CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set 12

Question. A solid consists of a circular cylinder surmounted by a right circular cone. The height of the cone is \( h \). If the total volume of the solid is 3 times the volume of the cone, then the height of the cylinder is :
(a) \( \frac{2}{3}h \)
(b) \( \frac{3}{2}h \)
(c) \( h \)
(d) \( 2h \)
Answer: (a) \( \frac{2}{3}h \)
Explanation :
Let \( x \) be the height of cylinder.
Since, volume of the total solid is equal to 3 times the volume of the cone.
So,
\( \frac{1}{3} \pi r^2 h + \pi r^2 x = 3\left(\frac{1}{3} \pi r^2 h\right) \)
\( \Rightarrow \frac{1}{3} \pi r^2 h + \pi r^2 x = \pi r^2 h \)
\( \Rightarrow \pi r^2 x = \pi r^2 h - \frac{1}{3} \pi r^2 h \)
\( \Rightarrow \pi r^2 x = \frac{2}{3} \pi r^2 h \)
\( \Rightarrow x = \frac{2}{3}h \)

Question. The number of solid spheres, each of diameter 6 cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is :
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (c) 5
Explanation :
Diameter of each sphere = 6 cm
\ Radius \( (r_1) = 3 \text{ cm} \)
Volume \( = \frac{4}{3} \pi r_1^3 = \frac{4}{3} \pi \times (3)^3 \text{ cm}^3 = 36\pi \text{ cm}^3 \)
Diameter of cylinder = 4 cm
\ Radius \( (r_2) = 2 \text{ cm} \)
and height \( (h) = 45 \text{ cm} \)
\ Volume \( = \pi r_2^2 h = \pi \times (2)^2 \times 45 \text{ cm}^3 = 180\pi \text{ cm}^3 \)
\ Number of sphere required \( = \frac{180\pi}{36\pi} = 5 \)

Question. If two solid-hemispheres of same base radius \( r \) are joined together along their bases, then curved surface area of this new solid is :
(a) \( 4\pi r^2 \)
(b) \( 6\pi r^2 \)
(c) \( 3\pi r^2 \)
(d) \( 8\pi r^2 \)
Answer: (a) \( 4\pi r^2 \)
Explanation :
Because curved surface area of a hemisphere is \( 2\pi r^2 \) each and here, we join two solid hemispheres along their bases of radius \( r \), from which we get a solid sphere.
Hence, the curved surface area of new solid \( = 2\pi r^2 + 2\pi r^2 = 4\pi r^2 \)

Question. The area of the base of a rectangular tank is \( 6500 \text{ cm}^2 \) and the volume of water contained in it is \( 2.6 \text{ m}^3 \). The depth of water in the tank is :
(a) 3.5 m
(b) 4 m
(c) 5 m
(d) 8 m
Answer: (b) 4 m
Explanation :
Area of the base of the rectangular tank \( = 6500 \text{ cm}^2 = \frac{6500}{10000} \text{ m}^2 = \frac{65}{100} \text{ m}^2 \)
Let the depth of the water be \( h \text{ metres} \).
So, \( \frac{65}{100} \times h = 2.6 \)
\( \Rightarrow h = \frac{2.6 \times 100}{65} = 4 \text{ m} \)
Hence, the depth of the water is 4m.

Question. If the height of a bucket in the shape of frustum of a cone is 16 cm and the diameters of its two circular ends are 40 cm and 16 cm, then its slant height is :
(a) 20 cm
(b) \( 12\sqrt{5} \text{ cm} \)
(c) \( 8\sqrt{13} \text{ cm} \)
(d) 16 cm
Answer: (a) 20 cm
Explanation :
Height of the frustum, \( h = 16 \text{ cm} \)
Radii of the circular ends, \( R \) and \( r \) are:
\( R = \frac{40}{2} = 20 \text{ cm} \)
and \( r = \frac{16}{2} = 8 \text{ cm} \)
The slant height of the frustum, \( l = \sqrt{(R - r)^2 + h^2} \)
\( \Rightarrow l = \sqrt{(20 - 8)^2 + 16^2} = \sqrt{12^2 + 16^2} \)
\( \Rightarrow l = \sqrt{144 + 256} = \sqrt{400} = 20 \text{ cm} \)

Question. The curved surface area of a right circular cone of height 15cm and base diameter 16 cm is :
(a) \( 60\pi \text{ cm}^2 \)
(b) \( 68\pi \text{ cm}^2 \)
(c) \( 120\pi \text{ cm}^2 \)
(d) \( 136\pi \text{ cm}^2 \)
Answer: (d) \( 136\pi \text{ cm}^2 \)
Explanation :
Height of cone, \( h = 15 \text{ cm} \)
Base radius, \( r = \frac{16}{2} = 8 \text{ cm} \)
Slant height, \( l = \sqrt{r^2 + h^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ cm} \)
C.S.A. of cone \( = \pi rl = \pi \times 8 \times 17 = 136\pi \text{ cm}^2 \)

Question. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2mm. The length of the wire is:
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m
Answer: (c) 36 m
Explanation :
Diameter of sphere = 6 cm
\ Radius \( (r) = \frac{6}{2} = 3 \text{ cm} \)
Volume \( = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times (3)^3 \text{ cm}^3 = 36\pi \text{ cm}^3 \)
Diameter of wire = 2 mm
\ Radius \( (R) = 1 \text{ mm} = \frac{1}{10} \text{ cm} \)
Let \( h \) be its length, then \( \pi R^2 h = 36\pi \)
\( \Rightarrow \pi \times \left(\frac{1}{10}\right)^2 h = 36\pi \)
\( \Rightarrow \frac{h}{100} = 36 \Rightarrow h = 3600 \text{ cm} = 36 \text{ m} \)

Question. During conversion of a solid from one shape to another, the volume of the new shape will :
(a) Decrease
(b) Increase
(c) Remain unaltered
(d) Be doubled
Answer: (c) Remain unaltered
Explanation :
During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

Question. A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with a hemisphere stucked at each end. The length of the entire capsule is 2 cm. The capacity of the capsule is : [NCERT Exemplar]
(a) \( 0.33 \text{ cm}^3 \)
(b) \( 0.34 \text{ cm}^3 \)
(c) \( 0.35 \text{ cm}^3 \)
(d) \( 0.36 \text{ cm}^3 \)
Answer: (d) \( 0.36 \text{ cm}^3 \)
Explanation :
Radius of the capsule = 0.25 cm
Let the length of the cylindrical part of the capsule be \( x \text{ cm} \).
So, \( 0.25 + x + 0.25 = 2 \Rightarrow x = 1.5 \text{ cm} \)
Capacity of the capsule = \( 2 \times (\text{Volume of the hemisphere}) + (\text{Volume of the cylinder}) \)
\( = 2 \times \left(\frac{2}{3} \pi r^3\right) + (\pi r^2 h) \)
\( = 2 \times \left(\frac{2}{3} \times \frac{22}{7} \times (0.25)^3\right) + \left(\frac{22}{7} \times (0.25)^2 \times 1.5\right) = 0.36 \text{ cm}^3 \)

Question. The diameter of a sphere is 14 cm. Its volume is :
(a) \( 1428 \text{ cm}^3 \)
(b) \( 1439 \text{ cm}^3 \)
(c) \( 1437.3 \text{ cm}^3 \)
(d) \( 1440 \text{ cm}^3 \)
Answer: (c) \( 1437.3 \text{ cm}^3 \)
Explanation :
Diameter = 14 cm
So, the radius = 7 cm
Volume of the sphere \( = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (7)^3 = 1437.3 \text{ cm}^3 \)

Question. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is :
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m
Answer: (c) 36 m
Explanation :
Diameter of sphere = 6 cm
\ Radius \( (r) = \frac{6}{2} = 3 \text{ cm} \)
Volume \( = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times (3)^3 = 36\pi \text{ cm}^3 \)
Diameter of wire = 2 mm
\ Radius \( (r_2) = \frac{2}{2} \text{ mm} = 1 \text{ mm} = \frac{1}{10} \text{ cm} \)
Let \( h \) be its length, then \( \pi (r_2)^2 h = 36\pi \)
\( \Rightarrow \pi \times \left(\frac{1}{10}\right)^2 h = 36\pi \Rightarrow h = 3600 \text{ cm} = 36 \text{ m} \)

Question. The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is :
(a) 27 : 20
(b) 20 : 27
(c) 4 : 9
(d) 9 : 4
Answer: (b) 20 : 27
Explanation :
Let the radii of the two cylinders be \( 2x \) and \( 3x \), and the heights of the two cylinders be \( 5y \) and \( 3y \) respectively.
Ratio of the volume of the cylinders \( = \frac{\pi (2x)^2 (5y)}{\pi (3x)^2 (3y)} = \frac{4x^2 \times 5y}{9x^2 \times 3y} = \frac{20}{27} \)
That is, the ratio of their volume is 20 : 27.

Question. The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is :
(a) 1 : 2
(b) 2 : 3
(c) 9 : 16
(d) 16 : 9
Answer: (d) 16 : 9
Explanation :
Let radii be \( r_1 \) and \( r_2 \).
Given, \( \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{64}{27} \Rightarrow \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{4}{3}\right)^3 \Rightarrow \frac{r_1}{r_2} = \frac{4}{3} \)
Ratio of their surface areas \( = \frac{4 \pi r_1^2}{4 \pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)
Hence, \( S_1 : S_2 = 16 : 9 \)

Question. A metallic sphere of radius 10.5cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of such cones is :
(a) 63
(b) 126
(c) 21
(d) 130
Answer: (b) 126
Explanation :
Radius of metallic sphere = 10.5 cm
\ Volume of the sphere \( = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times (10.5)^3 = 4630.5\pi \dots \text{(i)} \)
Now, Radius of the cone = 3.5 cm and height of the cone = 3 cm
\ Volume of the cone \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times (3.5)^2 \times 3 = 36.75\pi \dots \text{(ii)} \)
Number of cone \( = \frac{\text{Volume of sphere}}{\text{Volume of cone}} = \frac{4630.5\pi}{36.75\pi} = 126 \)

Question. Water flows at the rate of 10 metre per minute from a cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cm and depth 24 cm?
(a) 48 minutes 15 sec
(b) 51 minutes 12 sec
(c) 52 minutes 1 sec
(d) 55 minutes
Answer: (b) 51 minutes 12 sec
Explanation :
The radius of cylindrical pipe \( r = \frac{5}{2} \text{ mm} = 0.25 \text{ cm} \).
The volume per minute of water flow from the pipe \( = \pi \times (0.25)^2 \times 1000 = 62.5\pi \text{ cm}^3/\text{min} \).
The radius of cone \( = \frac{40}{2} = 20 \text{ cm} \) and depth \( = 24 \text{ cm} \).
Volume of cone \( = \frac{1}{3} \pi \times (20)^2 \times 24 = 3200\pi \text{ cm}^3 \).
Time taken \( = \frac{3200\pi}{62.5\pi} = 51.2 \text{ min} = 51 \text{ min } 12 \text{ sec} \).

Question. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is :
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 12 cm
Answer: (c) 6 cm
Explanation :
Let \( r \) be the radius of sphere. Given, \( 4\pi r^2 = 2\pi rh \).
\( 4\pi r^2 = 2\pi \times 6 \times 12 \Rightarrow 4r^2 = 144 \Rightarrow r^2 = 36 \Rightarrow r = 6 \text{ cm} \).

Question. If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere, the diameter of the sphere is :
(a) 12 cm
(b) 24 cm
(c) 30 cm
(d) 36 cm
Answer: (b) 24 cm
Explanation :
Sum of volume of 3 spheres \( = \frac{4}{3}\pi (6^3 + 8^3 + 10^3) = \frac{4}{3}\pi (216 + 512 + 1000) = \frac{4}{3}\pi \times 1728 \).
Let \( R \) be the radius of new sphere. \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \times 1728 \Rightarrow R^3 = 1728 \Rightarrow R = 12 \text{ cm} \).
Diameter \( = 2R = 24 \text{ cm} \).

Question. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is :
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 12 cm
Answer: (c) 6 cm

Question. If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere, the diameter of the sphere is :
(a) 12 cm
(b) 24 cm
(c) 30 cm
(d) 36 cm
Answer: (b) 24 cm

Question. The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is :
(a) \( \frac{4}{3} \pi \)
(b) \( \frac{10}{3} \pi \)
(c) \( 5 \pi \)
(d) \( \frac{20}{3} \pi \)
Answer: (a) \( \frac{4}{3} \pi \)

Question. A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the water rises by :
(a) 4.5 cm
(b) 3 cm
(c) 4 cm
(d) 2 cm
Answer: (a) 4.5 cm

Question. 12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is :
(a) \( \sqrt{3} \) cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Answer: (d) 4 cm

Question. What is the formula required to use for T.S.A. of an article which is made by dragging out a hemisphere from each end of a soild cylinder?
(a) C.S.A. of the cylinder – 2(C.S.A. of the hemisphere)
(b) C.S.A. of the cylinder + C.S.A. of the hemisphere
(c) C.S.A. of the cylinder + 2(C.S.A. of the hemisphere)
(d) C.S.A. of the cylinder – C.S.A. of the hemisphere
Answer: (c) C.S.A. of the cylinder + 2(C.S.A. of the hemisphere)

Very Short Answer Type Questions

Question. Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere ?
Answer: Let radius of hemisphere be \( r \) units.
Volume of hemisphere = S.A. of hemisphere (Given)
\( \frac{2}{3} \pi r^3 = 3 \pi r^2 \)
\( \Rightarrow r = \frac{9}{2} \) or diameter = 9 units

Question. If the total surface area of a solid hemisphere is 462 cm\(^2\), find its volume. (Use \( \pi = \frac{22}{7} \))
Answer: Given, total surface area of a solid hemisphere = 462 cm\(^2\)
Let its radius be \( r \).
Thus \( 3 \pi r^2 = 462 \)
\( \Rightarrow 3 \left( \frac{22}{7} \right) r^2 = 462 \)
\( \Rightarrow r^2 = 49 \Rightarrow r = 7 \) cm
So, volume of the hemisphere \( = \frac{2}{3} \pi r^3 = \frac{2}{3} \left( \frac{22}{7} \right) (7)^3 = 718.67 \) cm\(^3\).

Question. Two circular pieces of equal radii and maximum area touching each other are cut from a rectangular cardboard of dimensions 28 cm × 14 cm. Find the area of the remaining cardboard.
Answer: Given, dimensions of cardboard = 28 cm × 14 cm
Area = 392 cm\(^2\)
As the radii of the circles are equal and they have maximum area, their diameters are equal to the width of the cardboard.
Thus \( d = 14 \) cm or \( r = 7 \) cm
Area of the two circles \( = 2 \pi r^2 = 2 \left( \frac{22}{7} \right) (7)^2 = 308 \) cm\(^2\)
Area of the remaining portion = Area of cardboard – Area of two circles \( = 392 - 308 = 84 \) cm\(^2\).

Question. A solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed. (Use \( \pi = \frac{22}{7} \))
Answer: Given, Radius of sphere \( = 10.5 \) cm, Radius of cone \( = 3.5 \) cm, Height of cone \( = 3 \) cm
Volume of sphere \( = \frac{4}{3} \pi (10.5)^3 = 4851 \) cm\(^3\)
Volume of 1 cone \( = \frac{1}{3} \pi (3.5)^2 \times 3 = \frac{22}{7} \times (3.5)^2 = 38.5 \) cm\(^3\)
Number of cones \( = \frac{\text{Volume of sphere}}{\text{Volume of 1 cone}} = \frac{4851}{38.5} = 126 \).

Question. The volume of a hemisphere is 2425.5 cm\(^3\). Find its curved surface area. (Use \( \pi = \frac{22}{7} \))
Answer: Given, volume \( = 2425.5 \) cm\(^3\)
Let radius be \( r \). \( \frac{2}{3} \pi r^3 = 2425.5 \)
\( \frac{2}{3} \times \frac{22}{7} \times r^3 = \frac{24255}{10} \)
\( \Rightarrow \frac{44}{21} r^3 = \frac{4851}{2} \Rightarrow r^3 = \frac{441 \times 21}{8} = \frac{21^3}{2^3} \)
\( \Rightarrow r = \frac{21}{2} \) cm
Curved surface area \( = 2 \pi r^2 = 2 \times \frac{22}{7} \times \left( \frac{21}{2} \right)^2 = 693 \) cm\(^2\).

Question. Two cubes, each of volume 27 cm\(^3\) are joined end to end to form a solid. Find the surface area of the resulting cuboid.
Answer: Volume of each cube \( = 27 \) cm\(^3 \Rightarrow a^3 = 27 \Rightarrow a = 3 \) cm.
For resulting cuboid: length \( l = 3 + 3 = 6 \) cm, width \( w = 3 \) cm, height \( h = 3 \) cm.
Surface area \( = 2(lw + wh + hl) = 2(6 \times 3 + 3 \times 3 + 6 \times 3) = 2(18 + 9 + 18) = 2(45) = 90 \) cm\(^2\).

Question. A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.
Answer: Volume of cone \( = \frac{1}{3} \pi (5)^2 (20) \) cm\(^3\).
Let radius of sphere be \( r \). Volume of sphere \( = \text{Volume of cone} \)
\( \frac{4}{3} \pi r^3 = \frac{1}{3} \pi (5)^2 (20) \Rightarrow 4r^3 = 5^2 \times 20 \Rightarrow r^3 = 125 \Rightarrow r = 5 \) cm.
Diameter \( = 2r = 10 \) cm.

Question. Two cubes each of side 4 cm are joined end to end to form a solid. Find the surface area of the resulting cuboid.
Answer: Side of each cube \( a = 4 \) cm.
For cuboid: \( l = 8 \) cm, \( w = 4 \) cm, \( h = 4 \) cm.
Surface area \( = 2(8 \times 4 + 4 \times 4 + 4 \times 8) = 2(32 + 16 + 32) = 2(80) = 160 \) cm\(^2\).

Question. The dimensions of a metallic cuboid are 100 cm × 80 cm × 64 cm. It is melted and recast into a cube. Find the surface area of the cube.
Answer: Volume of cuboid \( = 100 \times 80 \times 64 = 512000 \) cm\(^3\).
Let side of cube be \( a \). \( a^3 = 512000 \Rightarrow a = 80 \) cm.
Total surface area of cube \( = 6a^2 = 6(80)^2 = 6 \times 6400 = 38400 \) cm\(^2\).

Question. A conical vessel, whose internal radius is 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which the water rises in the cylindrical vessel.
Answer: Volume of water \( = \frac{1}{3} \pi (5)^2 (24) \).
Let height in cylindrical vessel be \( h \). Volume in cylinder \( = \pi (10)^2 h \).
\( \pi (100) h = \frac{1}{3} \pi (25)(24) \Rightarrow 100h = 25 \times 8 \Rightarrow 100h = 200 \Rightarrow h = 2 \) cm.

Question. A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is filled into small cylindrical bottles, each of diameter 3 cm and height 4 cm. How many bottles are needed to empty the bowl ?
Answer: Volume of hemisphere \( = \frac{2}{3} \pi (9)^3 \).
Radius of cylinder \( = 1.5 \) cm, height \( = 4 \) cm. Volume of bottle \( = \pi (1.5)^2 (4) \).
Number of bottles \( = \frac{\frac{2}{3} \pi \times 729}{\pi \times 2.25 \times 4} = \frac{486}{9} = 54 \).

Question. A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls so formed.
Answer: Volume of cone \( = \frac{1}{3} \pi (12)^2 (24) \).
Radius of spherical ball \( = 3 \) cm. Volume of one ball \( = \frac{4}{3} \pi (3)^3 \).
Number of balls \( = \frac{\frac{1}{3} \pi (144)(24)}{\frac{4}{3} \pi (27)} = \frac{144 \times 24}{4 \times 27} = 32 \).

Question. The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm respectively. It is melted and recast into a solid cone of base diameter 14 cm. Find the height of the cone so formed.
Answer: Internal radius \( r_1 = 3 \) cm, external radius \( r_2 = 5 \) cm.
Volume of shell \( = \frac{2}{3} \pi (5^3 - 3^3) = \frac{2}{3} \pi (125 - 27) = \frac{2}{3} \pi (98) \).
Radius of cone \( R = 7 \) cm. Volume of cone \( = \frac{1}{3} \pi (7)^2 h \).
\( \frac{1}{3} \pi (49) h = \frac{2}{3} \pi (98) \Rightarrow 49h = 196 \Rightarrow h = 4 \) cm.

Question. A hemispherical bowl of internal diameter 30 cm is full of liquid. The liquid is filled into small cylindrical bottles each of diameter 5 cm and height 6 cm. How many bottles are needed to empty the bowl ?
Answer: Volume of bowl \( = \frac{2}{3} \pi (15)^3 \).
Radius of bottle \( = 2.5 \) cm, height \( = 6 \) cm. Volume of bottle \( = \pi (2.5)^2 (6) \).
Number of bottles \( = \frac{\frac{2}{3} \pi (3375)}{\pi (6.25)(6)} = \frac{2250}{37.5} = 60 \).

Question. A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed.
Answer: Radius of sphere \( = 10.5 \) cm. Volume \( = \frac{4}{3} \pi (10.5)^3 \).
Radius of cone \( = 1.75 \) cm, height \( = 3 \) cm. Volume of cone \( = \frac{1}{3} \pi (1.75)^2 (3) \).
Number of cones \( = \frac{\frac{4}{3} \pi (10.5)^3}{\frac{1}{3} \pi (1.75)^2 (3)} = \frac{4 \times 1157.625}{3 \times 3.0625} = 504 \).

Question. A spherical cannon ball of diameter 28 cm is melted and recast into a right circular conical mould with base diameter of 35 cm. Find the height of the cone.
Answer: Radius of sphere \( = 14 \) cm. Radius of cone \( = 17.5 \) cm.
Volume of cone = Volume of sphere
\( \frac{1}{3} \pi (17.5)^2 h = \frac{4}{3} \pi (14)^3 \Rightarrow (17.5)^2 h = 4(14)^3 \)
\( \Rightarrow h = \frac{4 \times 14 \times 14 \times 14}{17.5 \times 17.5} \approx 35.84 \) cm.

Question. Three solid metallic spheres of radii 3 cm, 4 cm and 5 cm respectively are melted to form a single solid sphere. Find the diameter of the resulting sphere.
Answer: Volume of resulting sphere = Sum of volumes of three spheres.
\( \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (3^3 + 4^3 + 5^3) = \frac{4}{3} \pi (27 + 64 + 125) = \frac{4}{3} \pi (216) \).
\( R^3 = 216 \Rightarrow R = 6 \) cm.
Diameter \( = 2R = 12 \) cm.

Question. A cylindrical pipe has inner diameter of 7 cm and water flows through it at 192.51 litres per minute. Find the rate of flow in the pipe in km/hr.
Answer: Inner radius \( r = 3.5 \) cm \( = 0.035 \) m.
Flow rate \( = 192.51 \) l/min \( = 192.51 \times 60 \) l/hr \( = 11550.6 \) l/hr \( = 11.55 \) m\(^3\)/hr.
Let flow rate be \( h \) m/hr. Volume \( = \pi r^2 h = 11.55 \).
\( \frac{22}{7} \times (0.035)^2 \times h = 11.55 \Rightarrow h = 3000 \) m/hr \( = 3 \) km/hr.

Question. The internal and external diameters of a hollow spherical shell are 6 cm and 10 cm respectively. It is melted and recast into a solid cylinder of base diameter 14 cm. Find the height of the cylinder so formed.
Answer: Internal radius \( r_1 = 3 \) cm, External radius \( r_2 = 5 \) cm.
Volume of shell \( = \frac{4}{3} \pi (5^3 - 3^3) = \frac{4}{3} \pi (125 - 27) = \frac{4}{3} \pi (98) \).
Radius of cylinder \( r = 7 \) cm. Volume \( = \pi r^2 h = \pi (49) h \).
\( \pi (49) h = \frac{4}{3} \pi (98) \Rightarrow 49h = \frac{392}{3} \Rightarrow h = \frac{8}{3} \approx 2.67 \) cm.