CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set 14

Question. A solid cylinder of diameter \( 12 \text{ cm} \) and height \( 15 \text{ cm} \) is melted and recast into \( 12 \) toys in the shape of a right-circular cone mounted on a hemisphere. Find the radius of the hemisphere if the total height of the cone is \( 3 \) times the radius.**
Answer: Sol. Given, Diameter of the cylinder \( = 12 \text{ cm} \)
\( \therefore \) Radius \( = 6 \text{ cm} \)
Height of the cylinder \( = 15 \text{ cm} \)
Number of toys \( = 12 \)
Let the radius of the hemisphere be \( r \text{ cm} \)
Thus, height of the cone \( = 3r \text{ cm} \)
Now, volume of cylinder \( = \pi(6)^2 15 \text{ cm}^3 \)
\( = (36) (15)\pi \text{ cm}^3 \)
\( = 540\pi \text{ cm}^3 \)
Now, total volume of \( 12 \) toys
\( = 12[\text{Volume of hemisphere } + \text{ Volume of cone}] \)
Thus, total volume of \( 12 \) toys
\( = 12 [\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2(3r)] \text{ cm}^3 \)
\( = 4 [3\pi r^3 + 2\pi r^3] \text{ cm}^3 \)
\( = 20\pi r^3 \text{ cm}^3 \)
Now, \( 20\pi r^3 = 540\pi \)
\( \Rightarrow r^3 = 27 \)
\( \Rightarrow r^3 = (3)^3 \)
\( \Rightarrow r = 3 \text{ cm} \)
Thus, radius of the hemisphere \( = 3 \text{ cm} \)
and height of the cone \( = (3)(3) \text{ cm} = 9 \text{ cm} \)
\( \therefore \) Total height of the toy
\( = (9 + 3) \text{ cm} = 12 \text{ cm} \).

Question. A well of diameter \( 4 \text{ m} \) is dug \( 14 \text{ m} \) deep. The earth taken out is spread evenly all around the well to form a \( 40 \text{ cm} \) high embankment. Find the width of the embankment.*
Answer: Sol. We have, diameter of well \( = 4 \text{ m} \)
\( \therefore \) Radius, \( r = 2 \text{ m} \)
and Height, \( h = 14 \text{ m} \).
Volume of earth taken out after digging the well
\( = \pi r^2h \)
\( = \frac{22}{7} \times 2 \times 2 \times 14 \)
\( = 176 \text{ m}^3 \)
Let \( x \) be the width of the embankment formed by the earth taken out.
Then, volume of embankment
\( \Rightarrow \frac{22}{7} [(2 + x)^2 - (2)^2] \times \frac{40}{100} = 176 \)
\( \Rightarrow \frac{22}{7} [4 + x^2 + 4x - 4] \times \frac{2}{5} = 176 \)
\( \Rightarrow x^2 + 4x = \frac{176 \times 5 \times 7}{22 \times 2} \)
\( \Rightarrow x^2 + 4x - 140 = 0 \)
\( \Rightarrow x^2 + 14x - 10x - 140 = 0 \)
\( \Rightarrow x(x + 14) - 10(x + 14) = 0 \)
\( \Rightarrow (x + 14) (x - 10) = 0 \)
\( \Rightarrow x = -14 \text{ or } 10 \)
\( x = -14 \text{ (neglect)} \)
\( \therefore x = 10 \)
Hence, width of embankment \( = 10 \text{ m} \).

Question. Water is flowing at the rate of \( 2.52 \text{ km/h} \) through a cylindrical pipe into a cylindrical tank, the radius of whose base in \( 40 \text{ cm} \). If the increase in the level of water in the tank, in half an hour is \( 3.15 \text{ m} \), find the internal diameter of the pipe.*
Answer: Sol. Let the internal radius of the pipe be \( x \text{ m} \).
Radius of base of tank \( = 40 \text{ cm} = \frac{2}{5} \text{ m} \)
Speed of water flowing through the pipe
\( = 2.52 \text{ km/hr} \)
\( = \frac{2.52}{2} \times 1000 \text{ m in half an hour} \)
\( = 1260 \text{ m in half an hour} \)
Volume of water flown through pipe in half an hour
\( = \pi r^2h \)
\( = \frac{22}{7} \times x \times x \times 1260 \)
\( = 3960 x^2 \)
Level of water raised in the tank \( = 3.15 \text{ m} = \frac{315}{100} \text{ m} \)
Volume of water in the tank with \( \frac{315}{100} \text{ m} \) height
\( = \pi \times \frac{2}{5} \times \frac{2}{5} \times \frac{315}{100} \)
Now,
\( \pi \times \frac{2}{5} \times \frac{2}{5} \times \frac{315}{100} = 3960 x^2 \)
\( \Rightarrow x^2 = \frac{22 \times 2 \times 2 \times 315}{7 \times 5 \times 5 \times 100 \times 3960} \)
\( \Rightarrow x^2 = \frac{4}{10000} \)
\( x = \frac{2}{100} \)
\( = 0.02 \text{ m} \)
Internal diameter of the pipe \( = 0.04 \text{ m} = 4 \text{ cm} \)

Question. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is \( 104 \text{ cm} \) and the radius of each of its hemispherical ends is \( 7 \text{ cm} \), find the cost of polishing its surface at the rate of ` \( 10 \) per \( \text{dm}^2 \). * [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, length of solid (Hemisphere \( + \) Cylinder \( + \) Hemisphere) \( = 104 \text{ cm} \)
Radius of the hemisphere \( = \text{ Radius of cylinder } = 7 \text{ cm} \)
Thus, height of the cylinder \( = [104 - 2(7)] \text{ cm} \)
\( = 90 \text{ cm} \)
Rate of polishing \( = \text{` } 10 \text{ per } \text{dm}^2 \)
Now, total surface area to be polished
\( = 2[\text{Curved surface area of a hemisphere}] + \text{ Curved surface area of the cylinder} \)
\( = [2\{2\pi(7)^2\} + 2\pi(7) (90)] \text{ cm}^2 \)
\( = 2\pi(7) [2(7) + 90] \text{ cm}^2 \)
\( = 2(\frac{22}{7}) (7) [14 + 90] \text{ cm}^2 \)
\( = 44 \times 104 \text{ cm}^2 \)
\( = 4576 \text{ cm}^2 \)
Now, \( 10 \text{ cm} = 1 \text{ dm} \)
Thus, \( 100 \text{ cm}^2 = 1 \text{ dm}^2 \)
So, \( 4576 \text{ cm}^2 = 45.76 \text{ dm}^2 \)
Thus, total cost of polishing the structure at ` \( 10 \text{ per } \text{dm}^2 = (10 \times 45.76) = \text{` } 457.60 \).

Question. A tent is in the form of a right circular cylinder of base radius \( 14 \text{ m} \) and height \( 3 \text{ m} \) is surmounted by a right circular cone of the same base radius. The total height of the tent is \( 13.5 \text{ m} \). Find the cost of the canvas used in making the tent at ` \( 80 \) per square metre and the cost of painting it at ` \( 2 \) per square metre.** [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, height of the cylindrical structure \( = 3 \text{ m} \)
Total height of the tent \( = 13.5 \text{ m} \)
Base radius of cylindrical structure \( = \) Base radius of conical structure \( = 14 \text{ m} \)
Rate of canvas \( = \text{` } 80 \text{ per } \text{m}^2 \)
Rate of painting \( = \text{` } 2 \text{ per } \text{m}^2 \)
Now, height of the conical structure \( = (13.5 - 3) \text{ m} = 10.5 \text{ m} \)
\( \therefore \) Total surface area of the tent
\( = \text{ Curved surface area of the (cylindrical structure + conical structure)} \)
\( = 2\pi rh + \pi rl \)
\( = \pi r (2h + l) \)
\( = \pi(14) [2(3) + \sqrt{(14)^2 + (10.5)^2}] \text{ m}^2 \)
\( = \pi(14) [6 + \sqrt{196 + 110.25}] \text{ m}^2 \)
\( = (14)\pi [6 + \sqrt{306.25}] \text{ m}^2 \)
\( = (14) \frac{22}{7} [6 + \sqrt{(17.5)^2}] \text{ m}^2 \)
\( = 44(6 + 17.5) \text{ m}^2 \)
\( = 44 \times 23.5 \text{ m}^2 = 1034 \text{ m}^2 \)
Thus, the cost of canvas \( = \text{` } (80 \times 1034) = \text{` } 82720 \)
and, the cost of painting \( = \text{` } (2 \times 1034) = \text{` } 2068 \).

Question. A container shaped like a cylinder having diameter \( 12 \text{ cm} \) and height \( 15 \text{ cm} \) is full of ice-cream. This ice-cream is to be filled into cones of height \( 12 \text{ cm} \) and diameter \( 6 \text{ cm} \) having a hemispherical shape on the top. Find the number of such cones that can be filled with ice-cream.** [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given,
Diameter of cylinder \( = 12 \text{ cm} \)
\( \therefore \) Radius of cylinder, \( r = 6 \text{ cm} \)
Height of cylinder, \( h = 15 \text{ cm} \)
Height of cone, \( H = 12 \text{ cm} \)
Diameter of cone \( = 6 \text{ cm} \)
\( \therefore \) Radius of cone, \( R = 3 \text{ cm} \)
Diameter of hemisphere \( = 6 \text{ cm} \)
\( \therefore \) Radius of hemisphere, \( r' = 3 \text{ cm} \)
Hence, volume of cylinder \( = \pi r^2h = \pi(6)^2 15 \text{ cm}^3 \)
\( = \pi(36) 15 \text{ cm}^3 = 540\pi \text{ cm}^3 \)
Volume of cone \( = \frac{1}{3} \pi R^2H = \frac{1}{3} \pi(3)^2 12 \text{ cm}^3 = 36\pi \text{ cm}^3 \)
Volume of hemisphere \( = \frac{2}{3} \pi(r')^3 = \frac{2}{3} \pi(3)^3 \text{ cm}^3 = 18\pi \text{ cm}^3 \)
\( \therefore \) Volume of icecream in each cone \( = (36\pi + 18\pi) \text{ cm}^3 = 54\pi \text{ cm}^3 \)
Thus, number of icecream cones \( = \frac{\text{Total volume of icecream in the cylinder}}{\text{Total volume of icecream in each cone}} = \frac{540\pi}{54\pi} = 10 \)
Hence, the number of icecream cones \( = 10 \).

Question. A gulabjamun when ready for eating contains sugar syrup of about \( 30\% \) of its volume. Find approximately how much syrup would be found in \( 45 \) such gulabjamuns if each of them is shaped like a cylinder with two hemispherical ends. The complete length of each of them is \( 5 \text{ cm} \) and the diameter is \( 2.8 \text{ cm} \). * [NCERT]
Answer: Sol. Given, Length of each gulabjamun \( = 5 \text{ cm} \)
Diameter of each gulabjamun \( = 2.8 \text{ cm} \)
Thus, Radius of each gulabjamun, \( r = 1.4 \text{ cm} \)
Total number of gulabjamuns \( = 45 \)
Percentage of syrup in each gulabjamun \( = 30\% \)
and length of the cylindrical part
\( = [5 - (1.4 + 1.4)] \text{ cm} = (5 - 2.8) \text{ cm} = 2.2 \text{ cm} \)
\( \therefore \) Volume of \( 45 \) gulabjamun \( = 45[\text{Volume of (cylindrical portion + 2 hemispherical ends)}] \)
\( = 45 [\pi(1.4)^2 2.2 + 2\{\frac{2}{3}\pi(1.4)^3\}] \text{ cm}^3 \)
\( = 45\pi(1.4)^2 [2.2 + \frac{4}{3}(1.4)] \text{ cm}^3 \)
\( = (45) \frac{22}{7} (1.4) (1.4) [2.2 + \frac{5.6}{3}] \text{ cm}^3 \)
\( = (45) \frac{44}{100} (14) [\frac{6.6 + 5.6}{3}] \text{ cm}^3 \)
\( = (45) \frac{44}{100} (14) [\frac{12.2}{3}] \text{ cm}^3 \)
\( = 1127.28 \text{ cm}^3 \)
Hence, volume of syrup in \( 45 \) gulabjamuns
\( = 30\% \text{ of } 1127.28 \text{ cm}^3 \)
\( = \frac{30}{100} \times 1127.28 \text{ cm}^3 = 338.18 \text{ cm}^3 \).

Question. A solid is in the form of a right-circular cone mounted on a hemisphere. The radius of the hemisphere is \( 2.1 \text{ cm} \) and the height of the cone is \( 4 \text{ cm} \). The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the cylinder is \( 5 \text{ cm} \) and its height is \( 9.8 \text{ cm} \), find the volume of water left in the tub. ** [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, radius of the cylinder \( = 5 \text{ cm} \)
Height of the cylinder \( = 9.8 \text{ cm} \)
Radius of the hemisphere and cone \( = 2.1 \text{ cm} \)
Height of cone \( = 4 \text{ cm} \)
Thus, volume of water in the tub
\( = \text{ Volume of the cylinder } \)
\( = \pi(5)^2 (9.8) \text{ cm}^3 \)
\( = \frac{22}{7} (25) (9.8) \text{ cm}^3 \)
\( = 22(35) \text{ cm}^3 = 770 \text{ cm}^3 \)
Now, volume of water displaced
\( = \text{ Volume of the solid } = \text{ Volume of (hemisphere + cone)} \)
\( = [\frac{2}{3} \pi(2.1)^3 + \frac{1}{3} \pi(2.1)^2 4] \text{ cm}^3 \)
\( = \frac{1}{3} \pi(2.1)^2 [2(2.1) + 4] \text{ cm}^3 \)
\( = \frac{22}{21} (2.1)^2 [4.2 + 4] \text{ cm}^3 \)
\( = (2.1) (2.2) (8.2) \text{ cm}^3 = 37.88 \text{ cm}^3 \)
Thus, the volume of the water left in the tub
\( = \text{ Volume of water in the tub } - \text{ Volume of water displaced } \)
\( = (770 - 37.88) \text{ cm}^3 = 732.12 \text{ cm}^3 \).

Question. A solid is in the form of a cylinder with hemi-spherical ends. The total height of the solid is \( 20 \text{ cm} \) and the diameter of the cylinder is \( 7 \text{ cm} \). Find the total volume of the solid. * [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. ABCD is a cylinder and BFC and AED are two hemisphere which has radius \( (r) = \frac{7}{2} \text{ cm} \).
Here, \( \text{AB} = 20 - 2 \times \frac{7}{2} \)
\( h = 13 \text{ cm}, r = \frac{7}{2} \text{ ( } r = \frac{d}{2} \text{ )} \)
Volume of cylinder \( = \pi r^2h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 13 = \frac{11 \times 13 \times 7}{2} = \frac{1001}{2} = 500.5 \text{ cm}^3 \)
Volume of two hemisphere \( = 2 \times \frac{2}{3} \pi r^3 = 2 \times \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} = \frac{49 \times 11}{3} = \frac{539}{3} = 179.67 \text{ cm}^3 \)
Total volume of solid \( = \text{ Volume of two hemisphere } + \text{ Volume of cylinder } \)
\( = 179.67 + 500.5 = 680.17 \text{ cm}^3 \).

Question. A farmer connects a pipe of internal diameter \( 25 \text{ cm} \) from a canal into a cylindrical tank which is \( 12 \text{ m} \) in diameter and \( 2.5 \text{ m} \) deep. If the water flows through the pipe at the rate of \( 3.6 \text{ km/hr} \), in how much time will the tank be completely filled ? Also find the cost of water if the irrigation department charges at the rate of ` \( 0.07 \text{ per } \text{m}^3 \).* [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, internal diameter of the pipe \( = 25 \text{ cm} = \frac{1}{4} \text{ m} \)
Diameter of the tank \( = 12 \text{ m} \)
Height of tank \( = 2.5 \text{ m} \)
Rate of flow of water \( = 3.6 \text{ km/hr} = \frac{3.6 \times 5}{18} \text{ m/s} = 1 \text{ m/s} \)
Thus, internal radius of pipe \( = 12.5 \text{ cm} = \frac{1}{8} \text{ m} \)
and Radius of tank \( = 6 \text{ m} \)
Now, volume of water flowing through the pipe in \( 1 \) second \( = \pi (\frac{1}{8})^2 1 \text{ m}^3 = \frac{\pi}{64} \text{ m}^3 \)
Volume of the tank \( = \pi(6)^2 (2.5) \text{ m}^3 = 90\pi \text{ m}^3 \)
Thus, time taken to fill the tank \( = \frac{\text{Volume of tank}}{\text{Volume of water flowing in 1 sec}} = \frac{90\pi}{\frac{\pi}{64}} = 90(64) \text{ sec} = 5860 \text{ sec} \)
\( = 96 \text{ min} = 1 \text{ hour and } 36 \text{ minutes} \).
Now, cost of water \( = \text{ Volume of water } \times \text{ Rate} = \text{` } [90\pi \times 0.07] = \text{` } [90 \times \frac{22}{7} \times \frac{7}{100}] = \text{` } 19.80 \).

Question. A juice-seller serves his customers using a glass whose inner diameter is \( 5 \text{ cm} \) but the bottom of the glass has a raised hemispherical portion that reduces its capacity. If the height of the glass is \( 10 \text{ cm} \), find the apparent and actual capacities of the glass. [Use \( \pi = 3.14 \)]*
Answer: Sol. Given, height of glass \( = 10 \text{ cm} \)
Diameter of glass \( = \) Diameter of the hemisphere \( = 5 \text{ cm} \)
Thus, radius of glass \( = \) Radius of the hemisphere \( = 2.5 \text{ cm} \)
Now, apparent capacity of the glass
\( = \pi(2.5)^2 \times 10 \text{ cm}^3 = 196.25 \text{ cm}^3 \)
And, actual capacity of the glass
\( = \) Apparent capacity of the glass \( - \) Volume of the hemisphere
\( = [196.25 - \frac{2}{3}\pi(2.5)^3] \text{ cm}^3 \)
\( = [196.25 - 32.71] \text{ cm}^3 \)
\( = 163.54 \text{ cm}^3 \)

Question. A hemispherical depression is cut out from one face of a cubical block of side \( 21 \text{ cm} \) such that the diameter of the hemisphere is equal to the edge of the cube. Find the volume and the total surface area of the remaining solid. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Given, side of the cube, \( a = 21 \text{ cm} \)
Diameter of the hemisphere \( = 21 \text{ cm} \)
Thus, radius of the hemisphere, \( r = 10.5 \text{ cm} \)
Volume of the cube \( = a^3 = (21)^3 \text{ cm}^3 = 9261 \text{ cm}^3 \)
Volume of the hemisphere
\( = \frac{2}{3}\pi r^3 = \frac{2}{3} \pi(10.5)^3 \text{ cm}^3 = 2425.5 \text{ cm}^3 \)
Thus, volume of the remaining block
\( = (9261 - 2425.5) \text{ cm}^3 = 6835.5 \text{ cm}^3 \)
Total surface area of the cube \( = 6(21)^2 \text{ cm}^2 \)
Curved surface area of the hemisphere \( = 2\pi r^2 = 2\pi(10.5)^2 \text{ cm}^2 \)
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
\( = [6(21)^2 + 2\pi(10.5)^2 - \pi(10.5)^2] \text{ cm}^2 \)
\( = [6(21)^2 + (\frac{22}{7})(10.5)^2] \text{ cm}^2 \)
\( = [2646 + 346.5] \text{ cm}^2 = 2992.5 \text{ cm}^2 \).

Question. Six tennis balls of diameter \( 62 \text{ mm} \) are placed in cylindrical tube (Fig.). Find the volume of the six balls and the internal volume of unfilled space in the tube and express this as a percentage of the volume of the tube.
Answer: Sol. Diameter of the tennis balls \( = 62 \text{ mm} \).
... Radius of the balls and tube is half the diameter
\( \therefore r = \frac{1}{2} \times 62 = 31 \text{ mm} \)
Volume of one ball \( = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 31 \times 31 \times 31 = 124,838.476 \text{ mm}^3 \)
\( = 124.838 \text{ cm}^3 \)
\( \therefore \) Volume of 6 tennis balls \( = 6 \times 124.838 = 749.028 \text{ cm}^3 \)
Height of tube \( (h) \) is 6 times the diameter of the ball
\( \therefore h = 6 \times 62 = 372 \text{ mm} \)
\( \therefore \) Volume of the tube \( = \pi r^2 h = \frac{22}{7} \times 31 \times 31 \times 372 \)
\( = 1123,546.29 \text{ mm}^3 = 1123.54 \text{ cm}^3 \)
\( \therefore \) Volume of unfilled space (shaded area) in the tube
\( = 1123.54 - 749.028 = 374.512 \text{ cm}^3 \)
Space as a percentage of the volume of the tube
\( = \frac{374.512}{1123.54} \times 100 = 33.33\% \)
Hence, volume of unfilled space is \( 33.3\% \) of tube.

Question. A sector of a circle of radius \( 6 \text{ cm} \) has an angle of \( 120^\circ \). It is rolled up so that the two bounding radii are joined together to form a cone. Find volume of cone and T.S.A. of the cone.
Answer: Sol. On rolling up the sector, we get a cone whose slant height \( = \) radius of sector \( = 6 \text{ cm} \).
Circumference of the base of cone \( = \) Length of sector
\( \Rightarrow 2\pi R = \frac{\theta}{360^\circ} \times 2\pi r = \frac{120^\circ}{360^\circ} \times 2 \times \pi \times 6 \)
\( \Rightarrow 2\pi R = 4\pi \text{ cm.} \)
\( \Rightarrow R = 2 \text{ cm} \)
T.S.A. of cone \( = \) C.S.A. of cone \( + \) Area of base of cone
\( = \text{Area of sector} + \pi(R)^2 \)
\( = \frac{\theta}{360^\circ} \pi l^2 + \pi \times (2)^2 \)
\( = \frac{120^\circ}{360^\circ} \times \pi \times (6)^2 + 4\pi \)
\( = 12\pi + 4\pi = 16\pi \text{ cm}^2 \)
Height of cone, \( h = \sqrt{l^2 - R^2} = \sqrt{6^2 - 2^2} \)
\( = \sqrt{36 - 4} = \sqrt{32} \)
\( = 4\sqrt{2} \text{ cm.} \)
Volume of cone \( = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi(2)^2 \times (4\sqrt{2}) \)
\( = \frac{16\pi\sqrt{2}}{3} \text{ cm}^3 \)

Question. A vessel is in the form of an inverted cone. Its height is \( 8 \text{ cm} \) and the radius of its top, which is open, is \( 5 \text{ cm} \). It is filled with water up to the brim. When lead shots, each of which is a sphere of radius \( 0.5 \text{ cm} \) are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Answer: Sol. We have, height of conical vessel, \( h = 8 \text{ cm} \) and its radius, \( r = 5 \text{ cm} \)
Now, Volume of cone \( = \) Volume of water in the cone
\( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 \)
\( = \frac{4,400}{21} \text{ cm}^3 \)
Now, volume of water flows out \( = \) Volume of lead shots
\( = \frac{1}{4} \times \text{Volume of water in the cone} \)
\( = \frac{1}{4} \times \frac{4,400}{21} = \frac{1,100}{21} \text{ cm}^3 \)
Now, radius of the lead shots \( R = 0.5 \text{ cm} = \frac{5}{10} \text{ cm} = \frac{1}{2} \text{ cm} \)
Volume of one spherical lead shot \( = \frac{4}{3} \pi R^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{11}{21} \text{ cm}^3 \)
\( \therefore \) Number of lead shots dropped in the vessel \( = \frac{\text{Volume of water flows out}}{\text{Volume of one lead shot}} \)
\( = \frac{1,100 / 21}{11 / 21} = \frac{1,100}{21} \times \frac{21}{11} = 100 \).

Question. A right cylindrical container of radius \( 6 \text{ cm} \) and height \( 15 \text{ cm} \) is full of ice-cream, which has to be distributed to \( 10 \) children in equal cones having hemispherical shape on the top. If the height of the conical portion is four times its base radius, find the radius of the ice-cream cone.☆
Answer: Sol. Let \( R \) and \( H \) be the radius and height of cylinder.
Given, \( R = 6 \text{ cm}, H = 15 \text{ cm} \).
Volume of ice-cream in the cylinder \( = \pi R^2 H = \pi \times 36 \times 15 = 540 \pi \text{ cm}^3 \)
Let the radius of cone be \( r \text{ cm} \)
Height of the cone \( (h) = 4r \)
Radius of hemispherical portion \( = r \text{ cm.} \)
\( \therefore \) Volume of ice-cream in cone \( = \text{Volume of cone} + \text{Volume of hemisphere} \)
\( = \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 = \frac{1}{3} \pi r^2 (h + 2r) \)
\( = \frac{1}{3} \pi r^2 (4r + 2r) \) (\… \( h = 4r \))
\( = \frac{1}{3} \times \pi r^2 \times 6r = 2\pi r^3 \)
Number of ice cream cones distributed to the children \( = 10 \)
\( 10 \times \text{Volume of ice-cream in each cone} = \text{Volume of ice-cream in cylindrical container} \)
\( \Rightarrow 10 \times 2\pi r^3 = 540\pi \)
\( \Rightarrow 20r^3 = 540 \)
\( \Rightarrow r^3 = 27 \Rightarrow r = 3 \)
Thus, the radius of the ice-cream cone is \( 3 \text{ cm} \).

Assertion and Reasoning Based Questions

Question. Assertion : A hemisphere of radius \( 7 \text{ cm} \) is to be painted outside on the surface. The total cost of painting at it ` \( 5 \) per \( \text{cm}^2 \) is ` \( 2300 \).
Reason : The total surface area hemisphere is \( 3\pi r^2 \).
(A) Both the Assertion and the Reason are correct and Reason is the correct explanation of the Assertion.
(B) Both the Assertion and the Reason are correct but Reason is not the correct explanation of the Assertion.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.
Answer: (C) Assertion is true but Reason is false.
Explanation :
Total surface area of the hemisphere is \( \pi r^2 + 2\pi r^2 = 3 \times \frac{22}{7} \times 7 \times 7 = 462 \text{ cm}^2 \)
Cost of painting \( = 462 \times 5 = \text{` } 2310 \)
So, assertion is false and reason is true.

Question. Assertion : The number of coins \( 1.75 \text{ cm} \) in diameter and \( 2 \text{ mm} \) thick from a melted cuboid (\( 10 \text{ cm} \times 5.5 \text{ cm} \times 3.5 \text{ cm} \)) is \( 400 \).
Reason : Volume of a cylinder of base radius \( r \) and height \( h \) is given by \( V = (\pi r^2 h) \) cubic units. And, area of a cuboid \( = (l \times b \times h) \) square units
(A) Both the Assertion and the Reason are correct and Reason is the correct explanation of the Assertion.
(B) Both the Assertion and the Reason are correct but Reason is not the correct explanation of the Assertion.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.
Answer: (A) Both the Assertion and the Reason are correct and Reason is the correct explanation of the Assertion.
Explanation :
Volume of cuboid \( l \times b \times h = 10 \times 5.5 \times 3.5 \text{ cm}^3 \)
Number of coins \( = \frac{\text{Volume of cuboid}}{\text{Volume of each coin}} = \frac{10 \times 5.5 \times 3.5}{\frac{22}{7} \times \frac{1.75}{2} \times \frac{1.75}{2} \times \frac{1}{5}} = 400 \)

Case Based Questions

Mathematics teacher of a school took her 10th standard students to show Red fort. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Red fort to students. Then the teacher said in this monument one can find combination of solid figures. There are 2 pillars which are cylindrical in shape. Also 2 domes at the corners which are hemispherical 7 smaller domes at the centre. Flag hoisting ceremony on Independence Day takes place near these domes.

Question. How much cloth material will be required to cover 2 big domes each of radius \( 2.5 \text{ metres} \)? (Take \( \pi = 22 / 7 \))
(a) \( 75 \text{ m}^2 \)
(b) \( 78.57 \text{ m}^2 \)
(c) \( 87.47 \text{ m}^2 \)
(d) \( 25.8 \text{ m}^2 \)
Answer: (b) \( 78.57 \text{ m}^2 \)
Explanation :
Radius of a dome, \( r = 2.5 \text{ m} \)
The dome is hemispherical in shape.
Then, cloth material required \( = 2 \times \text{Surface area of hemisphere} = 2 \times 2\pi r^2 = 4 \times \frac{22}{7} \times 2.5 \times 2.5 = 78.57 \text{ m}^2 \)

Question. The formula to find the volume of a cylindrical pillar :
(a) \( \pi r^2 h \)
(b) \( \pi r l \)
(c) \( \pi r (l + r) \)
(d) \( 2\pi r \)
Answer: (a) \( \pi r^2 h \)
Explanation :
The formula to find the volume of a cylindrical pillar is \( \pi r^2 h \)

Question. The lateral surface area of two pillars if height of the pillar is \( 7 \text{ m} \) and radius of the base is \( 1.4 \text{ m} \) is :
(a) \( 112.3 \text{ m}^2 \)
(b) \( 123.2 \text{ m}^2 \)
(c) \( 90 \text{ m}^2 \)
(d) \( 345.2 \text{ m}^2 \)
Answer: (b) \( 123.2 \text{ m}^2 \)
Explanation :
Height of each pillar, \( h = 7 \text{ m} \)
Radius of base, \( r = 1.4 \text{ m} \)
Lateral surface area or curved surface area of 2 pillars \( = 2 \times 2\pi r h = 4 \times \frac{22}{7} \times 1.4 \times 7 = 123.2 \text{ m}^2 \)

Question. The volume of a hemisphere if the radius of the base is \( 3.5 \text{ m} \) is :
(a) \( 85.9 \text{ m}^3 \)
(b) \( 80 \text{ m}^3 \)
(c) \( 98 \text{ m}^3 \)
(d) \( 89.83 \text{ m}^3 \)
Answer: (d) \( 89.83 \text{ m}^3 \)
Explanation :
Radius of hemisphere, \( r = 3.5 \text{ m} \)
Then, volume of a hemisphere, \( V = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (3.5)^3 = 89.83 \text{ m}^3 \)

Question. The ratio of sum of volumes of two hemispheres of radius \( 1 \text{ cm} \) each to the volume of a sphere of radius \( 2 \text{ cm} \) ?
(a) \( 1 : 1 \)
(b) \( 1 : 8 \)
(c) \( 8 : 1 \)
(d) \( 1 : 16 \)
Answer: (b) \( 1 : 8 \)
Explanation :
Volume of 2 hemispheres of radius \( 1 \text{ cm} = 2 \times \frac{2}{3} \pi r^3 = \frac{4}{3} \pi(1)^3 \)
Volume of sphere of radius \( 2 \text{ cm} = \frac{4}{3} \pi(2)^3 = \frac{32}{3} \pi \text{ cm}^3 \)
Then, required ratio \( = \frac{(4/3)\pi}{(32/3)\pi} = \frac{1}{8} = 1 : 8 \)

Isha is 10 years old girl. On the result day, Isha and her father Suresh were very happy as she got first position in the class. While coming back to their home, Isha asked for a treat from her father as a reward for her success. They went to a juice shop and asked for two glasses of juice. Aisha, a juice seller, was serving juice to her customers in two types of glasses. Both the glasses had inner radius 3 cm. The height of both the glasses was 10 cm. Give answer the following questions based on above conditions :

Question. The capacity of first glass is :
(a) 72 \( \pi \text{ cm}^2 \)
(b) 72 \( \pi \text{ cm}^2 \)
(c) 85 \( \pi \text{ cm}^2 \)
(d) 85 \( \pi \text{ cm}^2 \)
Answer: (b) 72\( \pi \text{ cm}^2 \)
Explanation :
Capacity of first glass \( = \pi r^2h - \frac{2}{3} \pi r^3 \)
\( = \pi r^2 (h - \frac{2}{3} r) \)
\( = \pi(3)^2 (10 - \frac{2}{3} \times 3) \)
\( = 9\pi(10 - 2) \)
\( = 72\pi \text{ cm}^3 \).

Question. The capacity of second glass is :
(a) 85.5 \( \pi \text{ cm}^3 \)
(b) 85\( \pi \text{ cm}^3 \)
(c) 85 \( \pi \text{ cm}^2 \)
(d) 85\( \pi \text{ cm}^3 \)
Answer: (a) 85.5\( \pi \text{ cm}^3 \).
Explanation :
Capacity of first glass \( = \pi r^2H - \frac{1}{3} \pi r^2h \)
\( = \pi r^2 (H - \frac{1}{3} h) \)
\( = \pi(3)^2 (10 - \frac{1}{3} \times 1.5) \)
\( = 9\pi(10 - 0.5) = 85.5\pi \text{ cm}^3 \).

Question. The ratio of the capacity of both types of glass is :
(a) 16 : 19
(b) 17 : 19
(c) 15 : 19
(d) 18 : 19
Answer: (a) 16 : 19
Explanation :
Ratio \( = \frac{\text{Capacity of first glass}}{\text{Capacity of second glass}} \)
\( = \frac{72}{85.5} = \frac{16}{19} = 16 : 19 \)

Question. Isha insisted to have the juice is first type of glass and her father decided to have the juice in second type of glass. Out of the two, Isha or her father Suresh, who got more quantities of juice
(a) Isha; 72\( \pi \text{ cm}^3 \)
(b) Suresh; 85.5\( \pi \text{ cm}^3 \)
(c) Suresh; 72\( \pi \text{ cm}^3 \)
(d) Suresh; 13.5\( \pi \text{ cm}^3 \)
Answer: (d) Suresh; 13.5\( \pi \text{ cm}^3 \)
Explanation :
From part (i) and (ii)
Suresh got more quantity of juice
\( = 85.5\pi \text{ cm}^3 - 72\pi \text{ cm}^3 \)
\( = 13.5\pi \text{ cm}^3 \).

Question. How much quantity of juice is purchased by Suresh from a juice seller ?
(a) 157.5\( \pi \text{ cm}^3 \)
(b) 1575\( \pi \text{ cm}^3 \)
(c) 157.5\( \pi \text{ cm}^3 \)
(d) none of the options
Answer: (a) 157.5\( \pi \text{ cm}^3 \).
Explanation :
Total quantity of juice purchased by Suresh
\( = (72\pi + 85.5\pi) \text{ cm}^3 \)
\( = 157.5\pi \text{ cm}^3 \).

Mathematics teacher of a school took her 10th standard students to show Gol Gumbaz. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Gul Gumbaz to students. Gol Gumbaz is the tomb of King Muhammad Adil Shah, Adil Shah Dynastry. Construction of the tomb, located in Vijayapura, Karanataka, India, was started in 1626 and completed in 1656. Then the teacher said in this monument one can find combination of solid figures. She pointed that there are cubical bases and hemispherical dome is at the top.

Question. The diagonal of the cubic portion of the Gol Gumbaz, if one side of cubical portion is 23 m is :
(a) 23 m
(b) 23 \( \sqrt{2} \) m
(c) 23 \( \sqrt{3} \) m
(d) 24 m
Answer: (c) 23 \( \sqrt{3} \) m
Explanation :
Diagonal of cubic portion \( = a \sqrt{3} \)
\( = 23 \sqrt{3} \) m
The dome is hemispherical in shape.

Question. If two solid hemispheres of same base radii 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 :
CSA of new solid \( = 2\pi r^2 + 2\pi r^2 = 4\pi r^2 \)

Question. A solid piece of iron taken out from the back of Gol Gumbaz in the form of a cuboid of dimensions 49 cm \( \times \) 33 cm \( \times \) 24 cm, is moulded to form a solid sphere. The radius of the sphere is :
(a) 19 cm
(b) 23 cm
(c) 25 cm
(d) 21 cm
Answer: (d) 21 cm
Explanation :
Volume of sphere \( = \) Volume of cuboid
\( \Rightarrow \frac{4}{3} \pi r^3 = l \times b \times h \)
\( \Rightarrow r^3 = \frac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22} \)
\( \Rightarrow r = 21 \text{ cm} \)

Question. The total surface area of a hemispherical dome having radius 7 cm is : (including base)
(a) 462 \( \text{ cm}^2 \)
(b) 294 \( \text{ cm}^2 \)
(c) 588 \( \text{ cm}^2 \)
(d) 154 \( \text{ cm}^2 \)
Answer: (a) 462 \( \text{ cm}^2 \)
Explanation :
TSA of hemispherical dome \( = 3\pi r^2 \)
\( = 3 \times \frac{22}{7} \times 7 \times 7 = 462 \text{ cm}^2 \)

Rasheed is very happy for his birthday celebration. He got lot of birthday gifts on his party. Out of all birthday gifts, he liked a playing top (lattu) most, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm.

Question. The height of the cone is :
(a) 5 cm
(b) 3 cm
(c) 3.25 cm
(d) 4 cm
Answer: (c) 3.25 cm
Explanation :
Height of the cone
\( = \) Height of the top
\( - \) height (radius) of the hemispherical part
\( = 5 - \frac{3.5}{2} = 5 - 1.75 \)
\( = 3.25 \text{ cm} \)

Question. The slant height of the cone is :
(a) 5 cm
(b) 3.7 cm
(c) 3.25 cm
(d) 4 cm
Answer: (b) 3.7 cm
Explanation :
Slant height of the cone
\( (l) = \sqrt{r^2 + h^2} \)
\( = \sqrt{(\frac{3.5}{2})^2 + (3.25)^2 \text{ cm}} \)
\( = 3.7 \text{ cm} \) (approx)

Question. The curved surface area of the cone is :
(a) 19.5 \( \text{ cm}^2 \)
(b) 20.35 \( \text{ cm}^2 \)
(c) 20.5 \( \text{ cm}^2 \)
(d) 19.25 \( \text{ cm}^2 \)
Answer: (b) 20.35 \( \text{ cm}^2 \)
Explanation :
CSA of cone \( = \pi r l = (\frac{22}{7} \times \frac{3.5}{2} \times 3.7) \text{ cm}^2 \)
\( = 20.35 \text{ cm}^2 \)

Question. The curved surface area of the hemisphere is :
(a) 19.5 \( \text{ cm}^2 \)
(b) 20.35 \( \text{ cm}^2 \)
(c) 20.5 \( \text{ cm}^2 \)
(d) 19.25 \( \text{ cm}^2 \)
Answer: (d) 19.25 \( \text{ cm}^2 \)
Explanation :
Curved surface area of the hemisphere
\( = 2\pi r^2 \)
\( = (2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}) \text{ cm}^2 \)
\( = 19.25 \text{ cm}^2 \)

Question. The total surface area of the toy is :
(a) 39.6 \( \text{ cm}^2 \)
(b) 39.35 \( \text{ cm}^2 \)
(c) 39.5 \( \text{ cm}^2 \)
(d) 39.85 \( \text{ cm}^2 \)
Answer: (a) 39.6 \( \text{ cm}^2 \)
Explanation :
Total Surface area of the toy
\( = \) CSA of hemisphere \( + \) CSA of cone
\( = 19.25 + 20.35 \text{ cm}^2 \)
\( = 39.6 \text{ cm}^2 \)

In the month of December 2020, it rained heavily throughout the day over the city of Hyderabad. Anil observed the raindrops as they reached him. Each raindrop was in the shape of a hemisphere surmounted by a cone of the same radius of 1 mm. Volume of one of such drops is 3.14 \( \text{ mm}^3 \). Anil collected the rain water in a pot having a capacity of 1099 \( \text{ cm}^3 \). [Use \( \sqrt{2} = 1.4 \)]. Based on the above situation, answer the following questions.

Question. The total height of the drop is :
(a) 1 mm
(b) 2 mm
(c) 3 mm
(d) 4 mm
Answer: (b) 2 mm
Explanation :
Radius of hemispherical part
\( = \) Radius of conical part, \( r = 1 \) mm.
Let the height of the conical part be \( h \) mm.
Volume of a rain drop
\( = \) Volume of hemispherical part
\( + \) Volume of conical part
\( \Rightarrow 3.14 = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h \)
\( \Rightarrow 3.14 = \frac{1}{3} \pi r^2 (2r + h) \)
\( \Rightarrow 3.14 = \frac{1}{3} \times 3.14 \times 1 \times 1 (2 \times 1 + h) \)
\( \Rightarrow 2 + h = 3 \)
\( \Rightarrow h = 1 \)
Height of a rain drop
\( = \) Height of hemispherical part
\( + \) Height of conical part
\( = (1 + 1) \) mm \( = 2 \) mm.

Question. The curved surface area of the drop is :
(a) 8.74 \( \text{ mm}^2 \)
(b) 9.12 \( \text{ mm}^2 \)
(c) 10.68 \( \text{ mm}^2 \)
(d) 12.54 \( \text{ mm}^2 \)
Answer: (c) 10.68 \( \text{ mm}^2 \)
Explanation :
Slant height of conical part,
\( l = \sqrt{r^2 + h^2} = \sqrt{1^2 + 1^2} \) mm
\( = \sqrt{2} \) mm \( = 1.4 \) mm.
CSA of a rain drop
\( = \) CSA of hemispherical part \( + \) CSA of conical part \( = 2\pi r^2 + \pi rl = \pi r (2r + l) \)
\( = 3.14 \times 1 \times (2 \times 1 + 1.4) \text{ mm}^2 \)
\( = (3.14 \times 3.4) \text{ mm}^2 \)
\( = 10.68 \text{ mm}^2 \)

Question. As the drop fell into the pot, it changed into a sphere. What was the radius of this sphere ?
(a) (3/4)\( ^{1/3} \)
(b) (4/3)\( ^{1/3} \)
(c) 3\( ^{1/3} \)
(d) 4\( ^{1/3} \)
Answer: (a) (3/4)\( ^{1/3} \)
Explanation :
Let the radius of the sphere be R mm.
Then, volume of sphere \( = \) Volume of a rain drop
\( \Rightarrow \frac{4}{3} \pi R^3 = 3.14 \)
\( \Rightarrow \frac{4}{3} \times 3.14 \times R^3 = 3.14 \)
\( \Rightarrow R^3 = \frac{3}{4} \)
\( \Rightarrow R = (\frac{3}{4})^{1/3} \)

Question. How many drops will fill the pot completely :
(a) 260000
(b) 280000
(c) 320000
(d) 350000
Answer: (d) 350000
Explanation :
Volume of pot \( = 1099 \text{ cm}^3 \)
\( = (1099 \times 10 \times 10 \times 10) \text{ mm}^3 \)
\( = 1099000 \text{ mm}^3 \).
Let n rain drops be needed to fill the pot.
Then, volume of n rain drops \( = \) Volume of pot
\( \Rightarrow n \times 3.14 = 1099000 \)
\( \Rightarrow n = \frac{1099000}{3.14} = 350000 \).

Question. The total surface area of a hemisphere of radius r is :
(a) 2/3 \( \pi r^3 \)
(b) 4/3 \( \pi r^3 \)
(c) 2\( \pi r^2 \)
(d) 3\( \pi r^2 \)
Answer: (d) 3\( \pi r^2 \)
Explanation :
Total surface area of a hemisphere of radius \( r \) is given by 3\( \pi r^2 \)