NCERT Solutions Class 10 Maths Chapter 13

Q.1 Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Ans.

$\begin{array}{l}\text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}volume of each cube}={\text{64 cm}}^3\\ \text{or}{\left(\text{edge}\right)}^3=\text{64}\\ \text{or edge\hspace{0.17em}of each cube}=4\text{cm}\\ \text{The two cubes are joined end to end to form a cuboid of}\\ \text{dimensions 4 cm, 4 cm and 8 cm.}\\ \therefore \text{surface area of the cuboid}=2(\mathrm{lb}+\mathrm{bh}+\mathrm{lh})\\ =2(4\times 4+4\times 8+4\times 8)\\ =160{\text{cm}}^2\end{array}$

Q.2 A vessel is 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.

Ans.

$\begin{array}{l}\text{Height of cylindrical part}=\mathrm{h}=13-7=6\text{cm}\\ \text{Inner surface area of the vessel}\\ =\mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\text{cylindrical part}\\ -\mathrm{Curved}\text{surface area of hemispherical part}\\ =2\mathrm{\pi rh}+2{\mathrm{\pi r}}^2=2\times \frac{22}{7}\times 7(6+7)=572{\text{cm}}^2\end{array}$

Q.3 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.

Ans.

$\begin{array}{l}\text{Radius of cone}=\text{Radius of hemisphere}=\mathrm{r}=3.5\text{cm}\\ \text{Height of hemisphere}=\text{Radius of hemisphere}=3.5\text{cm}\\ \text{Height of cone}=\mathrm{h}=15.5-3.5=12\\ \text{Slant height of cone}=\mathrm{l}=\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}=\sqrt{{12}^2+{(3.5)}^2}=\frac{25}{2}\text{cm}\\ \text{Total surface area of toy}=\text{CSA of conical part}\\ +\text{CSA of hemispherical part}\\ =\mathrm{\pi rl}+2{\mathrm{\pi r}}^2\\ =\frac{22}{7}\times 3.5\times \frac{25}{2}+2\times \frac{22}{7}\times {(3.5)}^2\\ =214.5{\text{\hspace{0.17em}cm}}^2\end{array}$

Q.4 A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Ans.

$\begin{array}{l}\text{From figure it is obvious that greatest diameter of the}\\ \text{hemisphere is 7 cm.}\\ \therefore \text{Radius of hemisphere}=\mathrm{r}=\frac{7}{2}\text{cm}\\ \text{Total surface area}=\text{Surface area of cubical part}\\ +\text{CSA of hemispherical part}\\ -\text{area of base of hemispherical part}\\ =6\times {\left(7\right)}^2+2{\mathrm{\pi r}}^2-{\mathrm{\pi r}}^2\\ =6\times {\left(7\right)}^2+\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\\ =332.5{\text{cm}}^2\end{array}$

Q.5 A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Ans.

$\begin{array}{l}\text{Diameter of hemisphere}=\text{edge of cube}=\mathrm{l}\\ \therefore \text{Radius of hemisphere}=\frac{\mathrm{l}}{2}\\ \text{Total surface area of solid}\\ =\text{Surface area of cubical part}\\ +\text{CSA of hemispherical part}\\ -\text{area of base of hemispherical part}\\ =6\times {\mathrm{l}}^2+2{\mathrm{\pi r}}^2-{\mathrm{\pi r}}^2\\ =6{\mathrm{l}}^2+\mathrm{\pi }{\frac{\mathrm{l}}{4}}^2\\ =\frac{1}{4}{\mathrm{l}}^2(24+\mathrm{\pi }){\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}unit}}^2\end{array}$

Q.6 A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the following figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.


Ans.

$\begin{array}{l}\text{Surface area of capsule}\\ =2\times \text{CSA of hemispherical part}\\ +\text{CSA of cylindrical part}\\ =4\mathrm{\pi }{\left(\frac{5}{2}\right)}^2+2\mathrm{\pi }\frac{5}{2}\times 9\\ =220{\text{mm}}^2\end{array}$

Q.7 From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Ans.

$\begin{array}{l}\text{Slant height of conical part}=\mathrm{l}=\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}=\sqrt{{(0.7)}^2+{(2.4)}^2}\\ =2.5\text{cm}\\ \text{Required surface area}\\ =\text{curved surface area of cylindrical portion}\\ +\text{curved surface area of conical portion}\\ +\text{\hspace{0.17em}\hspace{0.17em}area of cylindrical base}\\ =\text{2}\mathrm{\pi }\text{rh}+\mathrm{\pi rl}+\mathrm{\pi }{\text{r}}^2\\ =2\times \frac{22}{7}\times 0.7\times 2.4+\frac{22}{7}\times 0.7\times 2.5+\frac{22}{7}\times {(0.7)}^2\\ =17.60{\text{cm}}^2\approx 18{\text{cm}}^2\end{array}$

Q.8 A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the following figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.


Ans.

$\begin{array}{l}\text{Surface area of the given article}\\ =\text{curved surface area of cylindrical portion}\\ +2\times \text{curved surface area of hemispherical portion}\\ =\text{2}\mathrm{\pi }\text{rh}+2\times 2{\mathrm{\pi r}}^2\\ =2\times \frac{22}{7}\times 3.5\times 10+2\times 2\times \frac{22}{7}\times {(3.5)}^2\\ =374{\text{cm}}^2\end{array}$

Q.9 A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Ans.

$\begin{array}{l}\text{Volume of the given solid}=\text{Volume of cone}\\ +\text{Volume of hemisphere}\\ =\frac{1}{3}\times {\mathrm{\pi r}}^2\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^3\\ =\frac{1}{3}\times \mathrm{\pi }{\left(1\right)}^2\left(1\right)+\frac{2}{3}\mathrm{\pi }{\left(1\right)}^3=\mathrm{\pi }{\text{cm}}^3\end{array}$

Q.10 Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Ans.

$\begin{array}{l}\text{Volume of air present in the model}=\text{Volume of cylinder}\\ +2\times \text{Volume of cone}\\ ={\mathrm{\pi r}}^2\mathrm{h}+2\times \frac{1}{3}\times {\mathrm{\pi r}}^2\mathrm{h}\\ =\mathrm{\pi }{\left(\frac{3}{2}\right)}^2\left(8\right)+\frac{2}{3}\mathrm{\pi }{\left(\frac{3}{2}\right)}^2\left(2\right)\\ =66{\text{cm}}^2\end{array}$

Q.11 A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see the following figure)


Ans.

$\begin{array}{l}\text{Volume of one gulab jamun}=\text{Volume of cylinderical part}\\ +2\times \text{Volume of hemispherical part}\\ ={\mathrm{\pi r}}^2\mathrm{h}+2\times \frac{2}{3}\times {\mathrm{\pi r}}^3\\ =\mathrm{\pi }{(1.4)}^2(2.2)+\frac{4}{3}\mathrm{\pi }{(1.4)}^3\\ =25.05{\text{cm}}^3\\ \text{Volume of 45 gulab jamuns}=45\times 25.05{\text{cm}}^3=1,127.25{\text{cm}}^3\\ \text{Volume of sugar syrup}=30\mathrm{\%}\text{of the volume of 45 gulab jamuns}\\ =\frac{30}{100}\times 1,127.25\approx 338{\text{cm}}^3\end{array}$

Q.12 A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the following figure).


Ans.

$\begin{array}{l}\text{Volume of wood}=\text{Volume of cuboid}\\ -4\times \text{Volume cones}\\ =\mathrm{lbh}-4\times \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}\\ =15\times 10\times 3.5-\frac{4}{3}\mathrm{\pi }{\left(0.5\right)}^2\times 1.4\\ =523.53{\text{cm}}^3\end{array}$

Q.13 A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Ans.

$\begin{array}{l}\text{Let n lead shots were dropped in the vessel.}\\ \text{Volume of water spilled}=\text{Volume of dropped lead shots}\\ \text{or}\frac{1}{4}\times \text{Volume cone}=\mathrm{n}\times \frac{4}{3}{{\mathrm{\pi r}}_2}^3\\ \text{or}\frac{1}{4}\times \frac{1}{3}{{\mathrm{\pi r}}_1}^2\mathrm{h}=\mathrm{n}\times \frac{4}{3}{{\mathrm{\pi r}}_2}^3\\ \text{\hspace{0.17em}or}{5}^2\times 8=\mathrm{n}\times 16{(0.5)}^3\\ \text{\hspace{0.17em}or}\mathrm{n}=100\end{array}$

Q.14 A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)

Ans.

$\begin{array}{l}\text{Volume of the iron pole}\\ =\text{Volume of a cylinder of height 220 cm and radius 12 cm}\\ +\text{Volume of a cylinder of height 60 cm and radius 8 cm}\\ =\mathrm{\pi }{\left(12\right)}^2\times 220+\mathrm{\pi }{\left(8\right)}^2\times 60\\ =1,11,532.8{\text{cm}}^3\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Mass of 1 cm}}^3\text{iron}=8\text{gm}\\ \therefore \text{Mass of}1,11,532.8{\text{cm}}^3\text{iron}=8\times 1,11,532.8\text{gm}=892.262\text{kg}\end{array}$

Q.15 A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Ans.

$\begin{array}{l}\text{Volume of water left}\\ =\text{Volume of a cylinder}-\text{Volume of a soloid}\\ =\text{Volume of a cylinder}-\text{Volume of cone}-\text{Volume of hemisphere}\\ ={\mathrm{\pi r}}^2{\mathrm{h}}_1-\left(\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^3\right)\\ =\mathrm{\pi }{\left(60\right)}^2\times 180-\left[\frac{1}{3}\mathrm{\pi }{\left(60\right)}^2\times 120+\frac{2}{3}\mathrm{\pi }{\left(60\right)}^3\right]\\ =\mathrm{\pi }{\left(60\right)}^2\times 180-\frac{1}{3}\mathrm{\pi }{\left(60\right)}^2\times 120-\frac{2}{3}\mathrm{\pi }{\left(60\right)}^3\\ =\frac{1}{3}\mathrm{\pi }{\left(60\right)}^2\left(540-120-120\right)\\ =\frac{1}{3}\times \frac{22}{7}\times 3600\times 300\\ =\frac{22}{7}\times 360000{\text{cm}}^3\\ =\frac{22}{7}\times 0.36{\text{m}}^3\\ =1.131{\text{m}}^3\text{(approx.)}\end{array}$

Q.16 A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Ans.

$\begin{array}{l}\text{Volume of vessel}\\ =\text{Volume of sphere}+\text{Volume of cylinder}\\ =\frac{4}{3}{{\mathrm{\pi r}}_1}^3+{{\mathrm{\pi r}}_2}^2\mathrm{h}\\ =\frac{4}{3}\mathrm{\pi }{\left(4.25\right)}^3+\mathrm{\pi }{\left(1\right)}^2\times 8\\ =346.51{\text{cm}}^3\text{(approx.)}\\ \ne 345\\ \text{Hence the child is not correct.\hspace{0.17em}\hspace{0.17em}}\end{array}$

Q.17 A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Ans.

$\begin{array}{l}\text{Radius of sphere}={\mathrm{r}}_1=4.2\text{cm}\\ \text{Radius of cylinder}={\mathrm{r}}_2=6\text{cm}\\ \text{Height of cylinder}=\mathrm{h}\\ \text{Recasted cylinder and sphere will be of same volume.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of cylinder}=\text{Volume of sphere}\\ \text{or}{{\mathrm{\pi r}}_2}^2\mathrm{h}=\frac{4}{3}{{\mathrm{\pi r}}_1}^3\\ \text{or}{6}^2\mathrm{h}=\frac{4}{3}{(4.2)}^3\\ \text{or}\mathrm{h}=2.74\text{cm\hspace{0.17em}}\end{array}$

Q.18 Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Ans.

$\begin{array}{l}\text{Radius of 1st sphere}={\mathrm{r}}_1=6\text{cm}\\ \text{Radius of 2nd sphere}={\mathrm{r}}_2=8\text{cm}\\ \text{Radius of 3rd sphere}={\mathrm{r}}_3=10\text{cm}\\ \text{Let radius of the resulting sphere be}\mathrm{r}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of the resulting sphere}\\ =\text{Sum of volumes of the given 3 spheres}\\ \text{or}\frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}{{\mathrm{\pi r}}_1}^3+\frac{4}{3}{{\mathrm{\pi r}}_2}^3+\frac{4}{3}{{\mathrm{\pi r}}_3}^3\\ \text{or}{\mathrm{r}}^3={{\mathrm{r}}_1}^3+{{\mathrm{r}}_2}^3+{{\mathrm{r}}_3}^3=6^3+8^3+{\left(10\right)}^3\\ \text{or}\mathrm{r}=12\text{cm\hspace{0.17em}}\end{array}$

Q.19 A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Ans.

$\begin{array}{l}\text{Depth of the well}=\mathrm{d}=20\text{m}\\ \text{Radius of the well}=\mathrm{r}=\frac{7}{2}\text{m}\\ \text{Length of the platform}=\mathrm{l}=22\text{m}\\ \text{Width of the platform}=\mathrm{b}=14\text{m}\\ \text{Let height of the platform be}\mathrm{h}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of the platform}\\ =\text{Volume of the well}\\ \text{or}\mathrm{lbh}={\mathrm{\pi r}}^2\mathrm{d}\\ \text{or}22\times 14\mathrm{h}=\mathrm{\pi }{\left(\frac{7}{2}\right)}^2\times 20\\ \text{or}\mathrm{h}=2.5\text{m\hspace{0.17em}}\end{array}$

Q.20 A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Ans.

$\begin{array}{l}\text{Depth of the well}=\mathrm{d}=14\text{m}\\ \text{Radius of the well}=\mathrm{r}=\frac{3}{2}\text{m}=1.5\text{\hspace{0.17em}\hspace{0.17em}m}\\ \text{Width of the circular ring}=\mathrm{b}=4\text{m}\\ \text{Radius of the inner circle of the circular ring}\\ =\text{Radius of the well}=\mathrm{r}=\frac{3}{2}\text{m}=1.5\text{\hspace{0.17em}\hspace{0.17em}m}\\ \text{Radius of the outer circle of the circular ring}\\ ={\mathrm{r}}_1=\text{Radius of the well\hspace{0.17em}}+\text{Width of the circular ring}\\ =\frac{3}{2}+4=5.5\text{m}\\ \text{Shape of the embankment is cylindrical.}\\ \text{Let height of the embankment be}\mathrm{h}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of the embankment}\\ =\text{Volume of the well}\\ \text{or}{{\mathrm{\pi r}}_1}^2\mathrm{h}-{\mathrm{\pi r}}^2\mathrm{h}={\mathrm{\pi r}}^2\mathrm{d}\\ \text{or}{(5.5)}^2-{(1.5)}^2]\mathrm{h}={(1.5)}^2\times 14\\ \text{or}\mathrm{h}=1.125\text{m\hspace{0.17em}}\end{array}$

Q.21 A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Ans.

$\begin{array}{l}\text{Height of cylindrical container}=\mathrm{h}=15\text{cm}\\ \text{Radius of cylindrical container}=\mathrm{r}=\frac{12}{2}\text{cm}=6\text{cm}\\ \text{Height of cone}={\mathrm{h}}_1=12\text{cm}\\ \text{Radius of cone}={\mathrm{r}}_1=\frac{6}{2}\text{cm}=3\text{cm}\\ \text{Radius of hemisphere}=\text{Radius of cone}={\mathrm{r}}_1=3\text{cm}\\ \text{Volume of one ice-cream cone}\\ =\text{Volume of conical part}+\text{Volume of hemispherical part}\\ =\frac{1}{3}{{\mathrm{\pi r}}_1}^2{\mathrm{h}}_1+2{{\mathrm{\pi r}}_1}^2\\ =\frac{1}{3}\mathrm{\pi }\times 3^2\times 12+2\mathrm{\pi }\times 3^2=54\mathrm{\pi }\\ \text{Let the number of cones be}\mathrm{n}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of cylindrical container}\\ =\mathrm{n}\times \text{Volume of one ice-cream cone}\\ \text{or}{\mathrm{\pi r}}^2\mathrm{h}=\mathrm{n}\times 54\mathrm{\pi }\\ \text{or}{6}^2\times 15=54\mathrm{n}\\ \text{or}\mathrm{n}=10\end{array}$

Q.22 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Ans.

$\begin{array}{l}\text{Thickness of coin}=\mathrm{h}=2\text{mm}=0.2\text{cm}\\ \text{Radius of coin}=\mathrm{r}=\frac{1.75}{2}\text{cm}\\ \text{Cuboid is of the dimensions 5.5 cm × 10 cm × 3.5 cm.}\\ \text{Let number of coines be}\mathrm{n}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of cuboid}\\ =\mathrm{n}\times \text{Volume of one coin}\\ \text{or 5.5 cm × 10 cm × 3.5 cm}=\mathrm{n}\times {\mathrm{\pi r}}^2\mathrm{h}=\mathrm{n\pi }{\left(\frac{1.75}{2}\right)}^2(0.2)\\ \text{or}\mathrm{n}=400\end{array}$

Q.23 A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Ans.

$\begin{array}{l}\text{Height of cylindrical bucket}={\mathrm{h}}_1=32\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Radius of circular end of bucket}={\mathrm{r}}_1=\frac{18}{2}\text{cm}=9\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Height of conical heap}={\mathrm{h}}_2=24\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Let radius of circular end of conical heap\hspace{0.17em}\hspace{0.17em}be}{\mathrm{r}}_2\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume of sand in the cylindrical bucket}\\ =\text{Volume of sand in conical heap}\\ \text{or}{{\mathrm{\pi r}}_1}^2{\mathrm{h}}_1=\frac{1}{3}{{\mathrm{\pi r}}_2}^2{\mathrm{h}}_2\\ \text{or}{\left(\text{18}\right)}^2\times 32=\frac{1}{3}{{\mathrm{r}}_2}^2\times 24\\ \text{or}{\mathrm{r}}_2=36\text{cm}\\ \text{Slant height}=\sqrt{{24}^2+{36}^2}=12\sqrt{13}\text{cm}\\ \text{So, radius of circular end of conical heap}=36\text{cm}\\ \text{and slant height of conical heap}=12\sqrt{13}\text{cm}\end{array}$

Q.24 Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Ans.

$\begin{array}{l}\text{Canal is 6 m wide and 1.5 m deep.}\\ \therefore \text{Cross sectional area of the canal}=\text{6}\times 1.5=9{\text{m}}^2\\ \text{Speed of the water}=\text{10 km/h}=\frac{10,000}{60}\text{m/minute}\\ =\frac{500}{3}\text{\hspace{0.17em}\hspace{0.17em}m/minute}\\ \text{Volume of water flown out of canal in 1 minute}=\frac{500}{3}\times 9=1500{\text{m}}^3\\ \text{Volume of water flown out of canal in 30 minutes}=1500\times 30\\ =45000{\text{m}}^3\\ \text{Let the canal irrigate area A with 8 cm of standing water in}\\ \text{30 minutes.}\\ \text{So,}\\ \text{Volume of water flown out of canal in 30 minutes}=\text{A}\times {\text{0.08 m}}^3\\ \text{or,}45000{\text{m}}^3=\text{A}\times {\text{0.08 m}}^3\\ \text{or, A}=\frac{45000}{\text{0.08}}=562500{\text{m}}^2\\ \text{Therefore, area irrigated in 30 minutes is}562500{\text{m}}^2.\end{array}$

Q.25 A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Ans.

$\begin{array}{l}\text{Internal diameter of pipe}=\text{20 cm}\\ \therefore \text{Internal radius of pipe}=\frac{\text{20}}{2}\text{cm}=10\text{cm}=0.1\text{m}\\ \therefore \text{Cross sectional area of pipe}=\mathrm{\pi }\times {(0.1)}^2=0.01\mathrm{\pi }{\text{m}}^2\\ \text{Speed of the water}=\text{3 km/h}=\frac{3000}{60}\text{m/minute}\\ =50\text{\hspace{0.17em}\hspace{0.17em}m/minute}\\ \text{Volume of water flown out of pipe in 1 minute}=50\times 0.01\mathrm{\pi }=\frac{\mathrm{\pi }}{2}{\text{m}}^3\\ \text{Cylindrical tank is 2 m deep and 10 m in diameter.}\\ \text{Volume of cylindrical tank}={\mathrm{\pi r}}^2\mathrm{h}=50\mathrm{\pi }{\text{\hspace{0.17em}\hspace{0.17em}m}}^3\\ \text{Let the pipe fills the tank in}\mathrm{t}\text{minutes.}\\ \text{So,}\\ \text{Volume of water flown out of pipe in}\mathrm{t}\text{minutes}\\ =\text{Volume of cylindrical tank}\\ \text{or,}\frac{\mathrm{\pi }}{2}\mathrm{t}{\text{m}}^3=50\mathrm{\pi }{\text{\hspace{0.17em}\hspace{0.17em}m}}^3\\ \text{or,}\mathrm{t}=100\text{minutes}\\ \text{Therefore, pipe fills the tank in 100 minutes.}\end{array}$

Q.26 A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Ans.

$\begin{array}{l}\text{Radius of upper base of glass}={\mathrm{r}}_1=\frac{4}{2}=2\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Radius of lower base of glass}={\mathrm{r}}_2=\frac{2}{2}=1\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \mathrm{Capacity}\text{of glass}=\mathrm{Volume}\text{of frustum of cone}\\ =\frac{1}{3}\mathrm{\pi h}({{\mathrm{r}}_1}^2+{{\mathrm{r}}_2}^2+{\mathrm{r}}_1{\mathrm{r}}_2)\\ =\frac{14}{3}\mathrm{\pi }(2^2+1^2+2\times 1)=102\frac{2}{3}{\text{cm}}^3\end{array}$

Q.27 The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Ans.

$\begin{array}{l}\text{S}=\mathrm{l}=\\ \text{P i.e.,}\\ \text{2}\mathrm{\pi }{\text{r}}_1=18\text{cm and 2}\mathrm{\pi }{\text{r}}_2=6\text{cm}\\ {\text{i.e., r}}_1=\frac{9}{\mathrm{\pi }}{\text{cm and r}}_2=\frac{3}{\mathrm{\pi }}\text{cm}\\ \text{C}=\mathrm{\pi }\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)\mathrm{l}\\ =\mathrm{\pi }\left(\frac{9}{\mathrm{\pi }}+\frac{3}{\mathrm{\pi }}\right)\times \\ =48{\text{cm}}^2\end{array}$

Q.28 A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the following figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.


Ans.

$\begin{array}{l}\text{R}\mathbf{adius}\mathbf{on}\mathbf{the}\mathbf{open}\mathbf{side}={\mathrm{r}}_1=\mathbf{10}\mathbf{cm}\\ \text{R}\mathbf{adius}\mathbf{at}\mathbf{the}\mathbf{upper}\mathbf{base}={\mathrm{r}}_2=\mathbf{4}\mathbf{cm}\\ \text{S}\mathbf{lant}\mathbf{height}\mathbf{of}\mathbf{frustum}=\mathrm{l}=15\mathbf{cm}\\ \text{A}\mathbf{rea}\mathbf{of}\mathbf{material}\mathbf{used}\mathbf{for}\mathbf{making}\text{fez}\\ =\text{CSA}\mathbf{of}\mathbf{frustum}+\text{area of the}\mathbf{upper}\mathbf{base}\\ =\mathrm{\pi }({\mathrm{r}}_1+{\mathrm{r}}_2)\mathrm{l}+{{\mathrm{\pi r}}_2}^2=\mathrm{\pi }(10+4)\times 15+\mathrm{\pi }{4}^2=710\frac{2}{7}{\text{cm}}^2\end{array}$

Q.29

$\begin{array}{l}\text{A container, opened from the top and made up of a metal sheet,}\\ \text{is in the form of a frustum of a cone of height 16 cm with radii of its}\\ \text{lower and upper ends as 8 cm and 20 cm, respectively.}\\ \text{Find the cost of the milk which can completely fill the container, at}\\ \text{the rate of ₹ 20 per litre. Also find the cost of metal sheet used to}\\ {\text{make the container, if it costs ₹\hspace{0.17em}8 per 100 cm}}^{\text{2}}\text{. (Take}\mathrm{\pi }\text{= 3.14)}\end{array}$

Ans.

$\begin{array}{l}\text{Height of frustum=h=16 cm}\\ {\text{Radius of lower end\hspace{0.17em}=r}}_{\text{1}}\text{=8 cm}\\ {\text{Radius of upper end=r}}_{\text{2}}\text{=20 cm}\\ \text{Slant height of frustum=l=}\sqrt{{\left({\text{r}}_{\text{1}}{\text{-r}}_{\text{2}}\right)}^{\text{2}}{\text{+h}}^{\text{2}}}\\ \text{=}\sqrt{{\left(\text{8-20}\right)}^{\text{2}}{\text{+16}}^{\text{2}}}\text{\hspace{0.17em}=20 cm}\\ \text{Capacity of container=Volume of frustum}\\ \text{=}\frac{\text{1}}{\text{3}}{{\text{πh (r}}^{\text{2}}}_{\text{1}}{{\text{+r}}_{\text{2}}}^{\text{2}}{\text{+r}}_{\text{1}}{\text{r}}_{\text{2}}\text{)}\\ \text{=}\frac{\text{16}}{\text{3}}{\text{π (8}}^{\text{2}}{\text{+20}}^{\text{2}}\text{+8×20)}\\ \text{=10.45 litres}\\ \text{Cost of 10.45 litre milk=20×10.45= ₹ 209}\\ \text{Area of metal sheet used to make the container}\\ {\text{=π(r}}_{\text{1}}{\text{+r}}_{\text{2}}{{\text{)l+πr}}_{\text{1}}}^{\text{2}}\\ {\text{=π(8+20)×20+π×8}}^{\text{2}}\\ {\text{=624π cm}}^{\text{2}}\\ {\text{Cost of 624π cm}}^{\text{2}}\text{metal sheet=624×3.14×}\frac{\text{8}}{\text{100}}\text{=₹156.75}\end{array}$

Q.30 A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{AEO,}\\ \frac{\mathrm{EO}}{\mathrm{AO}}=\tan 30\mathrm{°}\\ \text{or}\mathrm{EO}=\frac{1}{\sqrt{3}}\mathrm{AO}=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}\text{cm}\\ \text{In}\mathrm{\Delta }\text{ABD,}\\ \frac{\mathrm{BD}}{\mathrm{AD}}=\tan 30\mathrm{°}\\ \text{or}\mathrm{BD}=\frac{1}{\sqrt{3}}\mathrm{AD}=\frac{20}{\sqrt{3}}=\frac{20\sqrt{3}}{3}\text{cm}\\ \text{R}\mathrm{adius}\text{of upper end of frustum}={\mathrm{r}}_1=\frac{10\sqrt{3}}{3}\mathrm{cm}\\ \text{R}\mathrm{adius}\text{of lower end of frustum}={\mathrm{r}}_2=\frac{20\sqrt{3}}{3}\mathrm{cm}\\ \text{Height of frustum}=\mathrm{h}=10\text{cm}\end{array}$

$\begin{array}{l}\text{Volume of}\mathrm{frustum}=\frac{1}{3}\mathrm{\pi h}({{\mathrm{r}}^2}_1+{{\mathrm{r}}_2}^2+{\mathrm{r}}_1{\mathrm{r}}_2)\\ =\frac{10}{3}\mathrm{\pi }({\left(\frac{10\sqrt{3}}{3}\right)}^2+{\left(\frac{20\sqrt{3}}{3}\right)}^2+\frac{10\sqrt{3}}{3}\times \frac{20\sqrt{3}}{3})\\ =\frac{22000}{9}{\text{cm}}^3\\ \text{R}\mathrm{adius}\text{of wire}=\mathrm{r}=\frac{1}{16}\times \frac{1}{2}=\frac{1}{32}\text{cm}\\ \text{Let length of wire be}\mathrm{l}.\\ \text{Volume of wire}=\text{Area of cross section}\times \text{length}=\mathrm{\pi }{\left(\frac{1}{32}\right)}^2\mathrm{l}\\ \text{\hspace{0.17em} Volume of}\mathrm{frustum}=\text{Volume of wire}\\ \text{or}\frac{22000}{9}=\frac{22}{7}\times {\left(\frac{1}{32}\right)}^2\mathrm{l}\\ \text{or}\mathrm{l}=7964.44\text{m}\end{array}$

Q.31 A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.

Ans.

$\begin{array}{l}\mathrm{One}\text{round of wire covers 3 mm height of cylinder.}\\ \text{Number of rounds}=\frac{\mathrm{Height}\text{of cylinder}}{\mathrm{Diameter}\mathrm{of}\mathrm{wire}}=\frac{12\text{cm}}{3\text{mm}}=\frac{12\text{cm}}{0.3\text{cm}}=40\\ \mathrm{Length}\text{of wire required in one round}=\mathrm{Circumference}\text{of cylinder}\\ =2\mathrm{\pi r}=2\mathrm{\pi }\times \frac{10}{2}=10\mathrm{\pi }\text{cm}\\ \text{Length of wire in 40 rounds}=40\times 10\mathrm{\pi }=40\times 10\times 3.14=1256\text{\hspace{0.17em}​\hspace{0.17em}}\mathrm{c}\text{m}\\ \text{Radius of wire}=\frac{0.3}{2}\text{cm}=0.15\text{cm}\\ \mathrm{Volume}\text{of wire}=\text{Area of cross section of wire}\times \mathrm{Length}\text{of wire}\\ =3.14\times {\left(0.15\right)}^2\times 1256\mathrm{c}{\text{m}}^3=88.7364{\text{cm}}^3\\ \mathrm{Mass}=\text{Volume}\times \text{Density}=88.7364\times 8.88=787.979232\text{gm = 788 gm (approx.)}\end{array}$

Q.32 A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)

Ans.

$\begin{array}{l}\text{The double cone formed by revolving right-angled}\mathrm{\Delta }\text{ABC}\\ \text{about hypotenus AC is shown above.}\\ \text{AC}=\sqrt{3^2+4^2}=5\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area of}\mathrm{\Delta }\text{ABC}=\frac{1}{2}\times \mathrm{AB}\times \mathrm{BC}=\frac{1}{2}\times 3\times 4=6{\text{cm}}^2\\ \text{or,}\frac{1}{2}\times \mathrm{AC}\times \mathrm{OB}=6\\ \text{or,}\frac{1}{2}\times 5\times \mathrm{OB}=6\\ \text{or,}\mathrm{OB}=\frac{12}{5}=2.4\text{cm}\\ \text{Volume of double cone}\\ =\text{Volume of cone\hspace{0.17em}1}+\text{Volume of cone 2}\\ =\frac{1}{3}{\mathrm{\pi r}}^2{\mathrm{h}}_1+\frac{1}{3}{\mathrm{\pi r}}^2{\mathrm{h}}_2\\ =\frac{1}{3}\mathrm{\pi }{(2.4)}^2({\mathrm{h}}_1+{\mathrm{h}}_2)=\frac{1}{3}\mathrm{\pi }{(2.4)}^2\times 5=30.14{\text{cm}}^3\\ \text{Surface area of double cone}=\text{Surface area of cone 1}\\ +\text{Surface area of cone 2}\\ ={\mathrm{\pi rl}}_1+{\mathrm{\pi rl}}_2\\ =3.14\times 2.4(3+4)=52.75{\text{cm}}^2\end{array}$

Q.33 A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Ans.

$\begin{array}{l}\text{Volume of cistern}=150\times 120\times 110=1980000{\text{cm}}^3\\ \text{Volume to be filled in cistern}=1980000-129600=1850400{\text{cm}}^3\\ \text{Let}\mathrm{n}\text{number of porous bricks were placed in cistern.}\\ \text{So, volume of}\mathrm{n}\text{bricks}=\mathrm{n}\times 22.5\times 7.5\times 6.5=1096.875\mathrm{n}\\ \text{Each brick absorbs one-seventeenth of its volume.}\\ \therefore \text{Volume absorbed by}\mathrm{n}\text{bricks}=\frac{\mathrm{n}}{17}(1096.875)\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Volume to be filled in cistern}+\text{Volume absorbed by}\mathrm{n}\text{bricks}\\ =\text{Volume of}\mathrm{n}\text{bricks}\\ \text{or,}1850400+\frac{\mathrm{n}}{17}(1096.875)=1096.875\mathrm{n}\\ \text{or,}\mathrm{n}=1792.41\\ \text{So, 1792 bricks can be put in the given cistern.}\end{array}$

Q.34 In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Ans.

$\begin{array}{l}\text{Rainfall=10 cm=0.1m}\\ {\text{Areaofthevalley=7280km}}^{\text{2}}\\ {\text{Vaolumeofrainfallinthevalley=7280×1000×1000×0.1m}}^{\text{3}}\\ {\text{=728000000 m}}^{\text{3}}\\ {\text{Volumeofwaterin3revers=3×1072000×75×3=723600000 m}}^{\text{3}}\\ \text{Thus,thetotalrainfallisapproximatelysameasthe}\mathrm{total}\\ \text{volumeof thethreerivers.}\end{array}$

Q.35 An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the following figure).


Ans.

$\begin{array}{l}\text{Height of frustum}={\mathrm{h}}_1=22-10=12\text{cm}\\ \text{Height of cylindrical part}={\mathrm{h}}_2=10\text{cm}\\ \text{R of lower end}={\mathrm{r}}_1=\frac{8}{2}=4\text{cm}\\ \text{R of end}={\mathrm{r}}_2=\frac{18}{2}=9\text{cm}\\ \text{S}=\mathrm{l}=\sqrt{{\left({\mathrm{r}}_1-{\mathrm{r}}_2\right)}^2+{{\mathrm{h}}_1}^2}\\ =\sqrt{{\left(4-9\right)}^2+{12}^2}=13\\ \mathrm{Areaoftinsheetrequired}\\ =\text{CSA of frustum part}+\text{CSA of cylindrical part}\\ =\mathrm{\pi }\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)\mathrm{l}+2{\mathrm{\pi r}}_1{\mathrm{h}}_2\\ =\frac{22}{7}\left(4+9\right)\times 13+2\times \frac{22}{7}\times 4\times 10\\ =782\frac{4}{7}{\text{cm}}^2\end{array}$

Q.36 Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Ans.

$\begin{array}{l}\text{Let ABC be a cone. A frustum DECB is cut by a plane parallel}\\ \text{to its base.}\\ \text{In}\mathrm{\Delta }\text{ADF and}\mathrm{\Delta }\text{ABG,}\\ \frac{\mathrm{DF}}{\mathrm{BG}}=\frac{\mathrm{AF}}{\mathrm{AG}}=\frac{\mathrm{AD}}{\mathrm{AB}}\\ \text{or}\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=\frac{{\mathrm{h}}_1-\mathrm{h}}{{\mathrm{h}}_1}=\frac{{\mathrm{l}}_1-\mathrm{l}}{{\mathrm{l}}_1}\\ \text{or}\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=1-\frac{\mathrm{h}}{{\mathrm{h}}_1}=1-\frac{\mathrm{l}}{{\mathrm{l}}_1}\\ \text{or}1-\frac{\mathrm{l}}{{\mathrm{l}}_1}=\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\\ \text{or}\frac{\mathrm{l}}{{\mathrm{l}}_1}=1-\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=\frac{{\mathrm{r}}_1-{\mathrm{r}}_2}{{\mathrm{r}}_1}\\ \text{or}\frac{{\mathrm{l}}_1}{\mathrm{l}}=\frac{{\mathrm{r}}_1}{{\mathrm{r}}_1-{\mathrm{r}}_2}\end{array}$

$\begin{array}{l}\text{CSA of frustum DECB}=\text{CSA of cone ABC}-\text{CSA of cone ADE}\\ ={\mathrm{\pi r}}_1{\mathrm{l}}_1-{\mathrm{\pi r}}_2({\mathrm{l}}_1-\mathrm{l})\\ ={\mathrm{\pi r}}_1\left(\frac{{\mathrm{lr}}_1}{{\mathrm{r}}_1-{\mathrm{r}}_2}\right)-{\mathrm{\pi r}}_2(\frac{{\mathrm{lr}}_1}{{\mathrm{r}}_1-{\mathrm{r}}_2}-\mathrm{l})\\ =\frac{{{\mathrm{\pi r}}_1}^2\mathrm{l}}{{\mathrm{r}}_1-{\mathrm{r}}_2}-{\mathrm{\pi r}}_2\left(\frac{{\mathrm{lr}}_2}{{\mathrm{r}}_1-{\mathrm{r}}_2}\right)\\ =\frac{{{\mathrm{\pi r}}_1}^2\mathrm{l}}{{\mathrm{r}}_1-{\mathrm{r}}_2}-\frac{{{\mathrm{\pi r}}_2}^2\mathrm{l}}{{\mathrm{r}}_1-{\mathrm{r}}_2}=\mathrm{\pi }({\mathrm{r}}_1+{\mathrm{r}}_2)\mathrm{l}\\ \text{CSA of frustum}=\mathrm{\pi }({\mathrm{r}}_1+{\mathrm{r}}_2)\mathrm{l}\\ \text{Total surface area of frustum}=\text{CSA of frustum}+\\ +\text{CSA of upper circular end}\\ +\text{CSA of lower circular end}\\ =\mathrm{\pi }({\mathrm{r}}_1+{\mathrm{r}}_2)\mathrm{l}+{{\mathrm{\pi r}}_1}^2+{{\mathrm{\pi r}}_2}^2\\ =\mathrm{\pi }[({\mathrm{r}}_1+{\mathrm{r}}_2)\mathrm{l}+{{\mathrm{r}}_1}^2+{{\mathrm{r}}_2}^2]\end{array}$

Q.37 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Ans.

$\begin{array}{l}\text{Let ABC be a cone. A frustum DECB is cut by a plane parallel}\\ \text{to its base.}\\ \text{In}\mathrm{\Delta }\text{ADF and}\mathrm{\Delta }\text{ABG,}\\ \frac{\mathrm{DF}}{\mathrm{BG}}=\frac{\mathrm{AF}}{\mathrm{AG}}=\frac{\mathrm{AD}}{\mathrm{AB}}\\ \text{or}\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=\frac{{\mathrm{h}}_1-\mathrm{h}}{{\mathrm{h}}_1}=\frac{{\mathrm{l}}_1-\mathrm{l}}{{\mathrm{l}}_1}\\ \text{or}\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=1-\frac{\mathrm{h}}{{\mathrm{h}}_1}=1-\frac{\mathrm{l}}{{\mathrm{l}}_1}\\ \text{or \hspace{0.17em}}\frac{\mathrm{h}}{{\mathrm{h}}_1}=\frac{{\mathrm{r}}_1-{\mathrm{r}}_2}{{\mathrm{r}}_1}\end{array}$

$\begin{array}{l}\text{Volume of frustum DECB}=\text{Volume of cone ABC}-\text{\hspace{0.17em}Volume of cone ADE}\\ =\frac{1}{3}{{\mathrm{\pi r}}_1}^2{\mathrm{h}}_1-\frac{1}{3}{{\mathrm{\pi r}}_2}^2({\mathrm{h}}_1-\mathrm{h})\\ =\frac{\mathrm{\pi }}{3}[{{\mathrm{r}}^2}_1{\mathrm{h}}_1-{{\mathrm{r}}_2}^2({\mathrm{h}}_1-\mathrm{h})]\\ =\frac{\mathrm{\pi }}{3}[{{\mathrm{r}}^2}_1\frac{{\mathrm{r}}_1\mathrm{h}}{{\mathrm{r}}_1-{\mathrm{r}}_2}-{{\mathrm{r}}_2}^2(\frac{{\mathrm{r}}_1\mathrm{h}}{{\mathrm{r}}_1-{\mathrm{r}}_2}-\mathrm{h})]\\ =\frac{\mathrm{\pi }}{3}[{{\mathrm{r}}^2}_1\frac{{\mathrm{r}}_1\mathrm{h}}{{\mathrm{r}}_1-{\mathrm{r}}_2}-{{\mathrm{r}}_2}^2\left(\frac{{\mathrm{r}}_2\mathrm{h}}{{\mathrm{r}}_1-{\mathrm{r}}_2}\right)]\\ =\frac{\mathrm{\pi h}}{3}\left[\frac{{{\mathrm{r}}_1}^3-{{\mathrm{r}}_2}^3}{{\mathrm{r}}_1-{\mathrm{r}}_2}\right]=\frac{\mathrm{\pi h}}{3}({{\mathrm{r}}^2}_1+{{\mathrm{r}}^2}_2+{\mathrm{r}}_1{\mathrm{r}}_2)\end{array}$

Q.38 A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)

Ans.

$\begin{array}{l}\text{Total canvas used}=\text{Total curved surface area of tent}\\ \text{Total curved surface area of tent}\\ =\text{curved surface area of cylindrical portion}\\ +\text{curved surface area of conical portion}\\ =\text{2}\mathrm{\pi }\text{rh}+\mathrm{\pi rl}\\ =2\times \frac{22}{7}\times 2\times 2.1+\frac{22}{7}\times 2\times 2.8\\ =44{\text{m}}^2\\ \mathrm{Cost}{\text{of 44 m}}^2\text{convas}=44\times 500=₹22000\end{array}$