Rd Sharma 2020 for Class 10 Math Chapter 14 - Surface Areas And Volumes

Answer:

We are given a solid sphere with radius cm.
From this sphere we have to make spherical balls of radius cm.

Let the no. of balls that can be formed be.

We know,
Volume of a sphere =.
So, volume of the bigger solid sphere= …… (a)

Volume of one smaller spherical ball = …… (b)

We know, the volume of the solid sphere should be equal to the sum of the volumes of the n spherical balls formed.

So, using (a) and (b), we get,

Therefore,

Hence, the no. of balls of radius that can be formed out of solid sphere of radius is 512.

Answer:

We are given a metallic block of dimension

We know that,

So, the volume of the given metallic block is

We want to know how many spherical bullets can be formed from this volume of the metallic block. It is given that the diameter of each bullet should be 5 cm.
We know,

Volume of a sphere

Here,

Let the no. of bullets formed be n.
We know that the sum of the volumes of the bullets formed should be equal to the volume of the metallic block.

Hence the no. of bullets that can be formed is 8400.

Answer:

We have one spherical ball of radius 3 cm

So, its volume …… (a)

It is melted and made into 3 balls.
The first ball has radius 1.5 cm

So, its volume …… (b)

The second ball has radius 2 cm

So, its volume …… (c)

We have to find the radius of the third ball.
Let the radius of the third ball be

The volume of this third ball …… (d)

We know that the sum of the volumes of the 3 balls formed should be equal to the volume of the given spherical ball.

Using equations (a), (b), (c) and (d)

Hence the diameter of the third ball should be 5 cm

Answer:

The brass volume that has to be drawn into a cylindrical wire is given is

We have to make a cylindrical wire out of it with diameter =0.25 cm

So the radius of this wire

We have to find the length of this wire.
Let the length of this wire be

We know that the volume of a cylinder.
We know, the volume of the cylinder should be equal to the volume of the given brass

Therefore, h = 448 m

Hence, the length of the cylindrical wire that can be formed is 448 m

Answer:

We are given a solid cylinder of, diameter = 2 cm

We have to recast it into a hollow cylinder of length = 16 cm

External Diameter = 20 cm and thickness = 2.5 mm=0.25 cm

We have to find the height of the solid cylinder that can be used to get a hollow cylinder of the desired dimensions.

Volume of a solid cylinder

So,

The volume of the given solid cylinder …… (a)

Here, height has to be found.

Volume of a hollow cylinder

Whereis the external radius and is the internal radius

External radius is given. Thickness of the hollow cylinder is also given.
So, we can find the internal radius of the hollow cylinder.
Thickness=R−r

So, the volume of the hollow cylinder…… (b)

From (a) and (b) we get,

Hence, the required height of the solid cylinder is h=79 cm

Answer:

A cylindrical vessel whose height is equal to its diameter is given.

It is filled with water.
We know that the volume of a cylinder

In this particular case,

Height is equal to the diameter, that is ,

The volume of cylindrical vessel becomes

The water from this vessel is transferred into two identical cylindrical vessels of

Diameter = 42 cm and, height h= 21 cm

Volume of each vessel

We know that the sum of the volumes of the two identical vessels must be equal to the volume of the given cylindrical vessel.

Therefore,

The diameter of the given cylinder is 42 cm

Answer:

We have 50 circular plates, each with diameter = 14 cm

That is, radius = 7 cm and thickness = 0.5 cm

These plates are stacked on top of one another.
So, the total thickness

This is clearly a cylindrical arrangement.
We know,

Total surface area of a cylinder

So, the total surface area of the given arrangement is

Answer:

We 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

Volume of this arrangement

Curved surface area

Hence and

Answer:

Given the diameter of the base of the the circular disc = 1.5 cm
Height = 0.2 cm
Volume of the circular disc = ${\mathrm{\pi r}}^2\mathrm{h}=\mathrm{\pi }\times {\left(\frac{1.5}{2}\right)}^2\times 0.2=\mathrm{\pi }\times {\left(0.75\right)}^2\times 0.2$              ...(i)
Height of the cylinder = 10 cm
Diameter = 4.5 cm
Volume of the cylinder = ${\mathrm{\pi R}}^2\mathrm{H}=\mathrm{\pi }{\left(\frac{4.5}{2}\right)}^2\times 10=\mathrm{\pi }\times {\left(2.25\right)}^2\times 10 ...\left(\mathrm{ii}\right)$
Now since the circular discs are used to make the cylinder so, let n be the number of circular discs required.
$$\begin{gathered} n\times \mathrm{Volume} \mathrm{of} \mathrm{circular} \mathrm{disc}=\mathrm{Volume} \mathrm{of} \mathrm{cylinder} \\ \Rightarrow \frac{\mathrm{Volume} \mathrm{of} \mathrm{cylinder}}{\mathrm{Volume} \mathrm{of} \mathrm{circular} \mathrm{disc}}=n \\ \Rightarrow \frac{\mathrm{\pi }\times {\left(2.25\right)}^2\times 10}{\mathrm{\pi }\times {\left(0.75\right)}^2\times 0.2}=n \\ \Rightarrow n=450 \end{gathered}$$
Hence, 450 metallic circular discs need to be melted to form the right circular cylinder.


 

Answer:

The dimensions of the solid rectangular lead piece is $66 \mathrm{cm}\times 42 \mathrm{cm}\times 21 \mathrm{cm}$.
Diameter of the spherical lead shots = 4.2 cm
Let n spherical lead shots be obtained from the rectangular piece. 
$$\begin{gathered} n\times \mathrm{volume} \mathrm{of} \mathrm{spherical} \mathrm{lead} \mathrm{shot}=\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{lead} \mathrm{piece} \\ \Rightarrow \frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{rectangular} \mathrm{lead} \mathrm{piece}}{\mathrm{volume} \mathrm{of} \mathrm{spherical} \mathrm{lead} \mathrm{shot}}=n \\ \mathit{\Rightarrow }\frac{66\times 42\times 21}{{\displaystyle \frac{4}{3}}{\mathit{\pi r}}^{\mathit{3}}}\mathit{=}n \\ \mathit{\Rightarrow }\frac{66\times 42\times 21}{{\displaystyle \frac{\mathrm{4}}{\mathrm{3}}\pi {\left(\frac{\mathrm{4}\mathrm{.}\mathrm{2}}{\mathrm{2}}\right)}^3}}\mathit{=}n \\ \mathit{\Rightarrow }\frac{58212}{38.808}\mathit{=}n \\ \mathit{\Rightarrow }n\mathit{=}1500 \end{gathered}$$
Hence, 1500 lead shots can be formed. 

DISCLAIMER: There is some error in the question given. Instead of 6 cm, there should be 66 cm. 
The result obtained is by taking 66 cm as the dimensions of the rectangular piece.
 

Answer:

Diameter 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.
$$\begin{gathered} n\times Volume of the spherical lead shots=Volume of the solid cube \\ \Rightarrow \frac{Volume of the solid cube}{Volume of the spherical lead shots}=n \\ \Rightarrow \frac{a^3}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}^3}=n \\ \Rightarrow \frac{{44}^3}{\frac{4}{3}\mathrm{\pi }\times {\left({\displaystyle \frac{4}{2}}\right)}^3}=n \\ \Rightarrow 2541=n \end{gathered}$$
Hence, 2541 spherical lead shots can be made
 

Answer:

The three cubes of metal are in the ratio 3 : 4 : 5. 
Let the edges of the cubes be 3x, 4x and 5x.
Volume of the three cubes will be 
$$\begin{gathered} V_1={\left(3x\right)}^3 \\ {V}_2={\left(4x\right)}^3 \\ {V}_3={\left(5x\right)}^3 \end{gathered}$$
Diagonal of the single cube = $12\sqrt{3} \mathrm{cm}$
We know diagonal of the cube = $a\sqrt{3}=12\sqrt{3}$
Hence, the side of the cube = 12 cm
Volume of the bigger cube $V_b={\left(12\right)}^3$
Volume of the three cubes = Volume of the single  
$$\begin{gathered} {\left(3x\right)}^3+{\left(4x\right)}^3+{\left(5x\right)}^3={\left(12\right)}^3 \\ \Rightarrow 27x^3+64x^3+125x^3=1728 \\ \Rightarrow 216x^3=1728 \\ \Rightarrow x^3=\frac{1728}{216}=8 \\ \Rightarrow x=2 \end{gathered}$$
Hence, the edges of the three cubes  will be $3\times \left(2\right), 4\times \left(2\right), 5\times \left(2\right)=6,8,10$ cm

Answer:

Radius of the solid metallic sphere, r = 10.5 cm
Radius of the cone, R = 3.5 cm
Height of the cone, H = 3 cm
Let the number of smaller cones formed be n. 
Volume of the metallic sphere, $V_s=\frac{4}{3}\mathrm{\pi }{\left(\mathrm{r}\right)}^3=\frac{4}{3}\mathrm{\pi }{\left(10.5\right)}^3$
Volume of the cone, $V_c=\frac{1}{3}\mathrm{\pi }{\left(\mathrm{R}\right)}^2\mathrm{H}=\frac{1}{3}\mathrm{\pi }{\left(3.5\right)}^2\times 3$
Let the number of cones thus formed be n.
$$\begin{gathered} n\times volume of the cone=volume of the sphere \\ \Rightarrow \frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{sphere}\left({\mathrm{V}}_{\mathrm{s}}\right)}{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cone}\left({\mathrm{V}}_{\mathrm{c}}\right)}=n \\ \Rightarrow \frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{\left(10.5\right)}^3}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{\left(3.5\right)}^2\times 3}=n \\ \Rightarrow 126=n \end{gathered}$$
Hence, 126 cones are thus formed.
 

Answer:

The radius of the metallic sphere is cmmm. Therefore, the volume of the metallic sphere is

Cubic mm

The metallic sphere is melted to produce a long wire of uniform cross section of radius mm. Let the length of the wire be l mm. Then, the volume of the wire is

Cubic mm

Since, the volume of the metallic sphere is equal to the volume of the wire, we have

Hence, the length of the wire ismmcm.

Hence

Answer:

Let the radius of the big metallic ball is 4r. Therefore, the volume of the big metallic ball is

The metallic sphere is melted to produce small balls of radius. Then, the volume of each of the small balls is

Since, the volume of the big metallic ball is equal to the sum of the volumes of the small balls, we have the number of produced small balls is

Hence, the number of small balls is

The surface area of the big ball is

The surface area of each of the small ball is

Therefore, the total surface area of the 64 small balls is

Now, we compute the following ratio

Hence, the total surface area of the small balls is equal to four times the surface area of the original big ball.

Answer:

The radius of the copper sphere is 3cm. Therefore, the volume of the copper sphere is

The copper sphere is melted to produce a right circular cone. The height of the right circular cone is 3cm. Let the base-radius of the right circular cone is r. Then, the volume of the right circular cone is

Since, the sphere is melted to recast the cone; the volumes of the sphere and the cone are equal. Hence, we have

Hence, the base-radius of the right circular cone is 6 cm.

Answer:

The radius of the copper rod is 0.5 cm and length is 8 cm. Therefore, the volume of the copper rod is

Let the radius of the wire is r cm. The length of the wire is 18 m=1800 cm. Therefore, the volume of the wire is

Since, the volume of the copper rod is equal to the volume of the wire; we have

$$\begin{gathered} V_1=V \\ \Rightarrow \mathrm{\pi }{r}^2\times 1800=\mathrm{\pi }\times {\left(0.5\right)}^2\times 8 \\ \Rightarrow r^2=\frac{0.25\times 8}{1800}=\frac{1}{900} \\ \Rightarrow r=\frac{1}{30}=0.033 \mathrm{cm} \end{gathered}$$

Hence, the radius of the wire is 0.033 cm = 0.33 mm.
So, thickness = $0.33\times 2=0.66 \mathrm{mm}$

Answer:

The internal and external radii of the hollow sphere are 3cm and 5cm respectively. Therefore, the volume of the spherical shell is

The spherical shell is melted to recast a solid cylinder of length cm. Let the radius of the solid cylinder is r cm. Therefore, the volume of the solid cylinder is

Since, the volume of the hollow spherical shell is equal to the volume of the solid cylinder; we have

Hence, the diameter of the solid cylinder is two times its radius, which is 14 cm.

Answer:

The dimension of the cuboid is 11cm10cm7cm. Therefore, the volume of the cuboid is

The radius and thickness of each coin arecm and 2mm = 0.2cm respectively. Therefore, the volume of each coin is

Since, the total volume of the melted coins is same as the volume of the cuboid; the number of required coins is

Answer:

The surface area of the metallic sphere is 616 square cm. Let the radius of the metallic sphere is r. Therefore, we have

Therefore, the radius of the metallic sphere is 7 cm and the volume of the sphere is

The sphere is melted to recast a cone of height 28 cm. Let the radius of the cone is R cm. Therefore, the volume of the cone is

Since, the volumes of the sphere and the cone are same; we have

Hence, the diameter of the base of the cone so formed is two times its radius, which is cm.

Answer:

Let the radius of the cone by r
Now, Volume cylindrical bucket = Volume of conical heap of sand
$$\begin{gathered} \Rightarrow \mathrm{\pi }{\left(18\right)}^2\left(32\right)=\frac{1}{3}\mathrm{\pi }{r}^2\left(24\right) \\ \Rightarrow {\left(18\right)}^2\left(32\right)=8r^2 \\ \Rightarrow r^2=18\times 18\times 4 \\ \Rightarrow r^2=1296 \\ \Rightarrow r=36 \mathrm{cm} \end{gathered}$$
Let the slant height of the cone be l.
Thus , the slant height is given by
$$\begin{gathered} l=\sqrt{{\left(24\right)}^2+{\left(36\right)}^2} \\ =\sqrt{576+1296} \\ =\sqrt{1872} \\ =12\sqrt{13} \mathrm{cm} \end{gathered}$$

Disclaimer: The answer given in the book for the slant height is not correct.

Answer:

Let the number of such cones formed be n
Now, Volume of solid metallic sphere = Volume of n solid cones

$$\begin{gathered} \Rightarrow \frac{4}{3}\times \frac{22}{7}\times {\left(5.6\right)}^3=n\times \frac{1}{3}\times \frac{22}{7}\times {\left(2.8\right)}^2\times 3.2 \\ \Rightarrow 4\times {\left(5.6\right)}^3=n\times {\left(2.8\right)}^2\times 3.2 \\ \Rightarrow n= 28 \end{gathered}$$

Answer:

Volume of solid cuboid of iron = Volume of cylindrical pipe
$$\begin{gathered} \Rightarrow lbh=\mathrm{\pi }h\left(R^2-r^2\right) \\ \Rightarrow 53\times 40\times 15=\frac{22}{7}\times h\left[{\left(\frac{8}{2}\right)}^2-{\left(\frac{7}{2}\right)}^2\right] \\ \Rightarrow 53\times 40\times 15=\frac{22}{7}\times h\left[4^2-{\left(3.5\right)}^2\right] \\ \Rightarrow 53\times 40\times 15=\frac{22}{7}\times h\times 3.75 \\ \Rightarrow h=2698.18 \mathrm{cm} \end{gathered}$$


Disclaimer: The answer given in the book is not correct.

Answer:

The internal and external radii of the hollow spherical shell are 3cm and 5cm respectively. Therefore, the volume of the hollow spherical shell is

The hollow spherical shell is melted to recast a cylinder of radius 7cm. Let, the height of the solid cylinder is h. Therefore, the volume of the solid cylinder is

Since, the volume of the solid cylinder is same as the volume of the hollow spherical shell, we have

Therefore, the height of the solid cylinder is

Answer:

The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is

The hollow spherical shell is melted to recast a cone of base- radius 4cm. Let, the height of the cone is h. Therefore, the volume of the cone is

Since, the volume of the cone is same as the volume of the hollow sphere, we have

Therefore, the height of the cone is

Answer:

The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is

The hollow sphere is melted to produce a right circular cone of base-radius 4cm. Let, the height and slant height of the cone be h cm and l cm respectively. Then, we have

The volume of the cone is

Since, the volume of the cone and hollow sphere are same, we have

Then, we have

Therefore, the height and the slant height of the cone are 14 cm and 14.56 cm respectively.

Answer:

The radius of the big spherical ball is 3cm. Therefore, the volume of the big spherical ball is

cubic cm

The radii of the 1^st and 2^nd small spherical balls are 1.5 cm and 2 cm respectively. Therefore, the volumes of the 1^st and 2^nd spherical balls are respectively

cubic cm,

cubic cm

Let, the radius of the 3^rd small spherical ball is r cm. Then, its volume is

cubic cm

Since, the big spherical ball is melted to produce the three small spherical balls; the volume of the big spherical ball is same as the sum of the volumes of the three small spherical balls. Therefore, we have

Therefore, the diameter of the 3^rd ball is

Answer:

Diameter of the circular pond is given = 40 m

So, the radius of this pond is 20 m

There is a path surrounding the pond. We are given the thickness of this path as 2 m

We have to grave this path with gravel. The depth of the path is also given 20 cm=0.2 m

This circular path can be viewed as a hollow cylinder of thickness 0.2 m and depth 0.2 m

We know,

Volume of a hollow cylinder

So the volume of the circular path with height 0.2 m

Hence, the volume of gravel required is

Answer:

Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is

m

The well is 16m deep. Thus, the height of the solid right circular cylinder ism.

Therefore, the volume of the solid right circular cylinder is

Let the height of the platform formed be x m. The length and the breadth of the platform are l=27.5m and b=7m respectively. Therefore, the volume of the platform is

Since, the well is spread to form the platform; the volume of the well is equal to the volume of the platform. Hence, we have

Hence, the height of the platform is 0.8 m = 80 cm.

Answer:

Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is

The well is 14m deep. Thus, the height of the solid right circular cylinder is m.

Therefore, the volume of the solid right circular cylinder is

Since, the embankment is to form around the right circular cylinder. Let the width of the embankment be x m. The height of the embankment is h = 40 cm = 0.4 m. Therefore, the volume of the platform is

Since, the well is spread to form the platform; the volume of the well is equal to the volume of the platform. Hence, we have

Hence, x = 5

Hence,

Answer:

The inner radius of the well is 4m and the height is 14m. Therefore, the volume of the Earth taken out of it is

The inner and outer radii of the embankment are 4m and 4+3=7m respectively. Let the height of the embankment be h. Therefore, the volume of the embankment is

Since, the volume of the well is same as the volume of the embankment; we have

Hence, the height of the embankment is

Answer:

The inner radius of the well ism and the height is 14m. Therefore, the volume of the Earth taken out of it is

The inner and outer radii of the embankment arem and 4+=m respectively. Let the height of the embankment be h. Therefore, the volume of the embankment is

Since, the volume of the well is same as the volume of the embankment; we have

Hence, the height of the embankment is

Answer:

We have the following figure

The length of each side of the cube is 9 cm. We have to find the volume of the largest right circular cone contained in the cube.

The diameter of the base circle is same as the length of the side of the cube. Thus, the diameter of the base circle of the right circular cone is 9 cm. Therefore, the radius of the base of the right circular cone iscm.

From the right angled triangle we have

h = 9 cm

Therefore, the volume of the solid right circular cone is

Hence largest volume of cone is

Answer:

The height and radius of the cylindrical bucket arecm andcm respectively. Therefore, the volume of the cylindrical bucket is

The bucket is full of sand and is emptied in the ground to form a conical heap of sand of height cm. Let, the radius and slant height of the conical heap be cm and cm respectively. Then, we have

The volume of the conical heap is

Since, the volume of the cylindrical bucket and conical hear are same, we have

Then, we have

Therefore, the radius and the slant height of the conical heap are 36 cm and 43.27 cm respectively.

Answer:

The fallen rains are in the form of a cuboid of height 1 cm, length 6 m = 600 cm and breadth 4 m = 400 cm. Therefore, the volume of the fallen rains is

The fallen rains are transferred into a cylindrical vessel of internal radius r₁ = 20 cm. Let, the height of the water in the cylindrical vessel is h₁ cm. Then, the volume of the water in the cylinder is

Since, the volume of the water in the cylinder is same as the volume of the rainfalls, we have

Therefore, the height of the water in the cylinder is 190.9 cm.

Answer:

The dimension of the roof is $22 \mathrm{m}\times 20 \mathrm{m}$.
Diameter of the cylinderical vessel = 2 m
Radius of the cylinderical vessel, R = 1 m
Height of the cylinderical vessel, H = 3.5 m
Let the height of the roof be h. 
Volume of water thus collected on the roof = $22\times 20\times h$
Volume of the cylinderical vessel =  $\mathrm{\pi }{\left(\mathrm{R}\right)}^2\mathrm{H}=\mathrm{\pi }{\left(\frac{2}{2}\right)}^2\times 3.5=\mathrm{\pi }\times 1\times 3.5=3.5\mathrm{\pi }$
Volume of water collected on the roof  = Volume of the cylinderical vessel 
$$\begin{gathered} 22\times 20\times h=3.5\mathrm{\pi } \\ \Rightarrow \mathrm{h}=\frac{3.5\mathrm{\pi }}{22\times 20}=0.025 \mathrm{m} \\ \Rightarrow \mathrm{h}=2.5 \mathrm{cm} \end{gathered}$$
 

Answer:

The base-radius and height of the conical flask are r and h respectively. Let, the slant height of the conical flask is l. Therefore, the volume of the water in the conical flask is

The water in the conical flask is poured into a cylindrical flask of base-radius mr. Let, the height of the water in the cylindrical flaks is h₁. Then, the volume of the water in the cylindrical flaks is

Since, the volume of the water in the cylindrical flaks is same as the volume of the water in the conical flaks, we have

Therefore, the height of the water in the cylinder is

Answer:

Suppose height of the rectangular tank is equal to h.
Length of the tank = 15 m
Breadth of the tank = 11 m
Further,
length of cylindrical tank = 5 m
Radius of  cylindrical tank = $\frac{21}{2}$m
To find out the least height of the tank, equate the volumes of two tanks.
$$\begin{gathered} 15\times 11\times h=\mathrm{\pi }{\left(\frac{21}{2}\right)}^2\times 5 \\ \Rightarrow h=\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}\times \frac{5}{15}\times \frac{1}{11} \\ \Rightarrow h=\frac{21}{2} \\ \Rightarrow h=10.5 \end{gathered}$$
Hence, the least height of the tank is equal to 10.5.

Answer:

The internal radius of the hemispherical bowl is 9cm. Therefore, the volume of the water in the hemispherical bowl is

The water in the hemispherical bowl is required to transfer into the cylindrical bottles each of radiuscm and height 4cm. Therefore, the volume of each of the cylindrical bottle is

Therefore, the required number of cylindrical bottles is

Hence

Answer:

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball into the tub, the height of the raised water is 6.75cm. Therefore, the volume of the raised water is

Let, the radius of the spherical ball is r. Therefore, the volume of the spherical ball is

Since, the volume of the raised water is same as the volume of the spherical ball, we have

Therefore, the radius of the spherical ball is

Answer:

The 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 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

cubic m

Since, the volume of the raised water is same as the volume of the water displaced by 500 persons, we have

Therefore, the water will be raised by

Answer:

The radius of the cylindrical jar is 6cm. The volume of the oil of height 2cm contained in the jar is

cubic cm

The radius of each small sphere is 1.5cm. Therefore, the volume of each small sphere is

cubic cm

Since, the volume of the raised water is same as the sum of the volumes of the immersed iron spheres, we have the number of immersed sphere is

Therefore, the number of iron spheres is

Answer:

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball of radius 9cm into the tub, the height of the raised water is h cm. Therefore, the volume of the raised water is

cubic cm

The volume of the spherical ball is

cubic cm

Since, the volume of the raised water is same as the volume of the spherical ball, we have

Therefore, the height of the raised water is

Answer:

The radius of each of the metallic sphere is 2cm. Therefore, the volume of each metallic sphere is

The total volume of the 16 spheres is

The internal dimension of the rectangular box is 16cm8cm8cm. Therefore, the volume of the rectangular box is

Therefore, the volume of the liquid is

Hence volume of liquid is

Answer:

The area of the base of the cuboid is 160 cm². After immersing three identical spheres the level of the water is increased by 2 cm. Therefore, the volume of the increased water is

Let the radius of each of the spheres is r cm. Then, the volume of each of the sphere is

The total volume of the three spheres is

Since, the volume of the increased water is equal to the total volume of the three spheres; we have

Hence, the radius of each of the sphere is 2.94 cm.

Answer:

Let the rise in the level of water in the vessel be h cm.
Now, Volume of 150 spherical marbles = Volume of water displaced in the vessel

$$\begin{gathered} \Rightarrow 150\times \frac{4}{3}\times \frac{22}{7}\times {\left(\frac{1.4}{2}\right)}^3=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2\times h \\ \Rightarrow 200\times {\left(0.7\right)}^3={\left(\frac{7}{2}\right)}^2\times h \\ \Rightarrow h=5.6 \mathrm{cm} \end{gathered}$$

Answer:

Let the number of the balls be n.
Volume of water flows out = Volume of n spherical bolls
$$\begin{gathered} \Rightarrow \frac{2}{5}\times \frac{1}{3}\mathrm{\pi }{R}^{2}h=n\times \frac{4}{3}\mathrm{\pi }{r}^3 \\ \Rightarrow \frac{2}{5}\times {\left(2.5\right)}^2\times 11=n\times 4{\left(\frac{0.5}{8}\right)}^3 \\ \Rightarrow 27.5=\frac{0.5}{8}n \\ \Rightarrow n=440 \end{gathered}$$

Answer:

Radius of the glass spheres, r = 2 cm
Dimensions of the cuboidal box = $16\mathrm{cm}\times 8\mathrm{cm}\times 8\mathrm{cm}$
volume of the spheres = $V_s=\frac{4}{3}\mathrm{\pi }{\left(\mathrm{r}\right)}^3=\frac{4}{3}\mathrm{\pi }{\left(2\right)}^3$
volume of the cuboidal box = $V_c=16\times 8\times 8=1024$
Volume of water in the cuboidal box = Volume of the cuboidal box − Volume of the 16 glass spheres
$$\begin{gathered} =1024-16\times \frac{4}{3}\mathrm{\pi }{\left(2\right)}^3 \\ =1024-536.6 \\ =487.6 {\mathrm{cm}}^3 \end{gathered}$$
Hence, the volume of the water in the cuboidal box = 487.6 cm³

Answer:

The inner radius of the cylindrical pipe r =1 cm.

Rate of flow of water = 80 cm/sec

volume of the water that flows through pipe in 1sec is ${\mathrm{\pi r}}^2\times 80=80\mathrm{\pi } {\mathrm{cm}}^3$ 

volume of the water that flows through pipe in half an hour $80\mathrm{\pi }\times 30\times 60=144000\mathrm{\pi } {\mathrm{cm}}^3$ 

radius of the base of the cylindrical tank is R = 40 cm

let the water level in the cylinderical tank after half an hour be h. 

volume of the raised water = $\mathrm{\pi }{\left(\mathrm{R}\right)}^2\mathrm{h}=\mathrm{\pi }{\left(40\right)}^2\mathrm{h}$

volume of the raised water in tank = volume of the water that flows through pipe

$$\begin{gathered} \Rightarrow \mathrm{\pi }{\left(40\right)}^2\mathrm{h}=144000\mathrm{\pi } \\ \Rightarrow \mathrm{h}=\frac{144000}{1600}=90 \mathrm{cm} \end{gathered}$$

Thus water level will rise by 90 cm in half an hour.

Answer:

The 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

Therefore, the volume of the flowing water in 30 min is

Thus, the irrigated area in 30 min of 8 cm=0.08 m standing water is

Answer:

The internal radius of the pipe is 10 cm=0.1 m. The water is flowing in the pipe at 3km/hr = 3000m/hr. Let the cylindrical tank will be filled in t hours. Therefore, the length of the flowing water in t hours is

Therefore, the volume of the flowing water is

The radius of the cylindrical tank is 5 m and the height is 2 m. Therefore, the volume of the cylindrical tank is

Since, we have considered that the tank will be filled in t hours; therefore the volume of

the flowing water in t hours is same as the volume of the cylindrical tank. Hence, we have

Hence, the tank will be filled in

Answer:

Given that, the cylindrical tank has the diameter of its base as 3 m and its height as 3.5 m.

Thus,
Volume of cylindrical tank = $\mathrm{\pi }{r}^{2}h$
                                            = $\frac{22}{7}\times {\left(\frac{3}{2}\right)}^2\times 3.5$
                                            = 24.75 m³
Now, 1 m³ = 1000 L
∴ 24.75 m³ = 24750 L
Half the capacity of tank = 12375 L

Now,
Time taken by the pipe to empty 225 litres = 1 minute
Time taken by the pipe to empty 1 litre = $\frac{1}{225}$ minutes

Therefore,
Time taken by the pipe to empty 12375 litres = $\frac{1}{225}\times 12375$
                                                                         = 55 minutes

Answer:

Increase in the level of water in half an hour, h = 3.15 m = 315 cm
Radius of the water tank, r = 40 cm
Volume of water that falls in the tank in half an hour = πr²h
                                                                             = π × (40)² × 315
                                                                             = 5,04,000 π cm³
Rate of flow of water = 2.52 km/h
Length of water column in half an hour = 2.52 ÷ 2 = 1.26 km = 1,26,000 cm
Let the internal diameter of the cylindrical pipe be d.
Volume of the water that flows through the pipe in half an hour = $\mathrm{\pi }{\left(\frac{d}{2}\right)}^2\times 126000$
We know
Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour
 
$$\begin{gathered} \Rightarrow \mathrm{\pi }{\left(\frac{d}{2}\right)}^2\times 126000=504000\mathrm{\pi } \\ \Rightarrow {\left(\frac{d}{2}\right)}^2=4 \\ \Rightarrow d=4 \mathrm{cm} \end{gathered}$$

Thus, the internal diameter of the pipe is 4 cm.

Answer:

Let the level of water in the pond rises by 21 cm in t hours.

Speed of water = 15 km/hr = 15000 m/hr
$$\begin{gathered} \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{pipe} = 14 \mathrm{cm} = \frac{14}{100}\mathrm{m} \\ \mathrm{So}, \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{pipe}, \mathrm{r} = \frac{14}{2\times 100}=\frac{7}{100}\mathrm{m} \end{gathered}$$

Volume of water flowing out of the pipe in 1 hour
$$\begin{gathered} ={\mathrm{\pi r}}^2\mathrm{h} \\ =\mathrm{\pi }\times {\left(\frac{7}{100}\right)}^2\times 15000 {\mathrm{m}}^3 \\ =231 {\mathrm{m}}^3 \end{gathered}$$

∴ Volume of water flowing out of the pipe in t hours = 231t m³.

Volume of water in the cuboidal pond
$$\begin{gathered} =50 \mathrm{m}\times 44 \mathrm{m}\times \frac{21}{100} \mathrm{m} \\ =462 {\mathrm{m}}^3 \end{gathered}$$

Volume of water flowing out of the pipe in t hours = Volume of water in the cuboidal pond

∴ 231t = 462

Thus, the water in the pond rise by 21 cm in 2 hours.

Answer:

Width of the canal = 300 cm = 3 m
Depth of the canal = 120 cm = 1.2 m
Speed of water flow = 20 km/h = 20000 m/h
Distance covered by water in 1 hour or 60 min = 20000 m
So, distance covered by the water in 20 min = $\frac{20}{60}\times 20000=\frac{20000}{3}m$
Amount of water irrigated in 20 mintues
$=3\times 1.2\times \frac{20000}{3}=24000$ m³
Area irrigated by this water if 8 cm of standing water is desired will be
$\frac{24000}{{\displaystyle \frac{8}{100}}}=300000 {\mathrm{m}}^2$

So, area irrigated will be 300000 m² or 30 hectors.

 

Answer:

Let the radius of the base of the cylinder be r cm
Let the height be h cm.
Now given that r + h = 37 cm
Total surface area = 1628 cm²             ...(i)
$$\begin{gathered} Total surface area of the cylinder = 2{\mathrm{\pi r}}^2+2\mathrm{\pi rh} \\ =2\mathrm{\pi r}\left(\mathrm{r}+\mathrm{h}\right) \\ =2\mathrm{\pi r}\times 37 \\ =74\mathrm{\pi r} ...\left(\mathrm{ii}\right) \end{gathered}$$
From equation (i) and (ii) we get
$$\begin{gathered} 74\mathrm{\pi r}=1628 \\ \Rightarrow \mathrm{r}=\frac{1628}{74\mathrm{\pi }}=7 {\mathrm{cm}}^3 \end{gathered}$$
Thus, the height will be 37 − 7 = 30 cm
Thus, the volume of the cylinder = ${\mathrm{\pi r}}^2\mathrm{h}=\mathrm{\pi }{\left(7\right)}^2\times 30=4620 {\mathrm{cm}}^3$
Hence the volume is 4620 cm³
 

Answer:

The height of the tent is 77dm = 7.7m. The height of the upper portion of the tent is
44dm = 4.4m. Therefore, the height of the cylindrical part isdm = 3.3m. The radius of the cylindrical part is m.

Let the slant height of the cone part is l m. Then, we have

Therefore, the slant height of the cone part is 18.3 m.

The curved surface area of the cylindrical part is

The curved surface area of the cone part is

Therefore, the total curved surface area of the tent is

The cost of canvas per m² is Rs 3.50. Hence, the total cost for canvas in Rs is

Hence total cost is

Answer:

The radius of the right circular cylinder is 7cm and the height is 14cm. Therefore, the radius of the largest sphere curved out from the cylinder is the minimum of the radius and half the height of the cylinder, which is 7cm. Therefore, the volume of the sphere is

Answer:

We consider the following figure as follows

Let the angle B is right angle and the sides of the triangle are AB = 4cm, BC = 3cm,
AC = 5cm.

When the triangle is revolved about the side AB, then the base-radius, height and slant height of the produced cone becomes BC, AB and AC respectively. Therefore, the volume of the produced cone is

In this case, the curved surface area of the cone is

When the triangle is revolved about the side BC, then the base-radius, height and slant height of the produced cone becomes AB, BC and AC respectively. Therefore, the volume of the produced cone is

In this case, the curved surface area of the cone is

Therefore, the difference between the volumes of the two cones so formed is

Hence the difference between the volumes is

And surface areas are

Answer:

Let the slant height of the cone be l.
Thus , the slant height is given by
$$\begin{gathered} l=\sqrt{{\left(\frac{14}{2}\right)}^2+{\left(24\right)}^2} \\ =\sqrt{49+576} \\ =\sqrt{625} \\ = 25 \mathrm{m} \end{gathered}$$
Now, the curved surface area of the tent is given by
$$\begin{gathered} \frac{22}{7}\times \frac{14}{2}\times 25 \\ =550 {\mathrm{m}}^2 \end{gathered}$$
The curved surface area will be equal to the area of the cloth
$$\begin{gathered} \Rightarrow 550=\mathrm{Length}\times 5 \\ \Rightarrow \mathrm{Length}=110 \mathrm{m} \end{gathered}$$
Now, the cost of cloth is given by
110 ⨯ 25
= Rs 2750

Answer:

Let the radius of the hemisphere be r cm.
Volume of hemisphere = $2425\frac{1}{2} {\mathrm{cm}}^3$
$$\begin{gathered} \Rightarrow \frac{2}{3}\mathrm{\pi }{r}^3=\frac{4851}{2} \\ \Rightarrow \frac{2}{3}\times \frac{22}{7}{r}^3=\frac{4851}{2} \\ \Rightarrow r^3=\frac{4851\times 3\times 7}{2\times 2\times 22} \\ \Rightarrow r^3=\frac{441\times 21}{2\times 2\times 2} \\ \Rightarrow r^3={\left(\frac{21}{2}\right)}^3 \\ \Rightarrow r=\frac{21}{2} \mathrm{cm} \end{gathered}$$
Now, the curved surface area of hemisphere is given by
$$\begin{gathered} 2\mathrm{\pi }{r}^2 \\ =2\times \frac{22}{7}\times {\left(\frac{21}{2}\right)}^2 \\ =693 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

The height of the hollow cylinder is 14 cm. Let the inner and outer radii of the hollow cylinder are r cm and R cm respectively. The difference between the outer and inner surface area of the hollow cylinder is

By the given condition, this difference is 88 square cm. Hence, we have

The volume of the metal used in making the cylinder is

By the given condition, the volume of the metal is 176 cubic cm. Hence, we have

Hence, we have two equations with unknowns R and r

Adding the two equations, we have

Then from the second equation, we have

Therefore, the outer and inner diameters of the hollow cylinder are 5cm and 3cm respectively.

Answer:

We are given the following hemi hollow sphere

The internal and external radii of the hollow hemispherical vessel arecm and cm respectively. Therefore, the total surface area of the hollow hemispherical vessel is

The cost of painting 1 square cm is 10 paise. Therefore the total cost of painting the vessel all over is

Hence total cost of painting is rupees

Answer:

We have the following figure to visualize the situation

Let the radius of the sphere is r. Therefore, the surface area of the sphere is

The circumscribed cylinder of the sphere must have radius r cm and height 2r cm. Therefore, the curved surface area of the cylinder is

Hence, S and S₁ are same. Thus the proof is complete.

Answer:

Let the radius of the hemisphere be r cm.
Total surface area of hemisphere = 462 cm²
$$\begin{gathered} \Rightarrow 3\mathrm{\pi }{r}^2=462 \\ \Rightarrow 3\times \frac{22}{7}\times {\left(r\right)}^2=462 \\ \Rightarrow r^2=49 \\ \Rightarrow r=7 \mathrm{cm} \end{gathered}$$
Now, the volume of hemisphere is given by
$$\begin{gathered} \frac{2}{3}\mathrm{\pi }{r}^3 \\ =\frac{2}{3}\times \frac{22}{7}{\left(7\right)}^3 \\ =\frac{2156}{3} \\ =718\frac{2}{3} {\mathrm{cm}}^3 \end{gathered}$$

Answer:

Diameter of the pipe = 5 mm = 0.5 cm
Radius of the pipe = $\frac{0.5}{2}\mathrm{cm}$ = $\frac{1}{4}cm$
Rate of flow of water through the pipe = 10 m/min = 1000 cm/min
Volume of water that flows out through the pipe in 1 min = ${\mathrm{\pi r}}^2\mathrm{h}=\mathrm{\pi }\times {\left(\frac{1}{4}\right)}^2\times 1000 {\mathrm{cm}}^3$
Volume of water flowing out through the pipe in t min = $\mathrm{\pi }{\left(\frac{1}{4}\right)}^2\times 1000t$
Diameter of the conical vessel = 40 cm
Radius = 20 cm
Height or depth = 24 cm
Volume of the conical vessel = $\frac{1}{3}{\mathrm{\pi R}}^2\mathrm{H}=\frac{1}{3}\mathrm{\pi }{\left(20\right)}^2\times 24=3200\mathrm{\pi }$
Time required to fill the vessel = $\frac{\mathrm{capacity} \mathrm{of} \mathrm{the} \mathrm{vessel}}{\mathrm{volume} \mathrm{of} \mathrm{water} \mathrm{flowing} \mathrm{per} \min }$
$$\begin{gathered} t=\frac{3200\mathrm{\pi }}{\mathrm{\pi }{\left({\displaystyle \frac{1}{4}}\right)}^2\times 1000} \\ t=51.2 \end{gathered}$$
So, the time required is 51.2 min = 51 min 12 sec



 

Answer:

Height of the cone, h = 120 cm
Radius of the cone, r = 60 cm

Height of the cylinder, H = 180 cm
Radius of the cylinder, R = 60 cm

Now,
Volume of the cylinder = ${\mathrm{\pi R}}^2\mathrm{H}=\mathrm{\pi }{\left(60\right)}^2\times 180 {\mathrm{cm}}^3$
Volume of the cone = $\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{1}{3}\mathrm{\pi }{\left(60\right)}^2\times 120$

Therefore,
Volume of water left in the cylinder = Volume of cylinder − volume of the cone 
                                                         $$\begin{gathered} =\mathrm{\pi }{\left(60\right)}^2\times 180-\frac{1}{3}\mathrm{\pi }{\left(60\right)}^2\times 120 \\ =\mathrm{\pi }{\left(60\right)}^2\left[180-40\right] \\ =\mathrm{\pi }\times 3600\left[140\right] \\ =\frac{22}{7}\times 3600\times 140 \\ =1584000 {\mathrm{cm}}^3 \\ =1.584 {\mathrm{m}}^3 \end{gathered}$$

Answer:

The heap of rice is in the form of a cone.
Diameter, d = 9 m
radius, r = $\frac{9}{2}\mathrm{m}$
height, h = 3.5 m
$$\begin{gathered} Volume, V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ =\frac{1}{3}\mathrm{\pi }{\left(\frac{9}{2}\right)}^2\times 3.5 \\ =74.25 {\mathrm{m}}^3 \end{gathered}$$
Thus, volume of rice = 74.25 m³
The canvas cloth required to cover the heap will be the curved surface area of the cone
$$\begin{gathered} l=\sqrt{h^2+r^2} \\ l=\sqrt{3.5^2+{\left(\frac{9}{2}\right)}^2} \\ l=\sqrt{12.25+20.25} \\ l=5.7 m \end{gathered}$$
$$\begin{gathered} CSA=\mathrm{\pi rl} \\ =\mathrm{\pi }\times \left(\frac{9}{2}\right)\times 5.7 \\ =80.62 {\mathrm{m}}^2 \end{gathered}$$
Hence, the canvas cloth required to cover the heap will be 80.62 m²

Answer:

The cylinderical bucket has the height H = 32 cm
radius, R = 18 cm
Volume of the cylinderical bucket will be
$V=\mathrm{\pi }{\left(\mathrm{R}\right)}^2\mathrm{H}=\mathrm{\pi }{\left(18\right)}^2\times 32$
Height of the conical heap, h = 24 cm
Volume of cylinderical bucket = Volume of the conical heap
$$\begin{gathered} \mathrm{\pi }{\left(\mathrm{R}\right)}^2\mathrm{H}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ \Rightarrow {\left(18\right)}^2\times 32=\frac{1}{3}{\mathrm{r}}^2\left(24\right) \\ \Rightarrow \frac{{\left(18\right)}^2\times 32\times 3}{24}={\mathrm{r}}^2 \\ \Rightarrow 1296={\mathrm{r}}^2 \\ \Rightarrow 36=\mathrm{r} \end{gathered}$$
Hence, the radius of the conical heap = 36 cm
$$\begin{gathered} \mathrm{Slant} \mathrm{height} l=\sqrt{h^2+r^2} \\ \Rightarrow l=\sqrt{{24}^2+{36}^2} \\ \Rightarrow l=\sqrt{576+1296} \\ \Rightarrow l=\sqrt{1872} \\ \Rightarrow l=43.2 cm \end{gathered}$$
Thus, the slant height l = 43.27 cm

Answer:

The radius of the hemispherical bowl, R = 9 cm
Radius of the cylinderical bottles, r = 1.5 cm
Height of the bottles, h = 4 cm
Let the number of bottles required be n. 
Volume of the hemispherical bowl = n × Volume of the cylinderical bottles
$$\begin{gathered} \frac{Volume of the hemispherical Bowl}{Volume of the cylinderical bottles}=n \\ \Rightarrow \frac{{\displaystyle \frac{2}{3}{\mathrm{\pi R}}^3}}{{\mathrm{\pi r}}^2\mathrm{h}}=n \\ \Rightarrow \frac{\frac{2}{3}{{\displaystyle \left(9\right)}}^3}{{{\displaystyle \left(1.5\right)}}^2{\displaystyle \left(4\right)}}=n \\ \Rightarrow 54=n \end{gathered}$$
Hence, the 54 bottles are required.

Answer:

Length of the pencil, h = 25 cm
circumference of the base = 1.5 cm
Curved surface area of the pencil which needs to be painted will be
$$\begin{gathered} CSA=circumference\times height \\ =1.5\times 25 c{m}^2 \\ =37.5 c{m}^2 \end{gathered}$$
= 0.375 dm²
Pencils manufactured in one day = 120000
So, the total area to be painted will be $120000\times 0.375 d{m}^2=45000 d{m}^2$
Cost of painting this area will be $45000\times 0.05=Rs 2250$
 

Answer:

Radius of conical vessel r = 5 cm

Height of conical vessel h = 24 cm

The volume of water = $\frac{3}{4}\times$volume of conical vessel.
$$\begin{gathered} =\frac{3}{4}\times \frac{1}{3}\pi r^{2}h \\ =\frac{3}{4}\times \frac{1}{3}\pi \times 25\times 24 \\ =150\mathrm{\pi } \end{gathered}$$

Let h' be the height of cylindrical vessel, which filled by the water of conical vessel,

Radius of cylindrical vessel = 10 cm

Clearly,

Volume of cylindrical vessel = volume of water
$$\begin{gathered} \pi {\left(10\right)}^{2}h=150\pi \\ \Rightarrow h=\frac{150\pi }{100\pi } \\ \Rightarrow h=1.5 \mathrm{cm} \end{gathered}$$

Thus, the height of cylindrical vessel is 1.5 cm.

Answer:

We have a right circular cylinder surmounted by a cone.

Diameter of cylinder = 24 m, Height if cylindrical portion = 11 m and the vertex of the cone is 16 meters above the ground. We have to find the area of canvas required for the tent.

Suppose curved area of the cone portion is.

From the above figure the slant height of the top is given by

Now, Let us suppose that the curved area of cylinder is

Therefore, the area of canvas is given by

Hence,

Answer:

Given:

Radius of the cylinder, height of the cylinder, , slant height of the cone.We have to find total surface area and volume of the rocket

Let us assume that the area of the cone is.

The area of the cylinder is given by

Total areais

Now, we are going to find the volume of the rocket V.

Volume of the cone is given by

Volume of the cylinder is

Total volume of the cone is given by

Hence, the area and the volume of the rocket is

Answer:

Given:

Height of the tent h = 77 dm = 7.7 m, diameter of cylinder

Height of the cylinder h₁ = 44 dm = 4.4 m, height of cone h₂ = 33 dm = 3.3 m

We have the following diagram

Radius

The curved area of cylinder is given by

$$\begin{gathered} S_1=2\mathrm{\pi rh} = 2\times \frac{22}{7}\times 18\times 4.4 \\ =497.82 {\mathrm{m}}^2 \end{gathered}$$

The slant height of the cone is

$$\begin{gathered} l=\sqrt{h^2+r^2} \\ =\sqrt{3.3^2+{18}^2} \\ =\sqrt{334.89} \\ =18.3 m \end{gathered}$$

The curved area of the cone is given by
$$\begin{gathered} S_2=\mathrm{\pi rl}=\frac{22}{7}\times 18\times 18.3 \\ =1035.25 {\mathrm{m}}^2 \end{gathered}$$

The total area of the canvas required is given as

S = S₁ + S₂
= 497.82 + 1035.25
=1533.07 m²

Therefore the cost of the canvas at the rate of Rs 3.5 per square meter is given by

$$\begin{gathered} 1533.07\times 3.5 \\ =Rs 5365.745 \end{gathered}$$

Hence the cost of the canvas is Rs 5365.745

Answer:

Given that, a toy is in the form of a cone surmounted on the hemisphere.

Diameter of the baseand the height of the cone, then we have to find the surface area of the toy.

We have the following figure

The radius of the base is

From the above figure, the slant height of the cone is

We know that when the surface area of the cone is, then

The surface area of the hemisphere is

Therefore the surface area of the toy is

Hence,

Answer:

We have the following diagram

For cone, we have

Curved surface area of the cone is given as

For cylindrical part, we have

Curved surface area of the cylinder is

The surface area of the hemisphere is

Total surface area of the solid is given by

Hence the total surface area of the solid is

Answer:

We have the following diagram

For cylindrical part, we have

Therefore, the curved surface area of the cylinder is given by

For conical part, we have

Therefore, the curved surface area of the conical part is

For hemisphere, we have

Therefore the surface area of the hemisphere is

The total surface area of the toy is

Hence, total surface area of the toy is

Answer:

To find the volume of the water left in the tube, we have to subtract the volume of the hemisphere and cone from volume of the cylinder.

For right circular cylinder, we have

The volume of the cylinder is

For hemisphere and cone, we have

Therefore the total volume of the cone and hemisphere is

The volume of the water left in the tube is

Hence, the volume of the water left in the tube is

Answer:

Given that:

Radius of the cylindrical base

Height of the cylindrical portion

Height of the conical portion

The volume of the cylinder is given by the following formula

The volume of the conical portion is

Therefore, the total volume of the circus tent is

Hence, the volume of the circus tent is

Answer:

To find the total capacity of the tank, we have to add the volume of the cylinder and cone.
 

Diameter of the cylinder,
 

Radius of the cylinder, $r=\frac{d}{2}=\frac{21}{2} \mathrm{cm}$

Height of the cylinder, $h_1=18\mathrm{cm}$

Also, radius of cone, r = $\frac{21}{2}\mathrm{cm}$

Height of the cone, $h_2=9\mathrm{cm}$

Now,

Total capacity of the tank

= Volume of the cylinder + Volume of 2 cones

$$\begin{gathered} =\mathrm{\pi }{r}^2{h}_1 + 2\times \frac{1}{3}\mathrm{\pi }{r}^2{h}_2 \\ =\mathrm{\pi }{r}^2 \left(h_1+\frac{2}{3}{h}_2\right) \\ =\frac{22}{7}\times {\left(\frac{21}{2}\right)}^2\times \left(18+\frac{2}{3}\times 9\right) \\ =\frac{22}{7}\times {\left(\frac{21}{2}\right)}^2\times 24 \\ = 8316 {\mathrm{cm}}^3 \end{gathered}$$

Hence the total capacity of the tank is 8316 cm³.

Answer:

Given that:

We have the following diagram

Slant height of cone is given by

The total surface area of the remaining part is given by

The volume of the remaining part is given by

Hence,

Answer:

Given that:

Radius of the base

Height of the cylinder

Height of the cone

Slant height of the cone

The total capacity of the tent is given by

The total area of canvas required for the tent is

Therefore, the total cost of the canvas is

Hence, the total capacity and cost is

Answer:

Given that:

Height of the cylinder

Radius of the cylinder and hemisphere are same and is given by

The volume of the cylinder is cylinder is

There are two hemispheres at each ends of the cylinder, therefore the volume of the two hemispheres is

Therefore, the total volume of the boiler is given by

Hence the volume of the boiler is

Answer:

Given that:

Radius of the same base

Height of the cylinder

The volume of the vessel is given by

The internal surface area of the solid is

Hence, the volume of the vessel and internal surface area of the solid is

Answer:

We have a solid composed of cylinder with hemispherical ends.

Radius of the two curved surfaces

Height of cylinder is h.

Total height of the body

So, total surface area is given by,

Change the units of curved surface area as,

Cost of polishing the surface is Rs 10 per.

So total cost,

Answer:

We have to find the volume of cork dust filled between the two vessels.

Radius of outer vessel

Radius of inner vessel

Height of the cylinder

So, volume of cork dust filled between the two vessels,

Volume of cork dust filled between the two vessels is 1980.

Answer:

We have to find the mass of the roller.

Radius of inner cylinder

Radius of outer cylinder

Length of the cylinder

So, volume of iron,

It is given that, 1 of iron has a mass of 7.8 gm.

So the mass of iron used,

Answer:

We 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

Total height of vessel

So, the inner surface area of a vessel,

Answer:

We have to find the total surface area of a toy which is a cone surmounted on a hemisphere.

Radius of hemisphere and the base of the cone

Height of the cone,

h = 15.5 − 3.5 = 12 cm
$$\begin{gathered} \mathrm{slant} \mathrm{height} \left(l\right)=\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2} \\ =\sqrt{{12}^2+3.5^2} \\ =\sqrt{156.25} \\ =12.5 \mathrm{cm} \end{gathered}$$

So, total surface area of toy,

$$\begin{gathered} S=\mathrm{\pi rl}+2{\mathrm{\pi r}}^2 \\ =\mathrm{\pi r}\left(\mathrm{l}+2\mathrm{r}\right) \\ =\frac{22}{7}\times 3.5\left(12.5+2\times 3.5\right) \\ =214.5 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

We have to find the outer and inner radius of a hollow pipe.

Radius of inner pipe be

Radius of outer cylinder be

Length of the cylinder

Difference between the outer and the inner surface area is 44

So,

So,

…… (1)

So, volume of metal used is 99 , so,

Use equation (1) in the above to get,

Therefore,

…… (2)

Solve equation (1) and (2) to get,

Answer:

We have to find the number of cones which can be filled using the ice cream in the cylindrical vessel.

Radius of the cylinder

Height of cylinder

Radius of cone and the hemisphere on it

Height of cone

Let ‘n’ number of cones filled. So we can write it as,

So,

Now put the values to get,

Therefore,

Answer:

We have to find the mass of a pole having a cylindrical base surmounted by a cone.

Radius of cone and cylinder

Height of cylinder

Height of cone

So volume of the pole is,

Put the values to get,

Mass of 1 of iron is 8 gm.

Therefore mass of the iron,

Answer:

We have to find the remaining volume of the cylinder when the toy is inserted into it. The toy is a hemisphere surmounted by a cone.

Radius of cone, cylinder and hemisphere

Height of cone

Height of the cylinder

So the remaining volume of the cylinder when the toy is inserted into it,

Put the values to get,

Answer:

We have to find the remaining volume of water left in the cylinder when the solid is inserted into it. The solid is a hemisphere surmounted by a cone.

Radius of cone, cylinder and hemisphere

Height of cone

Height of the cylinder

So the remaining volume of water left in the cylinder when the solid is inserted into it,

Put the values to get,

Answer:

We have a cylindrical vessel in which a cone is inserted. We have,

Radius of the cylinder

Radius of cone

Height of cylinder

Height of cone

(i) We have to find the volume of water displaced from the cylinder when cone is inserted.

So,

So volume of water displaced,

(ii) We have to find the volume of water remaining in the cylinder.

So volume of the water left in the cylinder,

Answer:

We have to find the remaining volume and surface area of a cubical box when a hemisphere is cut out from it.

Edge length of cube

Radius of hemisphere

Therefore volume of the remaining block,

So,

So, remaining surface area of the box,

Therefore,

Put the values to get the remaining surface area of the box,

Answer:

Solution:

Let the height of the conical part be h.

Radius of the cone = Radius of the hemisphere = r = 21 cm

The toy can be diagrammatically represented as

Volume of the cone =

Volume of the hemisphere =

According to given information:

Volume of the cone × Volume of the hemisphere

Thus, surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

Answer:

Height of cone = 9.5 − 3.5 = 6 cm
Volume of the solid = Volume of cone + Volume of hemisphere
$$\begin{gathered} =\frac{1}{3}\mathrm{\pi }{r}^{2}h+\frac{2}{3}\mathrm{\pi }{r}^3 \\ =\frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^2\times 6+\frac{2}{3}\times \frac{22}{7}\times {\left(3.5\right)}^3 \\ =77+89.83 \\ =166.83 {\mathrm{cm}}^3 \end{gathered}$$

Answer:

Volume of wood in the toy = Volume of cylinder − 2(Volume of hemisphere)
$$\begin{gathered} =\mathrm{\pi }{r}^{2}h-2\times \frac{2}{3}\mathrm{\pi }{r}^3 \\ =\frac{22}{7}\times {\left(3.5\right)}^2\times 10-2\times \frac{2}{3}\times \frac{22}{7}\times {\left(3.5\right)}^3 \\ =385-179.67 \\ =205.33 {\mathrm{cm}}^3 \end{gathered}$$

Answer:

The 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 the sphere = $\frac{7}{2}=3.5$
Volume of the wood left = Volume of cube − Volume of sphere
$$\begin{gathered} ={\left(\mathrm{Side}\right)}^3-\frac{4}{3}\mathrm{\pi }{r}^3 \\ =7^3-\frac{4}{3}\times \frac{22}{7}\times {\left(3.5\right)}^3 \\ =343-179.67 \\ =163.33 {\mathrm{cm}}^3 \end{gathered}$$

Answer:

Total surface area of the remaining solid

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

$$\begin{gathered} =2\mathrm{\pi }rh+\mathrm{\pi }rl+\mathrm{\pi }{r}^2 \\ =2\times \frac{22}{7}\times \frac{4.2}{2}\times 2.8+\frac{22}{7}\times \frac{4.2}{2}\times \sqrt{{\left(\frac{4.2}{2}\right)}^2+{\left(2.8\right)}^2}+\frac{22}{7}\times {\left(\frac{4.2}{2}\right)}^2 \\ =2\times \frac{22}{7}\times 2.1\times 2.8+\frac{22}{7}\times 2.1\times \sqrt{{\left(2.1\right)}^2+{\left(2.8\right)}^2}+\frac{22}{7}\times {\left(2.1\right)}^2 \\ =2\times \frac{22}{7}\times 2.1\times 2.8+\frac{22}{7}\times 2.1\times 3.5+\frac{22}{7}\times {\left(2.1\right)}^2 \\ =36.96+23.1+13.86 \\ =73.92 {\mathrm{cm}}^2 \end{gathered}$$

The total surface area of the remaining solid is 73.92 cm².

Answer:

The 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 = $\frac{21}{2}=10.5$ cm
Volume of the remaining solid  = Volume of cube − Volume of cone
$$\begin{gathered} ={\left(\mathrm{Side}\right)}^3-\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ ={\left(21\right)}^3-\frac{1}{3}\times \frac{22}{7}\times {\left(10.5\right)}^2\times 21 \\ =9261-2425.5 \\ =6835.5 {\mathrm{cm}}^3 \end{gathered}$$

Disclaimer: The answer given in the book is not correct.

Answer:

Volume of solid wooden toy = $166\frac{5}{6}$ cm³
⇒ Volume of cone + Volume of hemisphere = $166\frac{5}{6}$
$$\begin{gathered} \Rightarrow \frac{1}{3}\mathrm{\pi }{r}^{2}h+\frac{2}{3}\mathrm{\pi }{r}^3=\frac{1001}{6} \\ \Rightarrow \frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^2\times h+\frac{2}{3}\times \frac{22}{7}\times {\left(3.5\right)}^3=\frac{1001}{6} \\ \Rightarrow \frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^2\left(h+7\right)=\frac{1001}{6} \\ \Rightarrow \frac{38.5}{3}\left(h+7\right)=\frac{1001}{6} \\ \Rightarrow h+7=13 \\ \Rightarrow h=6 \mathrm{cm} \end{gathered}$$
Height of the solid wooden toy
= Height of cone + Radius of hemisphere
= 6 + 3.5
= 9.5 cm
Now, curved surface area of hemisphere = $2\mathrm{\pi }{r}^2$
$$\begin{gathered} =2\times \frac{22}{7}\times {\left(3.5\right)}^2 \\ =77 {\mathrm{cm}}^2 \end{gathered}$$
Cost of painting the hemispherical part of the toy = 10 ⨯ 77 = Rs 770

Answer:

Surface area of the remaining block
= Total Surface area of cubic block + Curved Surface area of cylinder − 2(Area of circular base)
$$\begin{gathered} =2\left(lb+bh+lh\right)+2\mathrm{\pi }rh-2\mathrm{\pi }{r}^2 \\ =2\left(15\times 10+10\times 5+15\times 5\right)+2\times \frac{22}{7}\times \frac{7}{2}\times 5-2\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^2 \\ =2\times 275+110-77 \\ =583 {\mathrm{cm}}^2 \end{gathered}$$

Answer:

let the total height of the building be H m.

let the radius of the base be r m. Therefore the radius of the hemispherical dome is r m.

 Now given that internal diameter = total height
$\Rightarrow 2r=H$

Total height of the building = height of the cylinder +radius of the dome
⇒ H = h + r
⇒ 2r = h + r
⇒ r = h

Volume of the air inside the building = volume of the cylinder+ volume of the hemisphere
$$\begin{gathered} \Rightarrow 41\frac{19}{21}={\mathrm{\pi r}}^2\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^3 \\ \Rightarrow \frac{880}{21}={\mathrm{\pi h}}^2\mathrm{h}+\frac{2}{3}{\mathrm{\pi h}}^3 \\ \Rightarrow \frac{880}{21}={\mathrm{\pi h}}^3\left(1+\frac{2}{3}\right) \\ \Rightarrow \frac{880}{21}={\mathrm{\pi h}}^3\left(\frac{5}{3}\right) \\ \Rightarrow \mathrm{h}=2 \mathrm{m} \end{gathered}$$

Hence, height of the building H = 2 × 2 = 4m

Answer:

The dimensions of the cuboid = 10 cm × 5 cm × 4 cm
Volume of the total cuboid = 10 cm × 5 cm × 4 cm = 200 cm³ 
Radius of the conical depressions, r = 0.5 cm
Depth, h = 2.1 cm
Volume of the conical depression = $\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{1}{3}\mathrm{\pi }{\left(0.5\right)}^2\left(2.1\right)=0.5495$ cm³
Edge of cubical depression, a = 3 cm
Volume of the cubical depression = $a^3=3^3=27 c{m}^3$
Volume of wood used to make the entire stand = Volume of the total cuboid − volume of conical depression − volume of cubical depression
$$\begin{gathered} =200-4\times 0.5495-27 \\ =170.802 c{m}^3 \end{gathered}$$ 

 

Answer:

Let the radius of the dome be r. 
Diameter be d.
Let the height of the building be H.
Given $d=\frac{2}{3}H$
$$\begin{gathered} \Rightarrow 2r=\frac{2}{3}H \\ \Rightarrow r=\frac{H}{3} \\ \Rightarrow 3r=H \end{gathered}$$
Also, h + r = H
$$\begin{gathered} \Rightarrow 3r=h+r \\ \Rightarrow 2r=h \\ \Rightarrow r=\frac{h}{2} \end{gathered}$$
Volume of air = Volume of air in the cylinder + Volume of air int he hemispherical dome
$$\begin{gathered} \Rightarrow {\mathrm{\pi r}}^2\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^3=67\frac{1}{21} \\ \Rightarrow \mathrm{\pi }{\left(\frac{\mathrm{h}}{2}\right)}^2\mathrm{h}+\frac{2}{3}\mathrm{\pi }{\left(\frac{h}{2}\right)}^3=\frac{1408}{21} \\ \Rightarrow h^3=\frac{11264}{176}=64 \\ \Rightarrow h=4 m \end{gathered}$$
Hence, the radius will be $r=\frac{h}{2}=\frac{4}{2}=2 m$
Height of the building, H = 3r = $3\times 2=6 m$

Answer:

The height of the cone, h = 4 cm
Diameter of the base, d = 8 cm
Radius of the cone, r = 4 cm
lateral side will be
$$\begin{gathered} l=\sqrt{h^2+r^2} \\ l=\sqrt{{\left(4\right)}^2+{\left(4\right)}^2}=4\sqrt{2} \end{gathered}$$
Volume of the toy = volume of hemisphere + volume of cone
$$\begin{gathered} \Rightarrow V=\frac{2}{3}{\mathrm{\pi r}}^3+\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ =\frac{{\mathrm{\pi r}}^2}{3}\left[2\mathrm{r}+\mathrm{h}\right] \\ =\frac{\mathrm{\pi }{\left(4\right)}^2}{3}\left[2\times 4+4\right] \\ =201.14 {\mathrm{cm}}^3 \end{gathered}$$
When the cube circumscribes the toy, then 
Volume of the cube = $a^3={\left(8\right)}^3=512 c{m}^3$
$\mathrm{Volume} \mathrm{of} \mathrm{cube}-\mathrm{volume} \mathrm{of} \mathrm{toy}=512-201.14=310.86 c{m}^3$
Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cone
$$\begin{gathered} =2\mathrm{\pi }{r}^2+\mathrm{\pi }rl \\ =2\mathrm{\pi }{\left(4\right)}^2+\mathrm{\pi }\times 4\times 4\sqrt{2} \\ =\mathrm{\pi }{\left(4\right)}^2\left[2+\sqrt{2}\right] \\ =171.68 {\mathrm{cm}}^2 \end{gathered}$$


 

Answer:

Total height of the tent above the ground = 27 m
Height of the cylinderical part, $h_1$= 6 m
Height of the conical part, $h_2$ = 21 m
Diameter = 56 m
Radius = 28 m
Curved surface area of the cylinder, CSA₁ = $2\mathrm{\pi }r{h}_1=2\mathrm{\pi }\times 28\times 6=336\mathrm{\pi }$
Curved surface area of the cylinder, CSA₂ will be
$$\begin{gathered} \mathrm{\pi }rl=\mathrm{\pi }r\left(\sqrt{h^2+r^2}\right)=\mathrm{\pi }\times 28\times \left(\sqrt{{21}^2+{28}^2}\right)=28\mathrm{\pi }\left(\sqrt{441+784}\right) \\ =28\mathrm{\pi }\times 35 \\ =980\mathrm{\pi } \end{gathered}$$
Total curved surface area = CSA of cylinder + CSA of cone
= CSA₁ + CSA₂
$$\begin{gathered} =336\mathrm{\pi }+980\mathrm{\pi } \\ =1316\mathrm{\pi } \\ =4136 {\mathrm{m}}^2 \end{gathered}$$
Thus, the area of the canvas used in making the tent is 4136 m² .

Answer:

The radii of the top and bottom circles are r₁ = 20 cm and r₂ = 10 cm respectively. The height of the bucket is h = 12 cm. Therefore, the volume of the bucket is

The slant height of the bucket is

The total surface area of the bucket is

The total cost of tin sheet used for making the bucket is

Answer:

The radii of the bottom and top circles are r₁ = 10 cm and r₂ = 6 cm respectively. The height of the frustum cone is h = 3 cm. Therefore, the volume of the bucket is

Hence

The slant height of the bucket is

cm

The total surface area of the frustum cone is

Square cm

Square cm

Hence

Answer:

The slant height of the frustum of the cone is l = 4 cm. The perimeters of the circular ends are 18 cm and 6 cm. Let the radii of the bottom and top circles are r₁ cm and r₂ cm respectively. Then, we have

The curved surface area of the frustum cone is

= 48 Square cm

Answer:

The height of the frustum of the cone is h = 16 cm. The perimeters of the circular ends are 44 cm and 33 cm. Let the radii of the bottom and top circles are r₁ cm and r₂ cm respectively. Then, we have

The slant height of the bucket is

The curved/slant surface area of the frustum cone is

Hence

The volume of the frustum of the cone is

Hence

The total surface area of the frustum cone is

= 860.25 Square cm

Hence

Answer:

The height of the conical bucket is h = 45 cm. The radii of the bottom and top circles are r₁ = 28cm and r₂ =7cm respectively.

The volume/capacity of the conical bucket is

Hence

Answer:

We have the following situation as shown in the figure
 
                                             V

Let VAB be a cone of height h₁ = VO₁ =20cm. Then from the symmetric triangles VO ₁A and VOA₁, we have

It is given that, volume of the cone VA₁O istimes the volume of the cone VAB. Hence, we have

Hence, the height at which the section is made is 20 − 4 = 16 cm.

Answer:

The height of the conical bucket is h = 24 cm. The radii of the bottom and top circles are r₁ = 15cm and r₂ = 5cm respectively.

The slant height of the bucket is

The curved surface area of the bucket is

cm²

Hence the curved surface area of the bucket is

Answer:

The height of the frustum cone is h = 12 cm. The radii of the bottom and top circles are r₁ = 12cm and r₂ = 3cm respectively.

The slant height of the frustum cone is

The total surface area of the frustum cone is

Hence

The volume of the frustum cone is

Hence

Answer:

The height of the frustum cone is h = 8 m. The radii of the end circles of the frustum are r₁ = 13m and r₂ =7m.

The slant height of the frustum cone is

The curved surface area of the frustum is

The base-radius of the upper cap cone is 7m and the slant height is 12m. Therefore, the curved surface area of the upper cap cone is

Hence, the total canvas required for the tent is

Hence total canvas is

Answer:

Radius, r₁ = 8 cm and r₂ = 20 cm
Height, h = 16 cm
Volume of milk that the container can hold
$$\begin{gathered} V=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_1^2+{\mathrm{r}}_1{\mathrm{r}}_2+{\mathrm{r}}_2^2\right) \\ \mathrm{V}=\frac{1}{3}\mathrm{\pi }\times 16\left(8^2+8\times 20+{20}^2\right) \\ \mathrm{V}=10459.42 {\mathrm{cm}}^3=10.45942 \mathrm{liters} \end{gathered}$$
Cost of milk will be $10.45942\times 44=Rs 460.21$

Answer:

Height of the bucket, h = 30 cm
Radii r₁ = 10 cm and r₂ = 20 cm
Capacity of the bucket,
$$\begin{gathered} V=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_1^2+{\mathrm{r}}_1{\mathrm{r}}_2+{\mathrm{r}}_2^2\right) \\ =\frac{1}{3}\mathrm{\pi }\times 30\left({10}^2+10\times 20+{20}^2\right) \\ =21980 {\mathrm{cm}}^3 \\ =21.980 \mathrm{liters} \end{gathered}$$

$$\begin{gathered} l=\sqrt{h^2+{\left(r_2-r_1\right)}^2} \\ l=\sqrt{{30}^2+{\left(20-10\right)}^2}=10\sqrt{10} \end{gathered}$$
Surface area of the bucket
$$\begin{gathered} S=CSA+area of the base \\ S=\mathrm{\pi }\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)\mathrm{l}+{{\mathrm{\pi r}}_1}^2 \\ \mathrm{S}=\mathrm{\pi }\left(10+20\right)10\sqrt{10}+\mathrm{\pi }{\left(10\right)}^2 \\ \mathrm{S}=2978.86+314=3292.86 {\mathrm{cm}}^2 \end{gathered}$$
Cost of milk which can completely fill the container at Rs 25/litre
= 21.980 × 25
= Rs 549.50

Answer:

Let the depth of the bucket is h cm. The radii of the top and bottom circles of the frustum bucket are r₁ =20cm and r₂ =12cm respectively.

The volume/capacity of the bucket is

Given that the capacity of the bucket isCubic cm. Thus, we have

Hence, the height of the bucket is

The slant height of the bucket is

The surface area of the used metal sheet to make the bucket is

Hence

Answer:

The height of the bucket is 20cm. The radii of the upper and lower circles of the bucket are r₁ =25cm and r₂ =10cm respectively.

The slant height of the bucket is

The surface area of the used aluminium sheet to make the bucket is

Therefore, the total cost of making the bucket is

Hence the total cost is

Answer:

The slant height of the frustum of a cone is l=10cm. The radii of the upper and lower circles of the bucket are r₁ =33cm and r₂ =27cm respectively.

The total surface area of the frustum of the cone is

Hence total surface area is

Answer:

The height of the bucket is h=16cm. The radii of the upper and lower circles of the bucket are r₁ =20 cm and r₂ = 8 cm respectively.

The slant height of the bucket is

The volume of the bucket is

Hence the volume of the bucket is

The surface area of the used metal sheet to make the bucket is

Therefore, the total cost of making the bucket is

Answer:

The height of the frustum of a cone is h=9cm. The radii of the upper and lower circles of the frustum of the cone are r₁ =30cm and r₂ =18cm respectively.

The slant height of the frustum of the cone is

The volume of the frustum of the cone is

The total surface area of the frustum of the cone is

Answer: