}_2^2\text{h}_2$
The volume of sphere $\text{V}_3=\frac{4}{3}\pi\text{r}^3$
The ratio of their volumes
$\text{V}_1:\text{V}_2:\text{V}_3=\pi\text{r}^2\text{h}:\pi\text{r}^2\text{h}:\frac{4}{3}\pi\text{r}^3$
$=1:\frac{1}{3}:\frac{4}{3}$
$\text{V}_1:\text{V}_2:\text{V}_3=3:1:4$
Hence, the required ratio are $3 : 1 : 4$
View full question & answer→Question 53 Marks
The height of a solid cylinder is $15\ cm$ and the diameter of its base is $7\ cm.$ Two equal conical holes each of radius $3\ cm,$ and height $4\ cm$ are cut off. Find the volume of the remaining solid.
AnswerDiameter of the base of a cylinder $= 7cm$
$\therefore$ Radius $(r_1) =\Big(\frac{7}{2}\Big)\text{cm}$
Height of cylinder $(h_1) = 15cm$
$\therefore$ Volume of cylinder $=\pi\text{r}_1^2\text{h}_1$
$=\frac{22}{7}\Big(\frac{7}{2}\Big)^2\times15\text{cm}^3$
$=\frac{22}{7}\times\frac{49}{4}\times15=\frac{1155}{2}\text{cm}^2$
Radius of each conical hole $(r_2) = 3cm$
anf height $(h_2) = 4cm$
Volume of 2 such conical holes
$=2\times\frac{1}{3}\pi\text{r}_2^2\text{h}_2$
$=\frac{2}{3}\times\frac{22}{7}\times(3)^2\times4=\frac{2}{3}\times\frac{22}{7}\times9\times4\text{cm}^3$
$=\frac{528}{7}\text{cm}^3$
$\therefore$ Volume of remaining solid $=\frac{1155}{2}-\frac{528}{7}$
$=\frac{8085-1056}{14}=\frac{7029}{14}\text{cm}^3$
$=502.07=502.1\text{cm}^3$
View full question & answer→Question 63 Marks
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is $9\ cm$.
AnswerEdge of cube $= 9\ cm$
$\therefore$ Diameter of cone $= 9\ cm$
And radius $(r) =\frac{9}{2}\text{cm}$
Hieght$(h) = 9\ cm$
$\therefore\text{volume}=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times\frac{9}{2}\times\frac{9}{2}\times9=190.93\text{cm}^3$ View full question & answer→Question 73 Marks
A cone of radius $4\ cm$ is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base. Compare the volumes of two parts.
AnswerLet h be the height of the given cone. One dividing the cone through the mid-point of its axis and parallel to its base into two parts, we obtain the following figure:
In two similar triagles $OAB$ and $DCB,$ we have $\frac{\text{OA}}{\text{CD}}=\frac{\text{OB}}{\text{BD}}$ This implies $\frac{4}{\text{r}}=\frac{\text{h}}{\frac{\text{h}}{2}}$
Therefore, $ r = 2$
Therefore,
$\frac{\text{volume of the smaller cone}}{\text{Volume of the frustum of the cone}}$
$=\frac{\frac{1}{3}\pi\times(2)^2\times\Big(\frac{\text{h}}{2}\Big)}{\frac{1}{3}\pi\times\Big(\frac{\text{h}}{2}\Big)[4^2+2^2+4\times2]}=\frac{1}{7}$
Therefore the ratio of volume of the smaller cone to the volume of the frustum of the cone is $1:7$ View full question & answer→Question 83 Marks
The surface area of a sphere is $616cm^2$. Find its radius.
AnswerThe surface area of sphere $= 616cm^2$
We know that
$4\pi\text{r}^2=616$
$\text{r}^2=\frac{616}{4\pi}$
Taking square root both the side
$\sqrt{\text{r}^2}=\sqrt{\frac{616}{4\pi}}$
$\text{r}=7\text{cm}$
View full question & answer→Question 93 Marks
Three cubes of a metal whose edges are in the ratio $3 : 4 : 5$ are melted and converted into a single cube whose diagonal is $12\sqrt{3}\text{cm}.$ Find the edges of the three cubes.
AnswerLet the edges of three cubes (in cm) be $3x, 4x$ and $5x,$ respectively.
Volume of the cubes after melting is $=(3 x)^3+(4 x)^3+(5 x)^3=216 \times 3 \mathrm{~cm}^3$
Let $a$ be the side of new cube so formed after melting. Therefore, $a^3=216 x^3$
So, $a = 6x, \text{Diagonal}=\sqrt{\text{a}+\text{a}^2+\text{a}^2}=\text{a}\sqrt{3}$
But it is given that diagonal of that diagonal of the new cube is $12\sqrt{3}\text{cm}.$
Therefore, $\text{a}\sqrt{3}=12\sqrt{3},$
i.e. $a = 12$
This gives $x = 2$
Therefore, edges of the three cubes are $6\ cm, 8\ cm$ and $10\ cm$ respectively.
View full question & answer→Question 103 Marks
$500$ persons have to dip in a rectangular tank which is $80m$ long and $50m$ broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is $0.04m^3?$
AnswerThe average displacement of water by a person is $0.04$ cubic $m.$ Hence, the total displacement of water in the rectangular tank by $500$ persons is $V = 500 × 0.04 = 20$ Cubic $m.$
The length and width of the rectangular tank are 80m and 50m respectively. Upon dipping in the tank, let the height of the raised water is be $h\ m$. Therefore, the volume of the raised
water is $V_1= 80 × 50 × h$
$= 4000h$ cubic $m$
Since, the volume of the raised water is same as the volume of the water displaced by $500$ persons, we have
$V_1= V$
$⇒ 4000h = 20$
$\Rightarrow\text{h}=\frac{20}{4000}$
$⇒ 0.005$
Therefore, the water will be raised by $0.005m$ or $0.5\ cm$
View full question & answer→Question 113 Marks
The radii of two cylinders are in the ratio $3 : 5$ and their heights are in the ratio $2 : 3.$ What is the ratio of their curved surface areas$?$
AnswerGiven that:
Ratio of radii of two cylinder
$r_1: r_2=3: 5$
Ratioof height of two cylinder
$h_1: h_2=2: 3$
Now, the ratio of their curved surface area
$\text{S}_1:\text{S}_2=2\pi\text{r}_1\text{h}_1:2\pi\text{r}_2\text{h}_2$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}_1\text{h}_1}{2\pi\text{r}_2\text{h}_2}=\frac{3}{5}\times\frac{2}{3}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{6}{15}=\frac{2}{5}$
$\Rightarrow \mathrm{~S}_1: \mathrm{S}_2=2: 5$
Hence, the ratio of their curved surface area are $2 : 5$
View full question & answer→Question 123 Marks
The difference between outside and inside surface areas of cylindrical metallic pipe $14\ cm$ long is $44m^2$. If the pipe is made of $99cm^3$ of metal, find the outer and inner radii of the pipe.
AnswerLet inner radius of pipe be $r_1$
Radius of outer cylinder be $r_2$
Length of cylinder $(h) = 14cm.$
Surface area of hollow cylinder $(r_2- r_1)=2\pi$
Given surface area of cylinder $= 44m^2$
$2\pi\text{Rh}-2\pi\text{r}\text{h}=44$
$2\pi\text{h}(\text{R}-\text{r})=44$
$2\times\frac{22}{7}\times14(\text{R}-\text{r})=44$
$\text{R}-\text{r}=\frac{44\times7}{2\times22\times14}$
$\text{R}-\text{r}=\frac{1}{2}\ ...(1)$
Now,
Volume of pipe $= 99cm^3$
$\pi\text{R}^2\text{h}-\pi_2\text{r}^2\text{h}=99$
$\pi\text{h}(\text{R}^2-\text{r}^2)=99$
$(\text{R}^2-\text{r}^2)=\frac{99}{\pi\times\text{h}}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99\times7}{22\times14}$
$(\text{R}+\text{r})(\text{R}-\text{r})=\frac{99}{22\times2}$
Putting value of $\text{R}-\text{r}=\frac{1}{2}$
$(\text{R}+\text{r})\frac{1}{2}=\frac{99}{22\times2}$
$(\text{R}+\text{r})=\frac{9}{2}\ ...(2)$
Adding equation $(1)$ and $(2)$
$\text{R}-\text{r}=\frac{1}{2}$
$\text{R}+\text{r}=\frac{9}{2}$
$2\text{R}=5$
$\text{R}=2.5$
And $\text{r}=\frac{9}{2}-\frac{5}{2}$
$\text{r}=2\text{cm}$
View full question & answer→Question 133 Marks
A hemisphere and a cone have equal bases. If their heights are also equal, then what is the ratio of their curved surfaces?
AnswerGiven that:
The base of the cone and hemisphere are equal.
Therefore,
Radius of the hemisphere = Height of the cone
Then, the salant height of the cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{\text{r}^2+\text{r}^2}$
$=\sqrt{2}\text{r}$
Curved surface area of hemisphere
$\text{S}_1=2\pi\text{r}^2$
And the curved surface area of the cone
$\text{S}_2=\pi\text{rl}$
Now,
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}^2}{\pi\text{rl}}=\frac{2\pi\text{r}^2}{\pi\text{r}\sqrt{2}\text{r}}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{\sqrt{2}}{1}$
Hence, the required ratio are $\sqrt{2}:1$
View full question & answer→Question 143 Marks
A solid metal sphere of $6\ cm$ diameter is melted and a circular sheet of thickness $1\ cm$ is prepared. Determine the diameter of the sheet.
AnswerDiameter of sphere $= 6\ cm$
Therefore,
Radius $= 3\ cm.$
Therefore,
Surface area of sphere
$=4\pi\text{r}^2$
$=4\times\pi\times(3)^2$
$=36\pi\ \text{cm}^2$
Area of the circular sheet $=\pi\text{r}^2$
Therefore,
Surface area of sphere $=$ area of the circular sheet
$=\pi\text{r}^2=36\pi$
$\text{r}=6\text{cm}$
Therefore,
Diameter of the sheet $= 2 × 6 = 12\ cm$
View full question & answer→Question 153 Marks
If a cone and a sphere have equal radii and equal volumes. What is the ratio of the diameter of the sphere to the height of the cone?
AnswerGiven that:
A cone and a sphere have equal radii and equal volume
Therefore,
Volume of cone = Volume of sphere
$\Rightarrow\frac{1}{3}\pi\text{r}^2\text{h}=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow r^2 h=4 r^3$
$⇒ h = 4r$
$⇒ h = 2 (2r)$
$⇒ h = 2d$
$\Rightarrow\frac{\text{h}}{\text{d}}=\frac{2}{1}$
$⇒ d : h = 1 : 2$
Hence, the required eatio are $1 : 2$
View full question & answer→Question 163 Marks
The slant height of the frustum of a cone is $4\ cm$ and the perimeters of its circular ends are $18\ cm$ and $6\ cm.$ Find the curved surface of the frustum.
AnswerPerimeter of the top of frustum $= 18cm$
$\therefore $ Radius $(r_1) =\frac{\text{C}}{2\pi}=\frac{18\times7}{2\times22}$
$=\frac{63}{22}\text{cm}$
and perimeter of the bottom $= 6cm$
$\therefore$ Radius $(r_2) =\frac{\text{C}}{2\pi}=\frac{18\times7}{2\times22}=\frac{21}{22}\text{cm}$
and slant height $(l) = 4cm$
curved surface area $=\pi (\text{r}_1+\text{r}_2) \text{l} $
$=\frac{22}{7}\Big(\frac{63}{22}+\frac{21}{22}\Big)\times4$
$= \frac{22}{7}\times\frac{84}{22}\times4=48\text{cm}^2$ View full question & answer→Question 173 Marks
A hollow sphere of internal and external radii $2\ cm$ and $4\ cm$ respectively is melted into a cone of base radius $4\ cm.$ Find the height and slant height of the cone.
AnswerThe internal and external radii of the hollow sphere are $2\ cm$ and $4\ cm$ respectively. Therefore, the volume of the hollow sphere is,
$\text{V}=\frac{4}{3}\pi\{(4)^3-(2)^3\}$
$=\frac{4}{3}\times\frac{22}{7}\times56$
$=\frac{32\times22}{3}$
The hollow sphere is melted to produce a right circular cone of base-radius $4\ cm.$ Let, the height and slant height of the cone be $h \ cm$ and $l \ cm$ respectively. Then, we have,
$\text{l}^2=(4)^2+\text{h}^2$
$\Rightarrow\text{l}^2=16+\text{h}^2$
The volume of the cone is,
$\text{V}_1=\frac{1}{3}\pi\text{r}^2_1\text{h}_1$
$=\frac{1}{3}\times\frac{22}{7}\times(4)^2\times\text{h}$
Since, the volume of the cone and hollow sphere are same, we have
$V_1= V$
$\Rightarrow\frac{1}{3}\times\frac{22}{7}\times(4)^2\times\text{h}=\frac{32\times22}{3}$
$\Rightarrow\frac{1}{7}\times(4)^2\times\text{h}=32$
$\Rightarrow\text{h}=\frac{32\times7}{16}$
$\Rightarrow\ =14$
Then, we have
$\text{l}^2=16+(14)^2$
$\Rightarrow\ =212$
$\Rightarrow\text{l}=14.56$
Therefore, the height and the slant height of the cone are $14\ cm$ and $14.56\ cm$ respectively.
View full question & answer→Question 183 Marks
A sphere of diameter $5\ cm$ is dropped into a cylindrical vessel partly filled with water. The diameter of the base of the vessel is $10\ cm.$ If the sphere is completely submerged, by how much will the level of water rise$?$
AnswerRadius of sphere $\text{r}=\frac{5}{2}\text{cm}$ radius of cylindrical vessel $\text{r}_1=\frac{10}{2}\text{cm}$
When the sphere is completely submerged into the vessel. The level of water will be raised let xbe height of level of raised water.
Therefore,
The volume of raised water in cylindrical vessel $=$ volume of sphere
$\pi\times(5)^2\times\text{x}=\frac{4}{3}\pi\Big(\frac{5}{2}\Big)^3$
$25\text{x}=\frac{4\times125}{3\times8}$
$\text{x}=\frac{4\times125}{3\times8\times25}$
$=\frac{5}{6}\text{cm}$
$\text{x}=\frac{5}{6}\text{cm}$
Hence, the level of water rise $=\frac{5}{6}\text{cm}$ View full question & answer→Question 193 Marks
If the areas of circular bases of a frustum of a cone are $4cm^2$ and $9cm^2$ respectively and the height of the frustum is $12\ cm$. What is the volume of the frustum$?$
AnswerGiven that:
Area of circular ends of frustum are
$\pi\text{r}_1^2=4$
$\Rightarrow\text{r}_1^2=\frac{4}{\pi}$
and
$\pi\text{r}_2^2=9$
$\text{r}^2_2=\frac{9}{\pi}$
The height of frustum $h = 12\ cm$
Now, the Volume of frustum
$\text{v}=\frac{\text{h}}{3}(\text{r}_1^2+\text{r}_2^2+\text{r}_1\text{r}_2)$
$=\frac{12}{3}\pi\Big(\frac{4}{\pi}+\frac{9}{\pi}+\sqrt{\frac{4}{\pi}}\times\sqrt{\frac{9}{\pi}}\Big)$
$=4\pi\Big(\frac{13}{\pi}+\sqrt{\frac{36}{\pi^2}}\Big)$
$=4\pi\Big(\frac{13}{\pi}+\frac{6}{\pi}\Big)$
$=4\times19$
$=76\text{cm}^2$
Hence, the Volume of frustum is $76cm^3$.
View full question & answer→Question 203 Marks
A cylinder and a cone are of the same base radius and of same height. Find the ratio of the value of the cylinder to that of the cone.
AnswerLet $r$ be the radius of base and h be the height of cylinder and cone
Therefore,
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{1}{\frac{1}{3}}$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{3}{1}$
$\Rightarrow\text{v}_1:\text{v}_2=3:1$
Hence, the required ratio are $3 : 1$
View full question & answer→Question 213 Marks
A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio $5 : 12,$ write the ratio of the total surface area of the cylinder to that of the cone.
AnswerLet $r = 5x$ and $h = 12x$ be the base radius and height of the cone and cylinder respectively.
Slant height of the cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(5\text{x})^2+(12\text{x}^2)^2}$
$=\sqrt{25\text{x}^2+144\text{x}^2}$
$\Rightarrow\text{l}=13\text{x}$
The total surface area of cylinder
$\text{S}_1=2\pi\text{r}(\text{h}+\text{r})$
The total surface area of cone
$\text{s}_2=2\pi\text{r}(\text{l}+\text{r})$
Now,
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\pi\text{r}(\text{h}+\text{r})}{\pi\text{r}(\text{l}+\text{r})}$
$=\frac{2(\text{h}+\text{r})}{(\text{l}+\text{r})}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2(12\text{x}+5\text{x})}{13\text{x}+5\text{x}}$
$\Rightarrow\frac{\text{S}_1}{\text{S}_2}=\frac{2\times17\text{x}}{18\text{x}}$
$\Rightarrow\text{S}_1:\text{S}_2=17:9$
Hence, the required ratio are $17 : 9$
View full question & answer→Question 223 Marks
A cylindrical tub of radius $5\ cm$ and length $9.8\ cm$ is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is $3.5\ cm$ and height of the cone outside the hemisphere is $5\ cm,$ find the volume of the water left in the tube.
AnswerGiven radius of cylinderical tube $(r) = 5\ cm.$
Height of cylindrical tube $(h) = 9.8\ cm$
Volume of cylinder $=\pi\text{r}^2\text{h}$
$\text{V}_1=\pi(5)^2(9.8)=770\text{\ cm}^3$
Given radius of hemisphere $(r) = 3.5\ cm$
Height of cone $(h) = 5\ cm$
Volume of hemisphere $=\frac{2}{3}\pi\text{r}^3$
$=\frac{2}{3}\times\pi(3.5)^3=89.79\text{\ cm}^3$
Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{\pi}{3}(3.5)^25=64.14\text{\ cm}^3$
Volume of cone $+$ Volume of hemisphere $(V_2) = 39.79 + 64 = 154\ cm^3$
View full question & answer→Question 233 Marks
A solid iron pole having cylindrical portion $110\ cm$ high and of base diameter $12\ cm$ is surmounted by a cone $9\ cm$ high. Find the mass of the pole, given that the mass of $1\ cm^3$ of iron is $8\ gm.$
AnswerDiamter of the base $= 12cm$
$\therefore$ Radius $(r) =\Big(\frac{12}{2}\Big)=6\text{cm}$
Height of the cylindrical portion $(h_1) = 110cm$
and height of conical portion $(h_2) = 9cm$
$\therefore$ Total volume of the pole
$=\pi\text{r}^2\text{h}_1+\frac{1}{3}\pi\text{r}^2\text{h}_2$
$=\pi\text{r}^2\Big(\text{h}_1+\frac{1}{3}\text{h}_2\Big)$
$=\frac{22}{7}\times(6)^2\Big(110+\frac{1}{3}\times9\Big)\text{cm}^2$
$=\frac{22}{7}+36(110+3)\text{cm}^3$
$=\frac{22}{7}\times36\times113\text{cm}^3$
$=\frac{89496}{7}\text{cm}^3$
Weight of $1cm^3$ of iron $= 8gm$
$\therefore$ Total weight $=\frac{89496}{7}\times8\text{gms}$
$=\frac{89496\times8}{7\times1000}\text{kg}=102.281\text{kg}$
$\therefore$ Mass of the pole $= 102.281\ kg$ View full question & answer→Question 243 Marks
$50$ circular plates each of diameter $14\ cm$ and thickness $0.5\ cm$ are placed one above the other to form a right circularcylinder. Find its total surface area.
AnswerGiven that $50$ circular plates each with diameter $= 14\ cm$
Radius of circular plates $(r) = 7\ cm$
Thickness of plates $= 0.5$
Since these plates are placed one above other so total thickness of plates $0.5 × 50 = 25\ cm$
Total surface area of a cylinder $=2\pi\text{rh}+2\pi\text{r}^2$
$=2\pi\text{rh}+2\pi\text{r}^2$
$=2\pi\text{r}(\text{h}+\text{r})$
$= 2\times\frac{22}{7}\times7(25+7)$
$T.S.A= 1408\ cm^2$
$\therefore$ Total surface area of circular plates is $1408\ cm^2$
View full question & answer→Question 253 Marks
The vertical height of a conical tent is $42dm$ and the diameter of its base is $5.4m.$ Find the number of persons it can accommodate if each person is to be allowed $29.16$ cubic dm.
AnswerRadius of conicaltent, $\text{r}=\frac{5.4}{2}$
$= 2.7m$
$= 27dm$
Height of conical tent $h = 42dm$
The volume of conical tent
$=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times27\times27\times42$
$=22\times27\times27\times2$
$=32076\text{dm}^3$
Since, each person is to be allowed $29.16dm^3$,
Therefore,
$=\frac{\text{Volume of conical tent}}{\text{place to be allow to each person}}$
$=\frac{32076}{29.16}$
$=\frac{3207600}{2916}$
$\text{No. of persons}=1100$
View full question & answer→Question 263 Marks
If the total surface area of a solid hemisphere is $462\ cm^2$, find its volume $\Big(\text{use}\ \pi=\frac{22}{7}\Big)$
AnswerLet the radius of the hemisphere be $r\ cm.$
Total surface area of hemisphere $= 462\ cm^2$
$\Rightarrow3\pi\text{r}^2=462$
$\Rightarrow3\times\frac{22}{7}\times(\text{r})^2=462$
$⇒ r^2= 49$
$⇒ r = 7\ cm$
Now, the volume of hemisphere is given by
$\frac{2}{3}\pi\text{r}^3$
$=\frac{2}{3}\times\frac{22}{7}(7)^3$
$=\frac{2156}{3}$
$=718\frac{2}{8}\text{cm}^3$
View full question & answer→Question 273 Marks
Water flows through a cylindrical pipe, whose inner radius is $1\ cm,$ at the rate of $80\ cm/sec$ in an empty cylindrical tank, the radius of whose base is $40\ cm.$ What is the rise of water level in tank in half an hour$?$
AnswerGiven, radius of tank, $r_1 =40\ cm$
Let height of water level in tank in half an hour $= h_1$
Also, given internal radius of cylindrical pipe, $r_1= 1\ cm$
and speed of water $= 80\ cm/s$ in $1$ water flow $= 80\ cm$
$\therefore$ In $30 ($min$)$ water flow $= 80 × 60 × 30 = 144000\ cm$
According to the question,
Volume of water in cylindrical tank $=$ Volume of water flow the circular pipe in half an hour
$\Rightarrow\pi\text{r}^2_1\text{h}_1=\pi\text{r}^2_1\text{h}_2$
$\Rightarrow40\times40\times\text{h}_1=1\times144000$
$\therefore\text{h}_1=\Big(\frac{144000}{40\text{x}40}\Big)=90\text{cm}$
Hence, the level of water in cylindrical tank rises $90\ cm$ in half an hour.
View full question & answer→Question 283 Marks
An iron pillar consists of a cylindrical portion $2.8m$ high and $20\ cm$ in diameter and a cone $42\ cm$ high is surmounting it. Find the weight of the pillar, given that $1$ cubic cm of iron weighs $7.5\ gm.$
AnswerVolume of cylindrical portion
$=\pi\text{r}^2\text{h}$
$=\frac{22}{7}\times\Big(\frac{20}{2}\Big)^2\times280$
$=88000\text{cm}^3$
Volume of conical portion
$\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\frac{22}{7}\times(10)^2\times42$
$=4400\text{cm}^3$
Total number
$= 88000 + 4400$
$= 92400$
So total height
$= 92400 × 7.5$
$= 693000\ gm$
$= 693\ kg$
View full question & answer→Question 293 Marks
A solid cone of base radius $10\ cm$ is cut into two parts through the mid-point of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone.
AnswerRadius of solid cone $(r) = 10\ cm$
Let total height $= h$
In $\triangle\text{AOB}$
$C$ is mid point of $AO$ and $CD || OB$
$\therefore\frac{\text{OB}}{\text{CD}}=\frac{\text{AO}}{\text{AC}}\Rightarrow\frac{10}{\text{CD}}=\frac{\text{h}}{\text{h}}$
$\Rightarrow\frac{10}{\text{CD}}=\frac{2}{1}\Rightarrow\text{CD}=\frac{10}{2}=5\text{cm}$
$\therefore\text{r}_2=5\text{cm}$
Volume of smaller cone
$=\frac{1}{3}\pi\text{r}^2_2\frac{\text{h}}{2}=\frac{1}{3}\pi\times5\times5\times\frac{\text{h}}{2}=\frac{25}{6}\pi\text{h}$
Volume of frustum $=\frac{1}{3}\pi\frac{\text{h}}{2}(\text{r}^2_1+\text{r}_1\text{r}_2+\text{r}_2^2)$
$=\frac{\text{h}\pi}{6}(10^2+10\times5+5^2)$
$\frac{\pi\text{h}}{6}(100+50+25)$
$=\frac{175}{6}\pi\text{h}$
Ratio between the upper part and lower part
$=\frac{25}{6}\pi\text{h}:\frac{175}{6}\pi\text{h}=1:7$ View full question & answer→Question 303 Marks
If the volumes of two cones are in the ratio $1 : 4$ and their diameters are in the ratio $4 : 5,$ then write the ratio of their height.
AnswerLet $r_1 r_2$ be the radii and $h_1, h_2$ be the height of two cones
It is given that the ratio of the volimes of two cones.
$V_1: V_2=1: 4$
and $2 r_1: 2 r_2=4: 5$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{4}{5}$
the ratio of Volume of two cones.
$\Rightarrow\frac{\text{V}_1}{\text{v}_2}=\frac{\frac{1}{3}\pi\text{r}_1^2\text{h}_1}{\frac{1}{3}\pi\text{r}_2^2\text{h}_2}$
$\Rightarrow\frac{1}{4}=\Big(\frac{4}{5}\Big)^\times\frac{\text{h}_1}{\text{h}_2}$
$\Rightarrow\frac{1}{4}=\frac{16}{25}\times\frac{\text{h}_1}{\text{h}_2}$
$\Rightarrow\text{h}_1:\text{h}_2=25:64$
Hence, the required ratio are $25 : 64$
View full question & answer→Question 313 Marks
A well with $10m$ inside diameter is dug $8.4m$ deep. Earth taken out of it is spread all around it to a width of $7.5m$ to form an embankment. Find the height of the embankment.
AnswerRadius of well
$\text{r}=\frac{10}{2}$
$= 5m$
Depth of well $h = 8.4m$
Clearly,
Volume of earth dugout
$=\pi(5)^2\times8.4$
$=\pi\times25\times8.4$
$=\frac{22\times25\times8.4}{7}\text{m}^3$
Let $h'$ be the height of embankment
Clearly,
Embankment forms a cylindrical shell whose inner and outer radius are $5m$ and $12.5m$ respectively.
$\therefore$ Volume of the embankment
$=\pi\{(12.5)^2-(5)^2\}\times\text{h}'$
$=\pi\times17.5\times7.5\times\text{h}'\ \text{m}^3$
But, volume of earth dugout = volume of the embankment
$\frac{22\times25\times8.4}{7}=\frac{22}{7}\times17.5\times7.5\times\text{h}$
$\text{h}=\frac{258\times8.4}{17.5\times7.5}$
$\text{h}'=1.6\text{m}$ View full question & answer→Question 323 Marks
The radii of the base of a cylinder and a cone are in the ratio $3 : 4$ and their heights are in the ratio $2 : 3.$ What is the ratio of their volumes$?$
AnswerLet $r_1$ and $r_2$ be the radii of the base of a cylinder and a cone.
The volume of cylinder $\text{V}_1=\pi\text{r}^2_1\text{h}_1...(1)$
The volume of cone $\text{V}_2=\pi\text{r}^2_2\text{h}_2...(2)$
Dividing $(i)$ by $(ii),$ the, we get
$\frac{\text{V}_1}{\text{V}_2}=\frac{\pi\text{r}^2_1\text{h}_1}{\frac{1}{3}\pi\text{r}^2_2\text{h}_2}$
$\frac{=3\times\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2\times\Big(\frac{\text{h}_1}{\text{h}_2}\Big)}{\Big(\frac{\text{r}_1}{\text{r}_2}=\frac{3}{4},\frac{\text{h}_1}{\text{h}_2}=\frac{2}{3},\ \text{given}\Big)}$
$\frac{\text{V}_1}{\text{V}_2}=3\times\Big(\frac{3}{4}\Big)^2\times\frac{2}{3}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{9}{8}$
$\text{V}_1:\text{V}_2=9:8$
View full question & answer→Question 333 Marks
Water in a canal $1.5m$ wide and 6m deep is flowing with a speed of $10\ km/hr$. How much area will it irrigate in $30$ minutes if $8\ cm$ of standing water is desired$?$
AnswerThe canal is $1.5\ m$ wide and $6\ m$ deep. The water is flowing in the canal at $10\ km/hr.$ Hence, in $30$ minutes, the length of the flowing standing water is
$=10\times\frac{30}{60}\text{km}$
$= 5\ km$
$= 5000\ m$
Therefore, the volume of the flowing water in $30$ min is
$V_1= 5000 × 1.5 × 6\ m^3$
Thus, the irrigated area in $30$ min of $8\ cm = 0.08\ m$ standing water is
$=\frac{5000\times1.5\times6}{0.08}$
$= 562500\ m^2$
View full question & answer→Question 343 Marks
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is $24\ m$. The height of the cylindrical portion is $11\ m$ while the vertex of the cone is $16\ m$ above the ground. Find the area of canvas required for the tent.
AnswerGiven diameter of cylinder $24\ m$
Radius $(r) =\frac{24}{2}=12\text{m}$
Given height of cylindrical part $(h_1) = 11\ m$
$\therefore$ Height of cone part $(h_2) = 5\ m$
Vertex of come above ground $= 11 + 5 = 16\ m$
Curved surface area of cone $(S_1) =\pi\text{rl}$
$=\frac{22}{7}\times12\times\text{l}$
Let $l$ be slant height of cone
$\text{l}=\sqrt{\text{r}^2+\text{h}^2_2}$
$\Rightarrow\text{l}=\sqrt{12^2+5^2}=13\text{m}$
$l = 13\ m$
$\therefore$ Curved surface area of cone (5) $=\frac{22}{7}\times12\times13\text{m}^2$
View full question & answer→Question 353 Marks
A cylindrical container is filled with ice-cream, whose diameter is $12\ cm$ and height is $15\ cm.$ the whole ice-cream is distributed to $10$ children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream.
AnswerVolume of cylindrical container
$=\pi\text{r}^2\text{h}$
$=\pi\times(6)^2\times15$
Amount of ice-cream distributed to $10$ children $=\frac{\pi\times(6)^2\times15}{10}$
Therefore,
Height of conical portion $=\ 2\ ×$ diameter of its bars
Let the diameter of bare $= r$
Height $= 2r$
Therefore,
Volume of the cones
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\text{h}+\frac{2}{3000}\pi\text{r}^3$
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2(\text{h}+2\text{r})$
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\Big(2\text{r}+2\times\frac{\text{r}}{2}\Big)$
$=\frac{1}{3}\pi\Big(\frac{\text{r}}{2}\Big)^2\times3$
$\text{r}=\frac{\pi\text{r}^3}{4}$
Therefore,
Volume of the cones $=$ amount distributed
$\frac{\pi\text{r}^3}{4}=\frac{\pi(6)^2\times15}{10}$
$\text{r}^3=\frac{4\times6\times6\times15}{10}=4\times6\times9$
$\text{r}=\sqrt[3]{6\times6\times6}=6\text{cm}$
View full question & answer→Question 363 Marks
The slant height of the frustum of a cone is $5\ cm.$ If the difference between the radii of its two circular ends is $4\ cm,$ write the height of the frustum.
AnswerLet $r_1$ and $r_2$ be the radius of frustum ends.
$r_1-r_2=4 \mathrm{~cm}$
Slant height of the frustum cone
$\mathrm{I}=5 \mathrm{~cm}$
$ \Rightarrow \mathrm{l}=\sqrt{\mathrm{h}^2+\left(\mathrm{r}_1-\mathrm{r}_2\right)^2} $
$ =5=\sqrt{\mathrm{h}^2+4^2}$
Squaring both the sides.
$ \Rightarrow 25=h^2+16 $
$ \Rightarrow h^2=25-16 $
$ \Rightarrow h^2=9 \mathrm{~cm}$
$ \Rightarrow h=3 \mathrm{~cm}$
Hence, the height of frustum is $3\ cm$
View full question & answer→Question 373 Marks
A container open at the top, is in the form of a frustum of a cone of height $24\ cm$ with radii of its lower and upper circular ends as $8\ cm$ and $20\ cm$ respectively. Find the cost of milk which can completely fill the container at the rate of $Rs. 21$ per litre.
AnswerUpper radius $(R) = 20\ cm$
Lower radius $(r) = 8\ cm$
Height $(h) = 24\ cm$
$\therefore$ Volume of frustum
$=\frac{\pi}{3}(\text{R}^2+\text{Rr}+\text{r}^2)\text{h}$
$=\frac{22}{7\times3}[20^2+20\times8+8^2]\times24\text{cm}^3$
$\frac{22}{21}[400+160+64]\times24\text{cm}^3$
$=\frac{22}{21}\times624\times24\text{cm}^3=\frac{329472}{21}\text{cm}^3$
Volume of milk in it
$=\frac{329472}{21}\times\frac{1}{1000}\text{l}$
$\frac{329.472}{21}\text{l}$
Rate of milk $= ₹\ 21$ per $l$
$\frac{329.472}{21}\times21$
$= ₹\ 329.47$ View full question & answer→Question 383 Marks
From a solid cylinder of height $2.5\ cm$ and diameter $4.2\ cm,$ a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.$\Big(\text{use}\ \pi=\frac{22}{7}\Big)$
AnswerDiameter of solid cylinder $= 4.2\ cm$
$\therefore$ Radius $(r) \Big(\frac{4.2}{2}\Big)=2.1\text{cm}$
Height $(h) = 2.8\ cm$
Slant height of cone
$=\sqrt{\text{r}^2+\text{h}^2}$
$=\sqrt{(2.1)^2+(2.8)^2}=\sqrt{4.41+7.84}$
$=\sqrt{12.25}=3.5\text{cm}$
Total surface area of remaining solid $=$ surface area of cylinder $+$ surface area of cone
$=2\pi\text{rh}+\pi\text{r}^2+\pi\text{rl}$
$=\pi\text{r}(2\text{h}+\text{r}+\text{l})$
$=\frac{22}{7}\times2.1(2\times2.8+2.1+3.5)\text{cm}^2$
$=6.6(5.6+2.1+3.5)\text{cm}^2$
$=6.6(11.2)=73.92\text{cm}^2$ View full question & answer→Question 393 Marks
The rain water from a roof of dimensions $22\ m × 20\ m$ drains into a cylindrical vessel having diameter of base $2\ m $ and height $3.5\ m.$ If the rain water collected from the roof just fills the cylindrical vessel, then find the rain in cm.
AnswerGiven, length of roof $= 22\ m$ and breadth of roof $= 20\ m$
Let the rainfall be a $cm.$
Volume of water on the roof $=22\times20\times\frac{\text{a}}{100}=\frac{22\text{a}}{5}\text{m}^3$
Also, we have radius of base of the cylinderical vessel $= 1\ m$
and height of the cylindrical vessel $= 3.5\ m$
$\therefore$ Volume of water in the cylindrical vessel when it is just full
$=\Big(\frac{22}{7}\times1\times1\times\frac{7}{2}\Big)=11\text{m}^3$
Now, volume of water on the roof $=$ volume of water in the vessel
$\Rightarrow\frac{22\text{a}}{5}=11$
$\therefore\text{a}=\frac{11\times5}{22}=2.5$
$\Big[\because$ volume of cylindricer = $\pi\times(\text{radius})^2\times\text{height}\Big]$
Hence, the rainfall is $2.5\ cm.$
View full question & answer→Question 403 Marks
A hemisphere of lead of radius $7\ cm$ is cast into a right circular cone of height $49\ cm.$ Find the radius of the base.
AnswerRadius of hemisphere $r = 7\ cm$
The volume of hemisphere
$=\frac{2}{3}\pi\text{r}^2$
$=\frac{2}{3}\pi\times(7)^3$
$=\frac{2}{3}\pi\times343$
$=\frac{686}{3}\pi\ \text{cm}^3$
Since,
The hemisphere cast into the right circular cone.
The height of cone $h = 49\ cm$
Let $x$ be the radius of cone.
Clearly,
Volume of cone $=$ volume of hemisphere
$\frac{1}{3}\pi\times49=\frac{686}{3}\pi$
$\text{x}^2=\frac{686\times3}{49\times3}$
$=14$
$\text{x}^2=14$
$\text{x}=\sqrt{14}$
$\text{x}=3.74\text{cm}$
Thus, the radius of cone $= 3.74\ cm$
View full question & answer→Question 413 Marks
A building is in the form of a cylinder surmounted by a hemi-spherical vaulted dome and contains $41\Big(\frac{19}{21}\Big)\text{m}^3$ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building$?$
AnswerLet total height of the building $=$ Internal diameter of the dome $= 2m$
Radius of building $($or dome$) =\Big(\frac{2\text{r}}{2}\Big)=\text{r}\ \text{m}$
Height of cylinder $= 2r - r = r\ m$
$\therefore$ Volume of the cylinder $=\pi\text{r}^2(\text{r})=\pi\text{r}^3\text{m}^3$
and volume of hemispherical dome cylinder $=\frac{2}{3}\pi\text{r}^3\text{m}^3$
$\therefore$ Total volume of the building $=$ volume of the cylinder $+$ volume of hemispherical dome.
$=\Big(\pi\text{r}^3+\frac{2}{3}\pi\text{r}^3\Big)\text{m}^3=\frac{5}{3}\pi\text{r}^3\text{m}^3$
According to the condition,
Volume of the building $=$ volume of the air
$\Rightarrow\frac{5}{3}\pi\text{r}^3=41\frac{19}{21}\Rightarrow\frac{5}{3}\pi\text{r}^3=\frac{880}{21}$
$\Rightarrow\text{r}^3=\frac{880\times7\times3}{21\times22\times5}=\frac{40\times21}{21\times5}=8$
$\Rightarrow\text{r}^3=8\Rightarrow\text{r}=2$
$\therefore$ Height of the building $= 2r = 2 × 2 = 4m.$ View full question & answer→Question 423 Marks
A hemisphere of lead of radius $7\ cm$ is cast into a right circular cone of height $49\ cm.$ Find the radius of the base.
AnswerRadius of hemisphere of lead $(r_1) = 7\ cm$
$\therefore\frac{2}{3}\pi\times7\times7\times7=\frac{686\pi}{3}\text{cm}^3$
Now volume of right circular cone $=\frac{686\pi}{3}\pi\ \text{cm}^3$
Height $(h) = 49\ cm$
Let $r_2$ be the radius, then
Volume $=\frac{1}{3}\pi\text{r}_2^2\text{h}$
$\Rightarrow686\pi=\frac{1}{3}\pi\text{r}_2^2\times49\Rightarrow\text{r}_2^2=\frac{686\pi\times3}{1\times49\pi}$
$\Rightarrow\text{r}_2^2=42\Rightarrow\text{r}_2=\sqrt{42}$
$\therefore$ Radius $=\sqrt{42}=6.480\text{cm}$
View full question & answer→Question 433 Marks
A sphere of maximum volume is cut-out from a solid hemisphere of radius $r,$ what is the ratio of the volume of the hemisphere to that of the cut-out sphere$?$
AnswerGiven that:
A sphere of maximum volume is cut from a solid hemisphere of radius $r.$
Therefore radius of sphere $=\frac{\text{r}}{2}$
The volume of sphere $=\frac{4}{3}\pi\Big(\frac{\text{r}}{2}\Big)^3$
$\Rightarrow\text{v}_1=\frac{1}{6}\pi\text{r}^3...(1)$
The volume of hemisphere
$\Rightarrow\text{v}_2=\frac{2}{3}\pi\text{r}^3...(2)$
Dividing Eq. $(1)$ by Eq. $(2)$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{\frac{1}{6}\pi\text{r}^3}{\frac{2}{3}\pi\text{r}^3}$
$\Rightarrow\frac{\text{v}_1}{\text{v}_2}=\frac{1}{4}$
$\Rightarrow\text{v}_2:\text{v}_1=4:1$
Hence the required ratio are $4 : 1$
View full question & answer→Question 443 Marks
A cylindrical vessel of diameter $14\ cm$ and height $42\ cm$ is fixed symmetrically inside a similar vessel of diameter $16\ cm$ and height $42\ cm.$ The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required$?$
AnswerDiameter of inner cylinder $= 14\ cm$
$\therefore$ Radius $(r) =\Big(\frac{14}{2}\Big)=7\text{ cm}$
Diameter of outer cylinder $= 16\ cm$
$\therefore$ Radius $(R) \Big(\frac{16}{2}\Big)=8\text{cm}$
Height $(h) = 42\ cm$
$\therefore$ space between the two cylinders
$=\pi\text{R}^2\text{h}-\pi\text{r}^2\text{h}$
$=\pi\text{h}(\text{R}^2-\text{r}^2)$
$=\frac{22}{7}\times42(8^2-7^2)\text{cm}^2$
$=22\times6\times(64-49)$
$=22\times6\times15\text{cm}^3$
$=1980\text{cm}^3$ View full question & answer→Question 453 Marks
A toy is in the form of a cone mounted on a hemisphere of radius $3.5\ cm.$ The total height of the toy is $15.5\ cm$ find the total surface area and volume of the toy.
AnswerRadius of the toy $(r) = 3.5\ cm$
Total height of the toy $= 15.5\ cm$
$\therefore$ Height of the conical part $= 15.5 – 3.5 = 12\ cm$
Slant height of the conical part $(l)$
$=\sqrt{\text{r}^2+\text{h}^2}=\sqrt{(3.5)^2+(12)^2}$
$=\sqrt{12.25+144}=\sqrt{156.25}=12.5\text{cm}$
- Now total surface area of the toy $=$ curved surface area of conical part $+$ curved surface area of hemispherical part
$=\pi\text{rl}+2\pi\text{r}^2=\pi\text{r}(\text{l}+2\text{r})$
$=\frac{22}{7}\times3.5(12.5+2\times3.5)\text{cm}^2$
$=214.5\text{cm}^2$
- Volume of the toy $=\frac{1}{3}\pi\text{r}^2\text{h}+\frac{2}{3}\pi\text{r}^3$
$=\frac{1}{3}\pi\text{r}^2(\text{h}+2\text{r})$
$=\frac{1}{3}\times\frac{22}{7}(3.5)^2(12+2\times3.5)\text{cm}^3$
$=\frac{1}{3}\times\frac{22}{7}\times12.25(12+7)\text{cm}^3$
$=\frac{22}{3}\times1.75\times19\text{cm}^3$
$=\frac{731.5}{3}=243.83\text{cm}^3$ View full question & answer→Question 463 Marks
A cylindrical vessel with internal diameter $10\ cm$ and height $10.5\ cm$ is full of water. A solid cone of base diameter $7\ cm$ and height $6\ cm$ is completely immersed in water. Find the value of water $(i)$ displaced out of the cylinder $(ii)$ left in the cylinder. $(\text{take}\ \pi=\frac{22}{7})$
AnswerInternal diameters of cylindrical vessel $= 10\ cm$
$\therefore$ Radius $(r) =\Big(\frac{10}{2}\Big)=5\text{cm}$
and height $(h) = 10.5\ cm$
$\therefore$ Volume of water filled in it
$=\pi\text{r}^2\text{h}=\pi\times(5)^2\times10.5\text{cm}^3$
$=\frac{22}{7}\times2.5\times\frac{105}{10}\text{cm}^3=825\text{cm}^3$
Diameter of the cone $= 7\ cm$
$\therefore$ Radius $(r) \frac{7}{2}\text{cm},$ Height $(h_1) = 6\ cm$
$\therefore$ Volume of cone $=\frac{1}{3}\pi\text{r}_1^2\text{h}_1$
$=\frac{1}{3}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times6\text{cm}^3=77\text{cm}^3$
- $\therefore$ Water displaced out of the cylinder $= 77\ cm^3$
- Water left in the vessal $= (825 - 77)\ cm^3= 748\ cm^3$
View full question & answer→Question 473 Marks
A bucket is in the form of a frustum of a cone and holds $15.25$ litres of water. The diameters of the top and bottom are $25\ cm$ and $20\ cm$ respectively. Find its height and area of tin used in its construction.
AnswerSince volume of frustum
$=\frac{\pi\text{h}}{3}(\text{R}^2+\text{Rr}+\text{r}^2)$
$=15250\text{cm}^3$
$\text{h}\times\frac{\pi}{3}\times\Bigg(\Big(\frac{25}{2}\Big)^2+(10)^2+\frac{25}{2}\times10\Bigg)$
$=15250$
$\text{h}\times\frac{\pi}{3}(156.25+22.5)$
$=15250$
$\text{h}=\frac{3\times7\times15250}{22\times381.25}$
$=38.18\text{cm}$
Area of the required
$=\pi(25.5+10)\sqrt{(12.5-10)+38.18}$
$=3017\text{cm}^2$
View full question & answer→Question 483 Marks
Two right circular cylinders of equal volumes have their heights in the ratio $1 : 2.$ What is the ratio of their radii$?$
AnswerLet $r_1$ and $r_2$ be the radii of two right circular cylinders and $h_1$ and $h_2$ be the heights.
Since,
Both the cylinder has the same volume.
Therefore,
$\pi\text{r}_1^2\text{h}_1=\pi\text{r}_2^2\text{h}_2$
$\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\frac{\text{h}_2}{\text{h}_1}$
$(\text{h}_1:\text{h}_2=1:2,\ \text{given})$
$\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\Big(\frac{2}{1}\Big)$
$\text{r}_1:\text{r}_2=\sqrt{2}:1$
View full question & answer→Question 493 Marks
A toy is in the form of a cone of radius $3.5\ cm$ mounted on a hemisphere of same radius. The total height of the toy is $15.5\ cm.$ Find the total surface area of the toy.
AnswerRadius of cone $(r) = 3.5\ cm$
Total height of the toy $= 15.5\ cm$
Height of the conical part $(h) = 15.5 – 3.5 = 12\ cm$
$\therefore$ Slant height of the cone $(l)$
$=\sqrt{\text{r}^2+\text{h}^2}=\sqrt{(3.5)^2+(12)^2}$
$=\sqrt{12.25+144}=\sqrt{156.25}=12.5\text{cm}$
Now total srrface area of the toy $=$ curved surface area of the conical part $+$ curved surface area of hemispherical part.
$=\pi\text{rl}+2\pi\text{r}^2=\pi\text{r}(\text{l}+2\text{r})$
$=\frac{22}{7}\times3.5\ (12.5+2\times3.5)\text{cm}^2$
$=\frac{22}{7}\times\frac{7}{2}(12.5+7)\text{cm}^2$
$=11(19.5)=214.5\text{cm}^2$ View full question & answer→Question 503 Marks
How many spherical lead shots of diameter $4\ cm$ can be made out of a solid cube of lead whose edge measures $44\ cm.$
AnswerDiameter of the spherical lead shots $= 4\ cm$
Edge length of the solid cube $(a) = 44\ cm.$
Let $n$ be the number of spherical lead shots made out of the solid cube.
$n\ ×$ Volume of the spherical lead shots $=$ volume of the solid cube
$\Rightarrow\frac{\text{volume of the solid cube}}{\text{volume of the spherical lead shots}}=\text{n}$
$\Rightarrow\frac{\text{a}^3}{\frac{4}{3}\pi\text{r}^3}=\text{n}$
$\Rightarrow\frac{44^3}{\frac{4}{3}\pi\times\Big(\frac{4}{2}\Big)^3}=\text{n}$
$\Rightarrow2541=\text{n}$
Hence, $2541$ spherical lead shot can be made.
View full question & answer→