CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set 13

Question. Two cubes each of volume \( 27 \text{ cm} \) are joined together. Find the surface area of the resulting solid?
(a) \( 109.4 \text{ cm}^2 \)
(b) \( 126 \text{ cm}^2 \)
(c) \( 150 \text{ cm}^2 \)
(d) \( 189.4 \text{ cm}^2 \)
Answer: (b) \( 126 \text{ cm}^2 \)
Explanation :
Volume of cube \( = a^3 = 27 \)
\( a = 3 \text{ cm} \)
Joining 2 cubes results in a cuboid.
Length of the cuboid \( (l) = 3 + 3 = 9 \text{ cm} \)
Height of the cuboid \( (h) = 3 \text{ cm} \)
Breadth of the cuboid \( (b) = 3 \text{ cm} \)
Surface area of cuboid \( = 2(lb + bh + hl) \)
\( = 2(27 + 9 + 27) \)
\( = 126 \text{ cm}^2 \)

Question. A cylinder and a cone are of the same base radius and height. Calculate the ratio of the volume of the cylinder and the cone.
Answer: Sol. Given, Radius of cylinder = Radius of cone \( = r \text{ cm} \)
and Height of cylinder = Height of cone \( = h \text{ cm} \)
Thus, \( \frac{\text{Volume of cylinder}}{\text{Volume of cone}} = \frac{\pi r^2 h}{\frac{1}{3} \pi r^2 h} \)
\( = \frac{3}{1} \)
Hence, Volume of cylinder : Volume of cone \( = 3 : 1 \).

Question. Two types of water tankers are available in a shop. One is in a cubic form of dimensions \( 1 \text{ m} \times 1 \text{ m} \times 1 \text{ m} \) and another is in the cylindrical form of height \( 1 \text{ m} \) and diameter \( 1 \text{ m} \). Calculate the volume of both the containers. (Use \( \pi = 3.14 \))
Answer: Sol. Dimensions of cubic tank,
\( l = 1 \text{ m}, b = 1 \text{ m}, h = 1 \text{ m} \)
\( \therefore \) Volume of cubic tank \( = lbh = 1 \times 1 \times 1 = 1 \text{ m}^3 \)
Height of cylindrical container \( = 1 \text{ m} \).
Radius \( = \frac{1}{2} \text{ m} \)
\( \therefore \) Volume of cylindrical tank \( = \pi r^2 h \)
\( = 3.14 \times \frac{1}{2} \times \frac{1}{2} \times 1 \)
\( = 0.785 \text{ m}^3 \)

Short Answer Type Questions

Question. A \( 5 \text{ m} \) wide cloth is used to make a conical tent of base diameter \( 14 \text{ m} \) and height \( 24 \text{ m} \). Find the cost of cloth used at the rate of ` \( 25 \) per m.
[Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, width of cloth \( = 5 \text{ m} \)
Base diameter of tent \( = 14 \text{ m} \)
\( \therefore \) Radius \( = \frac{14}{2} = 7 \text{ cm} \)
Height \( = 24 \text{ m} \)
Slant height of cone,
\( l' = \sqrt{h^2 + r^2} \)
\( = \sqrt{(24)^2 + (7)^2} \)
\( = \sqrt{576 + 49} = \sqrt{625} \)
\( = 25 \text{ cm} \)
Let the length of the cloth be \( l \text{ m} \).
\( \therefore \) Curved surface area of cone \( = \) Area of cloth
\( \Rightarrow \pi r l' = l \times b \)
\( \Rightarrow \frac{22}{7} \times 7 \times 25 = l \times 5 \)
\( \Rightarrow 22 \times 25 = l \times 5 \)
\( \Rightarrow l = \frac{22 \times 25}{5} \text{ m} \)
\( \Rightarrow l = 110 \text{ m} \)
Thus, total cost of the cloth \( = \text{` } (25 \times 110) = \text{` } 2750 \).

Question. A girl empties a cylindrical bucket, full of sand, of base radius \( 18 \text{ cm} \) and height \( 32 \text{ cm} \) on the floor, to form a conical heap of sand. If the height of this heap is \( 24 \text{ cm} \), find the slant height correct to two decimal places.**
Answer: Sol. Given, base radius of cylindrical bucket \( = 18 \text{ cm} \)
Height of cylindrical bucket \( = 32 \text{ cm} \)
Height of conical heap \( = 24 \text{ cm} \)
Let the radius of the cone be \( r \text{ cm} \).
Now, Volume of cylinder \( = \) Volume of cone
\( \Rightarrow \pi(18)^2 \times 32 = \frac{1}{3} \pi r^2 \times 24 \)
\( \Rightarrow r^2 = \frac{18 \times 18 \times 32 \times 3}{24} \)
\( = 18 \times 18 \times 4 \)
\( \Rightarrow r = \sqrt{18 \times 18 \times 2 \times 2} \)
\( = 18 \times 2 = 36 \text{ cm} \)
Thus, slant height of the cone \( = \sqrt{(24)^2 + (36)^2} \)
\( = \sqrt{576 + 1296} = \sqrt{1872} \)
\( = 43.27 \text{ cm} \).

Question. The largest possible sphere is carved out of a solid wooden cube of side \( 7 \text{ cm} \). Find the volume of the wood left. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Length of each side of the cube \( = 7 \text{ cm} \)
\( \therefore \) Largest possible diameter of the sphere \( = 7 \text{ cm} \)
\( \therefore \) Radius of the sphere \( = 3.5 \text{ cm} \)
Thus, volume of cube \( = (7)^3 \text{ cm}^3 = 343 \text{ cm}^3 \)
and volume of sphere \( = \frac{4}{3} \times \frac{22}{7} \times (3.5)^3 \text{ cm}^3 \)
\( = \frac{539}{3} \text{ cm}^3 \)
\( \therefore \) Remaining volume of wood \( = (343 - \frac{539}{3}) \text{ cm}^3 \)
\( = \frac{1029 - 539}{3} \text{ cm}^3 = 163.33 \text{ cm}^3 \).

Question. Due to heavy floods in a state, thousands were rendered homeless. \( 50 \) schools collectively offered to the state government to provide place and the canvas for \( 1500 \) tents to be fixed by the government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius \( 2.8 \text{ m} \) and height \( 3.5 \text{ m} \), with conical upper part of same base radius but of height \( 2.1 \text{ m} \). If the canvas used to make the tents costs ` \( 120 \) per sq. m, find the amount shared by each school to set-up the tents. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Radius of the base of cylinder \( (r) = 2.8 \text{ m} \)
Radius of the base of the cone \( (r) = 2.8 \text{ m} \)
Height of the cylinder \( (h) = 3.5 \text{ m} \)
Height of the cone \( (H) = 2.1 \text{ m} \)
Slant height of conical part \( (l) = \sqrt{r^2 + H^2} \)
\( = \sqrt{(2.8)^2 + (2.1)^2} = \sqrt{7.84 + 4.41} \)
\( = \sqrt{12.25} = 3.5 \text{ m} \)
Area of canvas used to make tent \( = \) CSA of cylinder \( + \) CSA of cone
\( = 2\pi rh + \pi rl \)
\( = 2 \times \frac{22}{7} \times 2.8 \times 3.5 + \frac{22}{7} \times 2.8 \times 3.5 \)
\( = 61.6 + 30.8 = 92.4 \text{ m}^2 \)
Cost of \( 1500 \) tents at ` \( 120 \) per sq. m \( = 1500 \times 120 \times 92.4 \)
\( = \text{` } 16,632,000 \)
Share of each school to set-up the tents \( = \frac{16632000}{50} = \text{` } 332,640 \).

Question. A housing society collects rain water from the roof of its building of area \( 22 \text{ m} \times 20 \text{ m} \) in a cylindrical vessel of diameter \( 2 \text{ m} \) and height \( 3.5 \text{ m} \) and then pumps this water into the main water tank so that everyone can use it. On a particular day, the rain water collected from the roof just fills the cylindrical vessel to the brim. Calculate the rainfall in cm.**
Answer: Sol. Given, Area of the roof \( = 22 \text{ m} \times 20 \text{ m} \)
Height of cylindrical vessel \( = 3.5 \text{ m} \)
Diameter of the vessel \( = 2 \text{ m} \)
\( \therefore \) Radius of the vessel \( = 1 \text{ m} \)
Let the total rainfall be \( h \text{ m} \).
Volume of rain water collected on the roof \( = \) Volume of the cylindrical vessel
\( \Rightarrow 22 \times 20 \times h = \pi(1)^2(3.5) \)
\( \Rightarrow 22 \times 20 \times h = \frac{22}{7}(3.5) \)
\( \Rightarrow 20h = 0.5 \)
\( \Rightarrow h = \frac{1}{40} \text{ m} = 2.5 \text{ cm} \).

Question. A hemispherical bowl of internal radius \( 9 \text{ cm} \) is full of water. Its contents are emptied in a cylindrical vessel of internal radius \( 6 \text{ cm} \). Find the height of the water in the cylindrical vessel.
Answer: Sol. Given, internal radius of hemispherical bowl \( = 9 \text{ cm} \)
Internal radius of cylindrical vessel \( = 6 \text{ cm} \)
Let the height of the water in the cylindrical vessel be \( h \text{ cm} \).
Now, Volume of the hemispherical vessel \( = \) Volume of water in the cylindrical vessel
\( \Rightarrow \frac{2}{3} \pi R^3 = \pi r^2 h \)
\( \Rightarrow \frac{2}{3} \pi (9)^3 = \pi (6)^2 h \)
\( \Rightarrow \frac{2}{3} (9)^3 = (6)^2 h \)
\( \Rightarrow h = \frac{2 \times 9 \times 9 \times 9}{3 \times 6 \times 6} = \frac{27}{2} \text{ cm} \)
\( \Rightarrow h = 13.5 \text{ cm} \)
Hence, the height of the water level in the cylindrical vessel \( = 13.5 \text{ cm} \).

Question. From a solid cylinder of height \( 7 \text{ cm} \) and base diameter of \( 12 \text{ cm} \), a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, Height of cylinder \( = \) Height of cone \( = 7 \text{ cm} \)
Base diameter of cylinder \( = \) Base diameter of cone \( = 12 \text{ cm} \)
Thus, base radius \( = 6 \text{ cm} \)
Thus, total surface area of the remaining solid \( = \) Total surface area of cylinder \( + \) Curved surface area of cone \( - \) Area of the base of the cylinder and cone
\( = 2\pi r(h + r) + \pi rl - \pi r^2 \)
\( = 2\pi \times 6(7 + 6) + \pi \times 6(\sqrt{6^2 + 7^2}) - \pi(6)^2 \text{ cm}^2 \)
\( = \pi [156 + 55.317 - 36] \text{ cm}^2 \)
\( = \pi [175.317] \text{ cm}^2 \)
\( = 550.997 \text{ cm}^2 \).

Question. The dimensions of a solid iron cuboid are \( 4.4 \text{ m} \times 2.6 \text{ m} \times 1.0 \text{ m} \). It is melted and recast into a hollow cylindrical pipe of \( 30 \text{ cm} \) inner radius and thickness \( 5 \text{ cm} \). Find the length of the pipe.
Answer: Sol. Volume of cuboid \( = 4.4 \times 2.6 \times 1 \text{ m}^3 \)
Inner radius of cylindrical pipe, \( r = 30 \text{ cm} = \frac{30}{100} \text{ m} \)
Thickness of pipe \( = 5 \text{ cm} \)
Outer radius of cylindrical pipe, \( R = 30 + 5 = 35 \text{ cm} = \frac{35}{100} \text{ m} \)
\( \therefore \) Volume of material used \( = \pi (R^2 - r^2)h \)
\( = \frac{22}{7} [(\frac{35}{100})^2 - (\frac{30}{100})^2] h \text{ m}^3 \)
\( = \frac{22}{7} (\frac{65}{100}) (\frac{5}{100}) h \text{ m}^3 \)
Now, \( \frac{22}{7} (\frac{65}{100}) (\frac{5}{100}) h = 4.4 \times 2.6 \)
\( \Rightarrow h = \frac{7 \times 4.4 \times 2.6 \times 100 \times 100}{22 \times 65 \times 5} \)
\( \Rightarrow h = 112 \text{ m} \).

Question. A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other. Their common diameter is \( 4.2 \text{ cm} \) and the heights of the cylindrical and conical portions are \( 12 \text{ cm} \) and \( 7 \text{ cm} \) respectively. Find the volume of the given toy. [Use \( \pi = \frac{22}{7} \)] 
Answer: Sol. Given, Diameter of hemisphere \( = \) Diameter of cone \( = 4.2 \text{ cm} \)
Thus, Radius of hemisphere \( = \) Radius of cylinder \( = \) Radius of cone \( = 2.1 \text{ cm} \)
Height of the cylinder \( = 12 \text{ cm} \)
Height of the cone \( = 7 \text{ cm} \)
Hence, volume of the given toy \( = \) Volume of (hemisphere \( + \) cylinder \( + \) cone)
\( = \frac{2}{3}\pi r^3 + \pi r^2 h + \frac{1}{3}\pi r^2 \times h \)
\( = [\frac{2}{3} \pi(2.1)^3 + \pi(2.1)^2(12) + \frac{1}{3} \pi(2.1)^2(7)] \text{ cm}^3 \)
\( = \pi(2.1)^2 [\frac{2}{3}(2.1) + (12) + \frac{1}{3}(7)] \text{ cm}^3 \)
\( = (\frac{22}{7})(\frac{21}{10})(\frac{21}{10}) [\frac{2}{3}(\frac{21}{10}) + (12) + \frac{7}{3}] \text{ cm}^3 \)
\( = 22(\frac{3}{10})(\frac{21}{10}) [\frac{7}{5} + 12 + \frac{7}{3}] \text{ cm}^3 \)
\( = 22(\frac{3}{10})(\frac{21}{10}) [\frac{21 + 180 + 35}{15}] \text{ cm}^3 \)
\( = 22(\frac{3}{10})(\frac{21}{10}) [\frac{236}{15}] \text{ cm}^3 \)
\( = 22(\frac{1}{10})(\frac{21}{10}) [\frac{236}{5}] \text{ cm}^3 \)
\( = 218.064 \text{ cm}^3 \).

Question. A solid is made up of a cube surmounted by a hemisphere. If the edges of the cube equal to \( 4.2 \text{ cm} \), find the total surface area. [Use \( \pi = \frac{22}{7} \)]  
Answer: Sol. Given, side of the cube \( = 5 \text{ cm} \)
Diameter of the hemisphere \( = 4.2 \text{ cm} \)
Hence, radius of the hemisphere \( = 2.1 \text{ cm} \)
Thus, total surface area of the solid
\( = \) Total surface area of cube \( + \) Curved surface area of hemisphere \( - \) Area of the base of the hemisphere
\( = [6(5)^2 + 2\pi(2.1)^2 - \pi(2.1)^2] \text{ cm}^2 \)
\( = [150 + \frac{22}{7}(2.1)^2] \text{ cm}^2 \)
\( = [150 + \frac{22}{7}(\frac{21}{10})(\frac{21}{10})] \text{ cm}^2 \)
\( = [150 + 22(\frac{3}{10})(\frac{21}{10})] \text{ cm}^2 \)
\( = [150 + 13.86] \text{ cm}^2 \)
\( = 163.86 \text{ cm}^2 \).

Question. A hemispherical bowl of internal diameter \( 36 \text{ cm} \) contains liquid. This liquid is filled into \( 72 \) cylindrical bottles of diameter \( 6 \text{ cm} \). Find the height of each bottle, if \( 10\% \) liquid is wasted in this transfer.
Answer: Sol. Internal diameter of hemispherical bowl \( = 36 \text{ cm} \)
\( \therefore \) Radius of hemispherical bowl \( (r) = 18 \text{ cm} \)
Volume of liquid, \( V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \pi \times (18)^3 \)
Now, Diameter of bottle \( = 6 \text{ cm} \)
\( \therefore \) Radius of bottle \( = 3 \text{ cm} \)
Now, Volume of cylindrical bottle \( = \pi R^2 h = \pi \times 3^2 h = 9\pi h \)
Volume of liquid transferred \( = \) Volume of liquid \( - 10\% \) volume of liquid
\( = \frac{2}{3} \pi (18)^3 - \frac{10}{100} (\frac{2}{3} \pi (18)^3) \)
\( = \frac{2}{3} \pi (18)^3 (1 - \frac{10}{100}) \)
\( = \frac{2}{3} \pi (18)^3 \times \frac{9}{10} \)
Volume of liquid transferred \( = \pi \times (18)^3 \times \frac{3}{5} \)
Number of cylindrical bottles \( = \frac{\text{Volume of liquid to be transferred}}{\text{Volume of a bottle}} \)
\( \Rightarrow 72 = \frac{\pi \times 18 \times 18 \times 18 \times \frac{3}{5}}{9\pi h} \)
\( h = \frac{27}{5} = 5.4 \text{ cm} \)
Hence, height of each bottle will be \( 5.4 \text{ cm} \).

Question. A cubical block of side \( 10 \text{ cm} \) is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of ` \( 5 \) per \( 100 \text{ sq. cm} \). [Use \( \pi = 3.14 \)]
Answer: Sol. Side of the cubical block \( (a) = 10 \text{ cm} \).
Since, the cube is surmounted by a hemisphere, therefore, the greatest diameter of the hemisphere should be equal to the side of the cube.
\( \therefore \) Diameter of the sphere \( = 10 \text{ cm} \)
\( \Rightarrow \) Radius of the sphere \( (r) = 5 \text{ cm} \)
Total surface area of solid \( = \) T.S.A. of the cube \( + \) C.S.A. of hemisphere \( - \) Inner cross-section area of hemisphere
\( = 6a^2 + 2\pi r^2 - \pi r^2 = 6a^2 + \pi r^2 \)
\( = 6(10)^2 + 3.14(5)^2 \)
\( = 600 + 25 \times 3.14 \)
\( = 600 + 78.5 = 678.5 \text{ cm}^2 \)
Cost of painting per square metre is ` \( 5 \)
Total cost for painting \( = \frac{678.5}{100} \times 5 = \text{` } 33.92 \)
Hence, total cost for painting will be ` \( 33.92 \).

Question. \( 504 \) cones each of diameter \( 3.5 \text{ cm} \) and height \( 3 \text{ cm} \) are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Diameter of each cone \( (d) = 3.5 \text{ cm} \)
\( \therefore \) Radius of each cone \( (r) = \frac{3.5}{2} = \frac{7}{4} \text{ cm} \)
Height of each cone \( (h) = 3 \text{ cm} \)
Volume of \( 504 \) cones \( = 504 \times \) Volume of one cone
\( = 504 \times \frac{1}{3} \pi r^2 h \)
\( = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 3 \)
Let radius of sphere be \( R \text{ cm} \).
\( \therefore \) Volume of sphere \( = \) Volume of \( 504 \) cones
\( \frac{4}{3} \times \frac{22}{7} \times R^3 = 504 \times \frac{1}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 3 \)
\( R^3 = \frac{504 \times 7 \times 7 \times 3}{4 \times 4 \times 4} \)
\( R = \sqrt[3]{\frac{3 \times 3 \times 7 \times 7 \times 7 \times 3}{2 \times 2 \times 2}} \)
\( R = \frac{21}{2} \text{ cm} \)
Now, surface area of sphere \( = 4\pi R^2 \)
\( = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \)
\( = 63 \times 22 = 1386 \text{ cm}^2 \)
Hence, surface area of sphere is \( 1386 \text{ cm}^2 \).

Question. The sum of the radius of base and height of a solid right circular cylinder is \( 37 \text{ cm} \). If the total surface area of the solid cylinder is \( 1628 \text{ sq. cm} \), find the volume of the cylinder. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Let the radius of base and height of a solid cylinder be \( r \) and \( h \) respectively.
Now, we have
\( r + h = 37 \text{ cm} \)...(i)
and T.S.A. of solid cylinder \( = 2\pi r (r + h) \)
\( = 1628 \text{ cm}^2 \)
\( \Rightarrow 2\pi r (37) = 1628 \) [Using (i)]
\( \Rightarrow r = \frac{1628}{37 \times 2 \times \frac{22}{7}} \)
\( r = 7 \text{ cm} \)
\( \therefore \) Volume of the cylinder \( = \pi r^2 h \)
\( = \frac{22}{7} \times 7 \times 7 \times 30 \) [Using eq. (i), \( h = 30 \)]
\( = 4620 \text{ cm}^3 \).

Question. A well of diameter \( 4 \text{ m} \) is dug \( 21 \text{ m} \) deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width \( 3 \text{ m} \) to form an embankment. Find the height of the embankment.
Answer: Sol. Given, diameter and height of cylindrical well are \( 4 \text{ m} \) and \( 21 \text{ m} \) respectively.
Now, the earth has been taken out to spread evenly all around.
Then, volume of earth dug out \( = \frac{22}{7} \times \frac{4}{2} \times \frac{4}{2} \times 21 = 264 \text{ m}^3 \)
and the volume of embankment of width \( 3 \text{ m} \) which forms a shape of circular ring
\( = \pi [(5)^2 - (2)^2] \times h \)
[... Outer radius \( = 2 + 3 = 5 \text{ cm} \)]
\( = \frac{22}{7} (25 - 4) \times h = 66 h \text{ m} \)
... Volume of earth dug out \( = \) Volume of embankment
\( \Rightarrow 264 = 66h \)
\( \Rightarrow h = \frac{264}{66} = 4 \text{ m} \)
Hence, the height of the embankment is \( 4 \text{ m} \).

Question. The internal and external radii of a hollow sphere are \( 3 \text{ cm} \) and \( 5 \text{ cm} \) respectively. The sphere is melted to form a solid cylinder of height \( \frac{8}{3} \text{ cm} \). Find the diameter and the curved surface area of the cylinder. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, external radius of sphere \( = 5 \text{ cm} \)
Internal radius of sphere \( = 3 \text{ cm} \)
Height of the solid cylinder \( = \frac{8}{3} \text{ cm} \)
Let the radius of the cylinder be \( r \text{ cm} \).
Thus, volume of hollow sphere \( = \) Volume of cylinder
\( \Rightarrow \frac{4}{3}\pi [(5)^3 - (3)^3] = \pi r^2 (\frac{8}{3}) \)
\( \Rightarrow (125 -

Question. The diameter of a metallic sphere is \( 6 \text{ cm} \). The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is \( 36 \text{ m} \), find its radius.*
Answer: Sol. Given, diameter of the sphere \( = 6 \text{ cm} \)
Thus, radius of the sphere \( = 3 \text{ cm} \)
Length of the wire \( = 36 \text{ m} = 3600 \text{ cm} \)
Let the radius of the wire be \( r \).
Hence,
Volume of the wire \( = \) Volume of the sphere
\( \Rightarrow \pi r^2(3600) = \frac{4}{3} \pi (3)^3 \)
\( \Rightarrow r^2(3600) = \frac{4}{3} (3)^3 \)
\( \Rightarrow r^2(3600) = 4(3)^2 \)
\( \Rightarrow r^2(3600) = 36 \)
\( \Rightarrow r^2(100) = 1 \)
\( \Rightarrow r^2 = (\frac{1}{10})^2 \)
\( \Rightarrow r = \frac{1}{10} \text{ cm} \)
\( \Rightarrow r = 0.1 \text{ cm} = 1 \text{ mm} \).

Question. The difference between the outer and inner curved surface area of a hollow right-circular cylinder \( 14 \text{ cm} \) long is \( 88 \text{ cm}^2 \). If the volume of metal used in making the cylinder is \( 176 \text{ cm}^3 \), find the outer and inner diameters of the cylinder. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, volume of the cylinder \( = 176 \text{ cm}^3 \)
Height of the cylinder \( = 14 \text{ cm} \)
and difference between outer and inner curved surface area \( = 88 \text{ cm}^2 \)
Now, let the outer radius be \( R \text{ cm} \) and the inner radius be \( r \text{ cm} \)
Thus, \( \frac{22}{7} (R^2 - r^2)14 = 176 \)
\( \Rightarrow (R^2 - r^2) = 4 \)
\( \Rightarrow (R - r) (R + r) = 4 \) …(i)
Now, outer curved surface area \( - \) Inner curved surface area \( = 88 \text{ cm}^2 \)
\( \Rightarrow 2\pi R(14) - 2\pi r(14) = 88 \)
\( \Rightarrow 2 \times \frac{22}{7} \times (14) (R - r) = 88 \)
\( \Rightarrow (R - r) = 1 \) …(ii)
Substituting equation (ii) in (i), we get
\( (R + r) = 4 \) …(iii)
Adding equations (ii) and (iii), we get
\( 2R = 5 \)
or \( R = 2.5 \text{ cm} \)
and \( r = 1.5 \text{ cm} \)
Thus, the outer diameter \( = 5 \text{ cm} \)
and the inner diameter \( = 3 \text{ cm} \).

Question. A sphere of diameter \( 12 \text{ cm} \), is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by \( 3\frac{5}{9} \text{ cm} \). Find the diameter of the cylindrical vessel.*
Answer: Sol. Given, diameter of sphere \( = 12 \text{ cm} \)
Then, radius of sphere \( (r) = \frac{12}{2} = 6 \text{ cm} \)
\( \therefore \) Volume of sphere \( = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \times \pi \times (6)^3 \text{ cm}^3 \)
Now, sphere is completely submerged in water and rise in water in cylindrical vessel is \( 3\frac{5}{9} \text{ cm} \).
Volume of sphere \( = \) Volume of change in water level in cylindrical vessel
\( \frac{4}{3} \pi \times (6)^3 = \pi r^2 \times \frac{32}{9} \)
\( r^2 = \frac{4 \times 6 \times 6 \times 6 \times 9}{3 \times 32} \)
\( r = \sqrt{81} \)
\( r = 9 \text{ cm} \).
\( \therefore \) Diameter of the cylindrical vessel is \( 18 \text{ cm} \).

Question. From a solid cylinder of height \( 10 \text{ cm} \) and base radius of \( 6 \text{ cm} \), a conical cavity of same height and same base diameter is hollowed out. Find the volume and the total surface area of the remaining solid. [Use \( \pi = 3.14 \)]
Answer: Sol. Given, Height of the cylinder \( = \) Height of the cone \( = 10 \text{ cm} \)
Base radius of the cylinder \( = \) Base radius of the cone \( = 6 \text{ cm} \)
Thus, total surface area of the remaining solid \( = \) Total surface area of the cylinder \( + \) Curved surface area of the cone \( - \) Area of the base of the cylinder and cone
\( = 2\pi \times 6(10 + 6) + \pi \times 6(\sqrt{6^2 + 10^2}) - \pi(6)^2 \text{ cm}^2 \)
\( = \pi [12(16) + 6(\sqrt{36 + 100}) - 36] \text{ cm}^2 \)
\( = \pi [192 + 6(\sqrt{136}) - 36] \text{ cm}^2 \)
\( = \pi [156 + 6(2\sqrt{34})] \text{ cm}^2 \)
\( = 12\pi [13 + \sqrt{34}] \text{ cm}^2 \)
\( = 12(3.14) [13 + \sqrt{34}] \text{ cm}^2 \)
\( = 709.55 \text{ cm}^2 \)
and Volume of the remaining solid \( = \) Volume of the cylinder \( - \) Volume of the cone
\( = [\pi(6)^2 10 - \frac{1}{3} \pi (6)^2 10] \text{ cm}^3 \)
\( = [\frac{2}{3} \pi (6)^2 10] \text{ cm}^3 \)
\( = [\frac{2}{1} \pi (12) 10] \text{ cm}^3 \)
\( = 753.6 \text{ cm}^3 \).

Question. Cylindrical vessel with internal diameter \( 10 \text{ cm} \) and height \( 10.5 \text{ cm} \) is full of water. A solid cone of base diameter \( 7 \text{ cm} \) and height \( 6 \text{ cm} \) is completely immersed in the vessel. Find the volume of water displaced and the volume remaining. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, diameter of cylindrical vessel \( = 10 \text{ cm} \)
Height of cylindrical vessel \( = 10.5 \text{ cm} \)
Diameter of cone \( = 7 \text{ cm} \)
\( \therefore \) Radius of cone \( = 3.5 \text{ cm} \)
Height of cone \( = 6 \text{ cm} \)
Total volume of water in the cylinder \( = \) Volume of cylinder
\( = \frac{22}{7} (5)^2 (10.5) \text{ cm}^3 \)
\( = \frac{22}{7} (25) (1.5) \text{ cm}^3 \)
\( = 33(25) \text{ cm}^3 = 825 \text{ cm}^3 \)
Volume of water displaced \( = \) Volume of the cone
\( = \frac{1}{3} (\frac{22}{7}) (3.5)^2 6 \text{ cm}^3 \)
\( = 77 \text{ cm}^3 \)
Thus, volume of water left behind \( = (825 - 77) \text{ cm}^3 = 748 \text{ cm}^3 \).

Question. Water in a canal, \( 5.4 \text{ m} \) wide and \( 1.8 \text{ m} \) deep, is flowing with a speed of \( 25 \text{ km/hour} \). How much area can it irrigate in \( 40 \text{ minutes} \), if \( 10 \text{ cm} \) of standing water is required for irrigation ? *☆
Answer: Sol. Volume of water flowing in \( 40 \text{ min} \)
\( = 5.4 \times 1.8 \times 25000 \times \frac{40}{60} \text{ m}^3 \)
\( = 162000 \text{ m}^3 \)
Height of standing water \( = 10 \text{ cm} = 0.10 \text{ m} \)
\( \therefore \) Area of be irrigated \( = \frac{162000}{0.10} \)
\( = 1620000 \text{ m}^2 \)
Given, Width of canal \( = 5.4 \text{ m} \)
Depth of canal \( = 1.8 \text{ m} \)
Speed of flowing water \( = 25 \text{ km/h} \)
\( \therefore \) Length of water in canal in \( 1 \text{ hr} = 25 \text{ km} = 25000 \text{ m} \)
Volume of water flown out from canal in \( 1 \text{ hr} = l \times b \times h \)
\( = 5.4 \times 1.8 \times 25000 = 243000 \text{ m}^3 \)
Volume of water in \( 40 \text{ min} = 243000 \times \frac{40}{60} = 162000 \text{ m}^3 \)
Thus, area irrigated with \( 10 \text{ cm} \) standing water in field
\( = \frac{\text{Volume}}{\text{Height}} = \frac{162000}{0.10} \text{ m}^2 \)
\( = 1620000 \text{ m}^2 = 162 \text{ hectare} \).

Long Answer Type Questions

Question. A hemispherical depression is cut out from one face of a cubical block of side \( 7 \text{ cm} \) such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid.* [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, side of the cube \( = 7 \text{ cm} \)
Diameter of the hemisphere \( = 7 \text{ cm} \)
Thus, radius of the hemisphere \( = 3.5 \text{ cm} \)
Total surface area of the cube \( = 6(7)^2 \text{ cm}^2 \)
Curved surface area of the hemisphere \( = 2\pi(3.5)^2 \text{ cm}^2 \)
Thus, total surface area of the remaining solid \( = \) Total surface area of cube \( + \) Curved surface area of the hemisphere \( - \) Area of the top face of the hemisphere
\( = [6(7)^2 + 2\pi(3.5)^2 - \pi(3.5)^2] \text{ cm}^2 \)
\( = [6(7)^2 + (\frac{22}{7})(3.5)^2] \text{ cm}^2 \)
\( = [294 + \frac{77}{2}] \text{ cm}^2 = 332.5 \text{ cm}^2 \).

Question. Sushant has a vessel in the shape of an inverted cone that is open at the top. Its height is \( 11 \text{ cm} \) and the radius of the top is \( 2.5 \text{ cm} \). It is full of water and metallic spherical balls of diameter \( 0.5 \text{ cm} \) are put in the vessel such that \( \frac{2}{5} \text{th} \) of the water flows out. Find the number of balls that were put in the vessel.*
Answer: Sol. Given, Height of the cone \( = 11 \text{ cm} \)
Radius of the top of the cone \( = 2.5 \text{ cm} \)
Diameter of each ball \( = 0.5 \text{ cm} \)
Thus, radius of each ball \( = 0.25 \text{ cm} \)
Volume of water that flows out \( = \frac{2}{5} \)(Volume of the cone)
Now, volume of the water in the cone \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \pi (2.5)^2 (11) \text{ cm}^3 \)
\( = \frac{11 \pi}{3} (2.5 \times 2.5) \text{ cm}^3 \)
Thus, the volume of water that flows out \( = \frac{11}{3} \pi (2.5 \times 2.5) (\frac{2}{5}) \text{ cm}^3 \)
\( = \) Total volume of all the spherical balls
Now, volume of 1 spherical ball \( = \frac{4}{3} \pi (0.25)^3 \text{ cm}^3 \)
\( = \frac{4}{3} \pi (0.25 \times 0.25 \times 0.25) \text{ cm}^3 \)
Hence, the number of spherical balls \( = \frac{(\frac{1}{3}) 11\pi (2.5 \times 2.5) (\frac{2}{5})}{(\frac{4}{3}) \pi (0.25 \times 0.25 \times 0.25)} \)
\( = \frac{11 (2.5 \times 2.5)}{10 (0.25 \times 0.25 \times 0.25)} = 440 \)
Thus, number of balls \( = 440 \).

Question. Two spheres of same metal weigh \( 1 \text{ kg} \) and \( 7 \text{ kg} \). The radius of the smaller sphere is \( 3 \text{ cm} \). The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.*
Answer: Sol. Given, radius of small sphere be \( r = 3 \text{ cm} \)
Then, Volume of small sphere \( = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3)^3 = 36 \pi \text{ cm}^3 \)
Density of small sphere \( = \frac{\text{Mass of sphere}}{\text{Volume of sphere}} \)
\( = \frac{1}{36 \pi} \text{ kg/cm}^3 \)
∵ Both spheres are made by same metal, then their densities will be same.
Let radius of bigger sphere \( = r' \) then,
Density of bigger sphere \( = \frac{\text{Mass of bigger sphere}}{\text{Volume of bigger sphere}} \)
\( \frac{1}{36 \pi} = \frac{7}{\frac{4}{3} \pi (r')^3} \)
\( (r')^3 = 189 \)
Then according to question, we have
Volume of bigger sphere \( + \) Volume of smaller sphere \( = \) Volume of new sphere.
\( \frac{4}{3} \pi (r')^3 + \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \)
\( \Rightarrow (r')^3 + r^3 = R^3 \)
\( \Rightarrow 189 + 27 = R^3 \)
\( \Rightarrow 216 = R^3 \)
\( \Rightarrow R = 6 \)
\( \Rightarrow D = 6 \times 2 = 12 \)
Radius of new sphere is \( 6 \text{ cm} \).
So, diameter is \( 12 \text{ cm} \).

Question. Three cubes of a metal whose edges are in the ratio of \( 3 : 4 : 5 \) are melted and converted into a single cube whose diagonal is \( 12\sqrt{3} \text{ cm} \). Find the edges of the three cubes.*
Answer: Sol. Given, diagonal of the larger cube \( = 12\sqrt{3} \text{ cm} \)
Ratio of the sides of the three smaller cube \( = 3 : 4 : 5 \)
Let the common factor between the edges of the smaller cubes be \( x \text{ cm} \).
Sum of ratios of the edges of the smaller cubes \( = 3 + 4 + 5 = 12 \)
Edge of first cube \( = \frac{3x}{12} \)
Edge of second cube \( = \frac{4x}{12} \)
and Edge of third cube \( = \frac{5x}{12} \)
Thus, volume of the first cube \( = (\frac{3x}{12})^3 = \frac{27x^3}{1728} \text{ cm}^3 \)
Volume of the second cube \( = (\frac{4x}{12})^3 = \frac{64x^3}{1728} \text{ cm}^3 \)
Volume of the third cube \( = (\frac{5x}{12})^3 = \frac{125x^3}{1728} \text{ cm}^3 \)
Now, diagonal of the larger cube \( = 12\sqrt{3} = a\sqrt{3} \)
Thus, side of the larger cube \( (a) = 12 \)
Volume of the larger cube \( = (12)^3 = 1728 \text{ cm}^3 \)
Thus, \( \frac{27x^3}{1728} + \frac{64x^3}{1728} + \frac{125x^3}{1728} = 1728 \)
\( \Rightarrow \frac{216x^3}{1728} = 1728 \)
\( \Rightarrow \frac{x^3}{8} = 1728 \)
\( \Rightarrow (\frac{x}{2})^3 = (12)^3 \)
\( \Rightarrow \frac{x}{2} = 12 \)
\( \Rightarrow x = 24 \text{ cm} \)
Thus, Edge of first cube \( = \frac{3 \times 24}{12} = 6 \text{ cm} \)
Edge of second cube \( = \frac{4 \times 24}{12} = 8 \text{ cm} \)
and Edge of third cube \( = \frac{5 \times 24}{12} = 10 \text{ cm} \).

Question. Water is flowing through a cylindrical pipe of internal diameter \( 2 \text{ cm} \) into a cylindrical tank of base radius \( 40 \text{ cm} \) at the rate of \( 0.4 \text{ m/s} \). Determine the rise in the water level in the tank in half an hour.*
Answer: Sol. Given, Internal diameter of cylindrical pipe \( = 2 \text{ cm} \)
Thus, radius of the cylindrical pipe \( = 1 \text{ cm} \)
Rate of flow of water through the pipe \( = 0.4 \text{ m/s} = 40 \text{ cm/s} \)
Base radius of the tank \( = 40 \text{ cm} \)
Time \( = 30 \text{ min} = (30 \times 60) \text{ sec} = 1800 \text{ sec} \)
Let the rise in water level \( = h \text{ m} \)
Now, volume of water that flows through the pipe in \( 1 \text{ sec} = \pi(1)^2 (40) \text{ cm}^3 = 40\pi \text{ cm}^3 \)
Thus, volume of water that flows through the pipe in \( 1800 \text{ sec} = (1800)40\pi \text{ cm}^3 \)
Thus, volume of water that collects in the tank in \( 1800 \text{ sec} = (1800)40\pi \text{ cm}^3 \)
\( \therefore \pi(40)^2h = (1800)40\pi \)
\( \Rightarrow (40)^2h = (1800)40 \)
\( \Rightarrow 40h = 1800 \)
\( \Rightarrow h = 45 \text{ cm} \)
Thus, rise in the water level in the tank in half an hour is \( 45 \text{ cm} \).

Question. A hemispherical tank full of water is emptied by a pipe at the rate of \( \frac{25}{7} \text{ l/s} \). How much time will it take to empty half the tank if the diameter of the base of the tank is \( 3 \text{ m} \) ?
Answer: Sol. Given, diameter of the tank \( = 3 \text{ m} \)
\( \therefore \) Radius \( = 1.5 \text{ m} \)
Rate of flow of water \( = \frac{25}{7} \text{ l/s} \)
Thus, volume \( = \frac{2}{3} \pi(1.5)^3 \text{ m}^3 = 2.25\pi \text{ m}^3 = 2250\pi \text{ l} \) [\( \because 1 \text{ m}^3 = 1000 \text{ l} \)]
Thus, half of the volume of the tank \( = 1125\pi \text{ l} \)
Thus, the time taken to empty half of the tank \( = \frac{1125\pi}{\frac{25}{7}} \text{ sec} \)
\( = \frac{1125 \times \frac{22}{7}}{\frac{25}{7}} \text{ sec} \)
\( = \frac{1125 \times 22}{25} \text{ sec} \)
\( = 990 \text{ sec} = 16 \text{ minutes } 30 \text{ seconds} \).

Question. A military tent of height \( 8.25 \text{ m} \) is in the form of a right circular cylinder of base diameter \( 30 \text{ m} \) and height \( 5.5 \text{ m} \) surmounted by a right circular cone of the same base radius. Find the length of the canvas used in making the tent, if the breadth of the canvas is \( 1.5 \text{ m} \).*
Answer: Sol. Given, total height of the tent \( = 8.25 \text{ m} \)
Height of the cylindrical structure \( = 5.5 \text{ m} \)
Base diameter of cylindrical structure \( = \) Base diameter of conical structure \( = 30 \text{ m} \)
Breadth of the canvas \( = 1.5 \text{ m} \)
Thus, height of the conical structure \( = (8.25 - 5.50) \text{ m} = 2.75 \text{ m} \)
Radius of cylindrical structure \( = \) Radius of conical structure \( = 15 \text{ m} \)
Thus, surface area of the tent \( = \) Curved surface area of the (cylindrical structure \( + \) conical structure)
\( = 2\pi rh + \pi rl \)
\( = 2\pi(15)(5.5) + \pi 15 \sqrt{(15)^2 + (2.75)^2} \)
\( = 15\pi [11 + 15.25] = 15\pi \times 26.25 = 1237.5 \text{ m}^2 \)
Thus, the length of canvas required \( = \frac{1237.5}{1.5} \text{ m} = 825 \text{ m} \).

Question. A solid iron pole consists of a cylinder of height \( 220 \text{ cm} \) and base diameter \( 24 \text{ cm} \), which is surmounted by another cylinder of height \( 60 \text{ cm} \) and radius \( 8 \text{ cm} \). Find the mass of the pole, given that \( 1 \text{ cm}^3 \) of iron has approximately \( 8 \text{ gm} \) mass.* (Use \( \pi = 3.14 \))
Answer: Sol. Let AB be the iron pole of height \( 220 \text{ cm} \) with base radius \( 12 \text{ cm} \) and there is an other cylinder CD of height \( 60 \text{ cm} \) whose base radius is \( 8 \text{ cm} \).
Volume of AB pole \( = \pi r_{1}^2 h_1 = 3.14 \times 12 \times 12 \times 220 \)
\( = 99475.2 \text{ cm}^3 \)
Volume of CD pole \( = \pi r_{2}^2 h_2 = 3.14 \times 8 \times 8 \times 60 \)
\( = 12057.6 \text{ cm}^3 \)
Total volume of the poles \( = 99475.2 + 12057.6 \)
\( = 111532.8 \text{ cm}^3 \)
It is given that,
Mass of \( 1 \text{ cm}^3 \) of iron \( = 8 \text{ gm} \)
Then mass of \( 111532.8 \text{ cm}^3 \) of iron \( = 111532.8 \times 8 \text{ gm} \)
Then, total mass of the pole is
\( = 111532.8 \times 8 \text{ gm} \)
\( = 892262.4 \text{ gm} \)
\( = 892.2624 \text{ kg} \)

Question. A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being \( 3.5 \text{ cm} \) and the total height of the solid is \( 9.5 \text{ cm} \). Find the volume of the solid.* [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, radius of hemisphere
\( = \) radius of cone \( = 3.5 \text{ cm} \)
Total height of the solid \( = 9.5 \text{ cm} \)
Height of the hemisphere
\( = \) Radius of the hemisphere
\( = 3.5 \text{ cm} \)
\( \therefore \) Height of the cone \( = (9.5 - 3.5) \text{ cm} = 6 \text{ cm} \)
Hence, volume of the solid
\( = \) Volume of the hemisphere \( + \) Volume of the cone
\( = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \pi r^2 (2r + h) \)
\( = \frac{1}{3} \pi (3.5)^2 [2(3.5) + (6)] \)
\( = \frac{1}{3} \pi (3.5)^2 [7 + 6] \)
\( = \frac{1}{3} (\frac{22}{7}) (3.5)^2 (13) \)
\( = \frac{77 \times 13}{6} \text{ cm}^3 \)
\( = 166.83 \text{ cm}^3 \).

Question. Water is flowing at the rate of \( 15 \text{ km/hr} \) through a pipe of diameter \( 14 \text{ cm} \) into a cuboidal pond which is \( 50 \text{ m} \) long and \( 44 \text{ m} \) wide. In what time will the level of water in the pond rise by \( 21 \text{ cm} \) ?**
Answer: Sol. Given, diameter of pipe \( = 14 \text{ cm} = 0.14 \text{ m} \)
\( \therefore \) Radius of the pipe \( = 0.07 \text{ m} \)
Length of pond \( = 50 \text{ m} \)
Breadth of pond \( = 44 \text{ m} \)
Height of water in the pond \( = 21 \text{ cm} = 0.21 \text{ m} \)
Rate of flow of water \( = 15 \text{ km/hr} \)
\( = \frac{15 \times 5}{18} \text{ m/sec} \)
\( = \frac{25}{6} \text{ m/sec} \)
Thus in \( 1 \) second, the volume of water that flows through the pipe
\( = \pi(0.07)^2 \frac{25}{6} \text{ m}^3 \)
\( = \frac{22}{7} (0.07)^2 \frac{25}{6} \text{ m}^3 \)
\( = \frac{77}{1200} \text{ m}^3 \)
Thus, volume of water flows into the pond in \( 1 \text{ sec.} \)
\( = \frac{77}{1200} \text{ m}^3 \)
\( \therefore \) Rise in water-level in the pond in \( 1 \text{ sec.} \)
\( = \frac{\frac{77}{1200}}{50 \times 44} \text{ m} \)
\( = \frac{7}{1200 \times 200} \text{ m} \)
Thus, the time taken to raise the water level by \( \frac{21}{100} \text{ m} \)
\( = \frac{\frac{21}{100}}{\frac{7}{1200 \times 200}} \text{ sec} \)
\( = \frac{21}{100} \times \frac{1200 \times 200}{7} \text{ sec} \)
\( = \frac{3}{1} \times \frac{1200 \times 2}{1} \text{ sec} \)
\( = 7200 \text{ sec} \)
\( = \frac{7200}{60} \text{ min} \)
\( = 120 \text{ min} = 2 \text{ hours} \).

Question. From a solid right circular cylinder of height \( 2.4 \text{ cm} \) and radius \( 0.7 \text{ cm} \), a right circular cone of same height and same radius is cut out. Find the total surface area of the remaining solid.*
Answer: Sol. Given, height of cylinder, \( h = 2.4 \text{ cm} \)
Radius of base, \( r = 0.7 \text{ cm} \)
and Slant height, \( l = \sqrt{h^2 + r^2} \)
\( = \sqrt{(2.4)^2 + (0.7)^2} \)
\( = \sqrt{5.76 + 0.49} \)
\( = \sqrt{6.25} \)
\( = 2.5 \text{ cm} \)
Total surface area of remaining solid
\( = \text{ CSA of cylinder } + \text{ CSA of cone } + \text{ Area of top } \)
\( = 2\pi rh + \pi rl + \pi r^2 \)
\( = \pi r [2h + l + r] \)
\( = \frac{22}{7} \times 0.7 [2 \times 2.4 + 2.5 + 0.7] \)
\( = 2.2 [4.8 + 2.5 + 0.7] \)
\( = 2.2 \times 8 \)
\( = 17.6 \text{ cm}^2 \)

Question. From a solid cylinder of height \( 15 \text{ cm} \) and diameter \( 16 \text{ cm} \), a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. ☆ [Take \( \pi = 3.14 \)]
Answer: Sol. Given, height of solid cylinder and cone \( (h) = 15 \text{ cm} \)
Diameter of cylinder and cone \( = 16 \text{ cm} \)
\( \therefore \) Radius of cylinder and cone \( (r) = 8 \text{ cm} \)
Thus, curved surface area of cylinder
\( = 2\pi rh \)
\( = 2 \times \pi \times 8 \times 15 \text{ cm}^2 \)
\( = 240 \pi \text{ cm}^2 \)
and curved surface area of cone
\( = \pi r \sqrt{r^2 + h^2} \)
\( = \pi (8) \sqrt{(8)^2 + (15)^2} \text{ cm}^2 \)
\( = \pi (8) \sqrt{64 + 225} \text{ cm}^2 \)
\( = \pi (8) \sqrt{289} \text{ cm}^2 \)
\( = \pi (8) (17) \text{ cm}^2 \)
\( = 136\pi \text{ cm}^2 \)
and the area of the top of cylinder
\( = \pi(8)^2 \text{ cm}^2 \)
\( = 64\pi \text{ cm}^2 \)
Hence, the total surface area of the remaining solid
\( = 240\pi + 136\pi + 64\pi \)
\( = 440\pi \text{ cm}^2 \)
\( = 440(3.14) \text{ cm}^2 \)
\( = 1381.6 \text{ cm}^2 \).

Question. Water is flowing at the rate of \( 6 \text{ km/hr} \) through a pipe of diameter \( 14 \text{ cm} \) into a rectangular tank which is \( 60 \text{ m} \) long and \( 22 \text{ m} \) wide. In what time will the level of water in the pond rise by \( 7 \text{ cm} \) ?*
Answer: Sol. Given, diameter of pipe \( = 14 \text{ cm} = 0.14 \text{ m} \)
\( \therefore \) Radius of the pipe \( = 0.07 \text{ m} \)
Length of tank \( = 60 \text{ m} \)
Breadth of tank \( = 22 \text{ m} \)
Height of water in the tank \( = 7 \text{ cm} = 0.07 \text{ m} \)
Rate of flow of water \( = 6 \text{ km/hr} \)
\( = \frac{6 \times 1000}{60} \text{ m/min} \)
\( = 100 \text{ m/min} \)
Volume of water flows through pipe in \( 1 \) minute
\( (V_1) = \frac{22}{7} \times (0.07)^2 \times 100 \)
\( = 1.54 \)
Volume of water in the tank with \( 7 \text{ cm} \) height
\( (V_2) = 60 \times 22 \times \frac{7}{100} \)
\( = 13.2 \times 7 \)
Time taken \( = \frac{V_2}{V_1} \)
\( = \frac{13.2 \times 7}{1.54} \)
\( = 60 \text{ min} \)
\( = 1 \text{ hr} \)

Question. A hollow sphere of internal and external diameters \( 4 \text{ cm} \) and \( 8 \text{ cm} \) respectively is melted to form a cone of base diameter \( 8 \text{ cm} \). Find the height and slant height of the cone.*
Answer: Sol. Given, external diameter of sphere \( = 8 \text{ cm} \)
Internal diameter of sphere \( = 4 \text{ cm} \)
Diameter of cone \( = 8 \text{ cm} \)
Thus, external radius of sphere \( = 4 \text{ cm} \)
Internal radius of sphere \( = 2 \text{ cm} \)
Base radius of cone \( = 4 \text{ cm} \)
Let the height of the cone be \( h \text{ cm} \).
Now, volume of hollow sphere \( = \) Volume of cone
\( \Rightarrow \frac{4}{3} \pi [(4)^3 - (2)^3] = \frac{1}{3} \pi (4)^2 h \)
\( \Rightarrow 64 - 8 = 4h \)
\( \Rightarrow h = 14 \text{ cm} \)
Let the slant height of the cone be \( l \text{ cm} \).
Thus, \( l = \sqrt{(4)^2 + (14)^2} \text{ cm} \)
\( = \sqrt{16 + 196} \text{ cm} \)
\( = \sqrt{212} \text{ cm} \)
\( = 2\sqrt{53} \text{ cm} \)
Thus, the height of the cone is \( 14 \text{ cm} \) and the slant height is \( 2\sqrt{53} \text{ cm} \).

Question. A toy is in the shape of a cone mounted on a hemisphere, the radius of each of them being \( 3.5 \text{ cm} \) and the total height of the solid is \( 15.5 \text{ cm} \). Find the volume and total surface area of the toy.** [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, radius of hemisphere \( = \) Radius of cone \( = 3.5 \text{ cm} \)
Total height of the solid \( = 15.5 \text{ cm} \)
... Height of the hemisphere
\( = \) Radius of the hemisphere \( = 3.5 \text{ cm} \)
\( \therefore \) Height of the cone \( = (15.5 - 3.5) \text{ cm} = 12 \text{ cm} \)
Hence, volume of the solid
\( = \) Volume of the hemisphere \( + \) Volume of the cone
\( = \frac{2}{3} \pi (3.5)^3 + \frac{1}{3} \pi (3.5)^2 (12) \)
\( = \frac{1}{3} \pi (3.5)^2 [2(3.5) + (12)] \)
\( = \frac{1}{3} \pi (3.5)^2 [7 + 12] \)
\( = \frac{1}{3} \pi (3.5)^2 [19] \)
\( = \frac{1}{3} (\frac{22}{7}) (3.5)^2 (19) \)
\( = \frac{77 \times 19}{6} = 243.83 \text{ cm}^3 \)
Total surface area of the toy
\( = \text{ Curved surface area of the (hemisphere + cone) } \)
\( = 2\pi r^2 + \pi rl \)
\( = 2\pi r^2 + \pi r \sqrt{r^2 + h^2} \)
\( = [2\pi (3.5)^2 + \pi(3.5) \sqrt{(3.5)^2 + (12)^2}] \)
\( = \pi(3.5) [2(3.5) + \sqrt{12.25 + 144}] \)
\( = \frac{22}{7}(3.5) [7 + \sqrt{156.25}] \)
\( = 11 [7 + 12.5] = 11 [19.5] \)
\( = 214.5 \text{ cm}^2 \).

Question. A cylindrical tub of radius \( 5 \text{ cm} \) and length \( 9.8 \text{ cm} \) is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is \( 3.5 \text{ cm} \) and the height of the cone is \( 5 \text{ cm} \), find the volume of water left in the tub.* [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, Radius of cylindrical tub \( = 5 \text{ cm} \)
Height of cylindrical tub \( = 9.8 \text{ cm} \)
Radius of hemisphere \( = 3.5 \text{ cm} \)
Height of cone \( = 5 \text{ cm} \)
Thus, volume of water in the tub
\( = \text{ Volume of the cylinder } \)
\( = \pi (5)^2 (9.8) \)
\( = \frac{22}{7} (25) (9.8) \)
\( = 22 \times 35 \)
\( = 770 \text{ cm}^3 \)
Now, volume of water displaced
\( = \text{ Volume of the solid } \)
\( = \text{ Volume of (hemisphere + cone) } \)
\( = \frac{2}{3} \pi(3.5)^3 + \frac{1}{3} \pi(3.5)^2 5 \)
\( = \frac{1}{3} \pi(3.5)^2 [2(3.5) + 5] \)
\( = \frac{22}{21} (3.5)^2 [7 + 5] \)
\( = \frac{22}{21} (3.5)^2 [12] \)
\( = 154 \text{ cm}^3 \)
Thus, the volume of the water left in the tub
\( = \text{ Volume of the water in the tub } - \text{ Volume of water displaced } \)
\( = (770 - 154) \text{ cm}^3 \)
\( = 616 \text{ cm}^3 \).