CBSE Class 10 Maths HOTs Surface Area and Volumes Set E

Very Short Answer Type Questions

Question. A rectangular sheet paper \( 40 \text{ cm} \times 22 \text{ cm} \) is rolled to form a hollow cylinder of height \( 40 \text{ cm} \). Find the radius of the cylinder.
Answer: Here, \( h = 40 \text{ cm} \), circumference \( = 22 \text{ cm} \)
\( 2 \pi r = 22 \)
\( r = \frac{22 \times 7}{2 \times 22} = \frac{7}{2} = 3.5 \text{ cm} \)

Question. A cylinder, a cone and a hemisphere have same base and same height. Find the ratio of their volumes.
Answer: Volume of cylinder : Volume of cone : Volume of hemisphere
\( = \pi r^2 h : \frac{1}{3} \pi r^2 h : \frac{2}{3} \pi r^3 \)
\( = \pi r^2 h : \frac{1}{3} \pi r^2 h : \frac{2}{3} \pi r^2 \times h \quad (h = r) \)
\( = 1 : \frac{1}{3} : \frac{2}{3} \)
or, \( 3 : 1 : 2 \)

Question. What is the ratio of the total surface area of the solid hemisphere to the square of its radius.
Answer: \( \frac{\text{Total surface area of hemisphere}}{\text{Square of its radius}} = \frac{3 \pi r^2}{r^2} = \frac{3 \pi}{1} \)
Total surface area of hemisphere : Square of radius \( = 3 \pi : 1 \)

Question. Two cubes each of volume \( 8 \text{ cm}^3 \) are joined end to end, then what is the surface area of resulting cuboid.
Answer: Side of the cube, \( a = \sqrt[3]{8} = 2 \text{ cm} \)
Now the length of cuboid \( l = 4 \text{ cm} \)
Breadth, \( b = 2 \text{ cm} \)
Height, \( h = 2 \text{ cm} \)
Surface area of cuboid \( = 2(l \times b + b \times h + h \times l) \)
\( = 2(4 \times 2 + 2 \times 2 + 2 \times 4) \)
\( = 2 \times 20 = 40 \text{ cm}^2 \)

Question. The radius of sphere is \( r \text{ cm} \). It is divided into two equal parts. Find the whole surface of two parts.
Answer: Whole surface of each part \( = 2 \pi r^2 + \pi r^2 = 3 \pi r^2 \)
Total surface of two parts \( = 2 \times 3 \pi r^2 = 6 \pi r^2 \)

Question. What is the volume of a right circular cylinder of base radius \( 7 \text{ cm} \) and height \( 10 \text{ cm} \)? Use \( \pi = \frac{22}{7} \).
Answer: Here \( r = 7 \text{ cm}, h = 10 \text{ cm} \),
Volume of cylinder \( = \pi r^2 h \)
\( = \frac{22}{7} \times (7)^2 \times 10 \)
\( = 1540 \text{ cm}^3 \)

Question. If the radius of the base of a right circular cylinder is halved, keeping the height same, find the ratio of the volume of the reduced cylinder to that of original cylinder.
Answer: \( \frac{\text{Volume of reduced cylinder}}{\text{Volume of original cylinder}} = \frac{\pi \times (\frac{r}{2})^2 h}{\pi r^2 h} \)
\( = \frac{1}{4} = 1 : 4 \)

Question. If the area of three adjacent faces of a cuboid are \( X, Y, \) and \( Z \) respectively, then find the volume of cuboid.
Answer: Let the length, breadth and height of the cuboid is \( l, b, \) and \( h \) respectively.
\( X = l \times b \)
\( Y = b \times h \)
\( Z = l \times h \)
\( XYZ = l^2 \times b^2 \times h^2 \)
Volume of cuboid \( = l \times b \times h \)
\( l^2 b^2 h^2 = XYZ \)
or, \( lbh = \sqrt{XYZ} \)

Question. The radii of two cylinders are in the ratio \( 2 : 3 \) and their heights are in the ratio \( 5 : 3 \), find the ratio of their volumes.
Answer: \( \frac{\text{Volume of 1st cylinder}}{\text{Volume of 2nd cylinder}} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} \)
\( = \left(\frac{r_1}{r_2}\right)^2 \times \frac{h_1}{h_2} \)
\( = \left(\frac{2}{3}\right)^2 \times \frac{5}{3} \)
\( = \frac{4}{9} \times \frac{5}{3} = \frac{20}{27} \)
\( = 20 : 27 \)

Question. Volume of two spheres are in the ratio \( 64 : 27 \), find the ratio of their surface areas.
Answer: \( \frac{\text{Volume of I sphere}}{\text{Volume of II nd}} = \frac{64}{27} \)
\( \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{64}{27} \)
\( \frac{r_1^3}{r_2^3} = \frac{4^3}{3^3} \)
\( \frac{r_1}{r_2} = \frac{4}{3} \)
Ratio of their surface areas
\( \frac{4 \pi r_1^2}{4 \pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)

Question. A solid metallic object is shaped like a double cone as shown in figure. Radius of base of both cones is same but their heights are different. If this cone is immersed in water, find the quantity of water it will displace.
Answer: Volume of the upper cone \( = \frac{1}{3} \pi r^2 h \)
Volume of the lower cone \( = \frac{1}{3} \pi r^2 H \)
Total volume of both the cones \( = \frac{1}{3} \pi r^2 h + \frac{1}{3} \pi r^2 H \)
\( = \frac{1}{3} \pi r^2 (h + H) \)
The quantity of water displaced will \( \frac{1}{3} \pi r^2 (h + H) \) cube units.

Question. Find the volume (in \( \text{cm}^3 \)) of the largest right circular cone that can be cut off from a cube of edge \( 4.2 \text{ cm} \).
Answer: Edge of the cube \( = 4.2 \text{ cm} \).
Height of the cone \( = 4.2 \text{ cm} \).
Radius of the cone \( = \frac{4.2}{2} = 2.1 \text{ cm} \).
Volume of the cone \( = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (2.1)^2 \times 4.2 \)
\( = 19.4 \text{ cm}^3 \)

Question. The circumference of the edge of a hemisphere bowl is \( 132 \text{ cm} \). When \( \pi \) is taken as \( \frac{22}{7} \), find the capacity of the bowl in \( \text{cm}^3 \).
Answer: Let \( r \) be the radius of bowl.
Circumference of bowl
\( 2 \pi r = 132 \)
\( r = \frac{132 \times 7}{2 \times 22} = 21 \text{ cm} \)
Capacity i.e volume of the bowl
\( \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 \)
\( = 19404 \text{ cm}^3 \)

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

Question. Find the number of solid sphere of diameter \( 6 \text{ cm} \) can be made by melting a solid metallic cylinder of height \( 45 \text{ cm} \) and diameter \( 4 \text{ cm} \).
Answer: Let the number of sphere = \( n \)
Radius of sphere = \( 3 \text{ cm} \), radius of cylinder = \( 2 \text{ cm} \)
Volume of spheres = Volume of cylinder
\( n \times \frac{4}{3} \pi r^3 = \pi r_1^2 h \)
\( n \times \frac{4}{3} \times \frac{22}{7} \times (3)^3 = \frac{22}{7} \times (2)^2 \times 45 \)
\( 36n = 180 \)
\( n = \frac{180}{36} = 5 \)
Number of solid sphere = 5.

Question. Three solid metallic spherical balls of radii \( 3 \text{ cm} \), \( 4 \text{ cm} \) and \( 5 \text{ cm} \) are melted into a single spherical ball, find its radius.
Answer: Let the radius of spherical ball = \( R \).
Volume of spherical ball = Volume of three balls
\( \frac{4}{3} \pi R^3 = \frac{4}{3} \pi [(3)^3 + (4)^3 + (5)^3] \)
\( R^3 = 27 + 64 + 125 = 216 \)
\( R = 6 \text{ cm} \)

Question. 12 solid spheres of the same size are made by melting a solid metallic cone of base radius 1 cm and height of 48 cm. Find the radius of each sphere.
Answer: No. of spheres = 12
Radius of cone, \( r = 1 \text{ cm} \)
Height of the cone = 48
Volume of 12 spheres = Volume of cone
Let the radius of sphere be \( R \).
\( 12 \times \frac{4}{3} \pi R^3 = \frac{1}{3} \pi r^2 h \)
\( 12 \times \frac{4}{3} \pi R^3 = \frac{1}{3} \pi \times (1)^2 \times 48 \)
\( R^3 = 1 \)
\( R = 1 \text{ cm} \)

Question. Three cubes of iron whose edges are 3 cm, 4 cm and 5 cm respectively are melted and formed into a single cube, what will be the edge of the new cube formed ?
Answer: Let the edge of single cube be \( x \text{ cm} \).
Volume of single cube = Volume of three cubes
\( x^3 = (3)^3 + (4)^3 + (5)^3 \)
\( = 27 + 64 + 125 = 216 \)
\( x = 6 \text{ cm} \)

Question. A solid sphere of radius r melted and recast into the shape of a solid cone of height r . Find the radius of the base of a cone.
Answer: Volume of sphere = Volume of cone
Let the radius of cone be \( R \text{ cm} \).
\( \frac{4}{3} \pi r^3 = \frac{1}{3} \pi R^2 \times r \)
\( 4r^3 = R^2 r \)
\( R^2 = 4r^2 \)
\( R = 2r \)

Question. If a cone is cut into two parts by a horizontal plane passing through the mid-points of its axis, find the ratio of the volume of the upper part and the cone.
Answer: As per question the figure is shown below.
Volume of upper cone = \( \frac{1}{3} \pi \left(\frac{r}{2}\right)^2 \times \frac{h}{2} \)
\( = \frac{1}{3} \pi \frac{r^2}{4} \times \frac{h}{2} \)
\( = \frac{1}{3} \pi \frac{r^2 h}{8} \)
Volume of full cone = \( \frac{1}{3} \pi r^2 h \)
\( \frac{\text{Volume of upper of cone}}{\text{Volume of cone}} = \frac{\frac{1}{3} \pi \times \frac{r^2 h}{8}}{\frac{1}{3} \pi r^2 h} = \frac{1}{8} \)
\( = 1 : 8 \)

Question. What is the frustum of a right circular cone of height 16 cm with radii of its circular ends as 8 cm and 20 cm has slant height equal to ?
Answer: As per question the figure is shown below.
Slant height of the frustum,
\( l = \sqrt{h^2 + d^2} \)
\( = \sqrt{(16)^2 + (12)^2} \)
\( = \sqrt{256 + 144} \)
\( = \sqrt{400} \)
\( = 20 \text{ cm.} \)

SHORT ANSWER TYPE QUESTIONS

Question. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
Answer: Diameter of hemisphere = Side of cubical block
\( 2r = 7 \)
or, \( r = \frac{7}{2} \)
Surface area of solid
= Surface area of the cube
- Area of base of hemisphere
+ curved surface area of hemisphere
\( = 6l^2 - \pi r^2 + 2\pi r^2 \)
\( = 6 \times 49 - 11 \times \frac{7}{2} + 77 = 332.5 \text{ cm}^2 \)

Question. The curved surface area of a cylinder is \( 264 \text{ m}^2 \) and its volume is \( 924 \text{ m}^3 \). Find the ratio of its height to its diameter.
Answer: Curved Surface area of cylinder \( = 2 \pi r h \)
Volume of cylinder \( = \pi r^2 h \)
\( \frac{\pi r^2 h}{2 \pi r h} = \frac{924}{264} \implies \frac{r}{2} = \frac{7}{2} \)
Thus \( r = 7 \text{ m} \) and substituting in \( 2 \pi r h = 264 \) we have
\( 2 \times \frac{22}{7} \times 7 \times h = 264 \)
\( h = 6 \text{ m} \)
Now \( \frac{h}{2r} = \frac{6}{14} = \frac{3}{7} \)
Hence, \( h : d = 3 : 7 \)

Question. If the total surface area of a solid hemisphere is 462 cm2, find its volume. Use \( \pi = \frac{22}{7} \)
Answer: Total surface area of hemisphere,
\( 3 \pi r^2 = 462 \text{ cm}^2 \)
\( 3 \times \frac{22}{7} \times r^2 = 462 \)
\( r^2 = \frac{462 \times 7}{22 \times 3} = 49 \)
\( r = 7 \text{ cm.} \)
Volume of hemisphere,
\( \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \)
\( = \frac{2156}{3} = 718.67 \text{ cm}^3 \)

Question. A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs.25 per meter.
Answer: Given, radius \( r = 7 \text{ m} \) and height \( h = 24 \text{ m} \)
Slant height of tent,
\( l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} \)
\( = \sqrt{625} = 25 \text{ m.} \)
Curves surface area
\( \pi rl = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^2 \)
Curves surface area of tent will be required area of cloth. Let \( x \) meter of cloth is required
\( 5x = 550 \) or, \( x = \frac{550}{5} = 110 \text{ m.} \)
Thus 110 m of cloth is required.
Cost of cloth \( = 25 \times 110 = Rs. 2750 \).

Question. Find the number of plates, 1.5 cm in diameter and 0.2 cm thick, that can be fitted completely inside a right circular of height 10 cm and diameter 4.5 cm.
Answer: Each one of the circular plate is also a cylinder.
Volume of plate \( V_p = \pi r^2 h = \pi \times (.75)^2 \times (.2) \)
\( = \frac{9\pi}{80} \text{ cm}^3 \)
Volume of right circular cylinder
\( V_c = \pi (2.25)^2 (10) = 405 \frac{\pi}{8} \text{ cm}^3 \)
Number of plates \( = \frac{\frac{405\pi}{8}}{\frac{9\pi}{80}} = \frac{405\pi}{9\pi} \times \frac{80}{8} \)
\( = 450 \text{ plates.} \)

Question. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the volume of the remaining solid to the nearest cm3. Use \( \pi = \frac{22}{7} \)
Answer: As per question the figure is shown below.
Volume of remaining solid
= Volume of cylinder - Volume of cone
\( = \pi r^2 h - \frac{1}{3} \pi r^2 h = \frac{2}{3} \pi r^2 h \)
\( = \frac{2}{3} \times \frac{22}{7} \times 0.7 \times 0.7 \times 2.4 \)
\( = 44 \times 0.1 \times 0.7 \times 0.8 \)
\( = 4.4 \times .56 = 2.464 \text{ cm}^3 \)

Question. A solid metallic cylinder of radius 3.5 cm and height 14 cm melted and recast into a number of small solid metallic ball, each of radius \( \frac{7}{12} \) cm. Find the number of balls so formed.
Answer: Let the number of recasted balls be \( N \)
Radius of cylinder \( R = 3.5 \text{ cm} \)
Height of cylinder \( h = 14 \text{cm} \)
Radius of recasted ball \( r = \frac{7}{12} \)
Volume of balls = Volume of cylinder
\( n \frac{4}{3} \pi r^3 = \pi R^2 h \)
\( n \times \frac{4}{3} \times \frac{7}{12} \times \frac{7}{12} \times \frac{7}{12} = 3.5 \times 3.5 \times 14 \)
\( n = \frac{3.5 \times 3.5 \times 14 \times 3 \times 12 \times 12 \times 12}{4 \times 7 \times 7 \times 7} \)
\( = 0.5 \times 0.5 \times 2 \times 3 \times 3 \times 12 \times 12 \)
\( = 648 \)
Hence, number of recasted balls = 648

Question. A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Answer: Radius of sphere \( = \frac{6}{2} = 3 \text{ cm} \)
Radius of cylinder vessel \( = \frac{12}{2} = 6 \text{ cm} \)
Let the level of water rise in cylinder be \( h \).
Volume of sphere \( = \frac{4}{3} \pi r^3 \)
\( = \frac{4}{3} \times \pi \times 3 \times 3 \times 3 \)
\( = 36\pi \text{ cm}^3 \)
Volume of sphere = Increase volume in cylinder
\( 36\pi = \pi(6)^2 \times h \)
\( h = 1 \text{ cm} \)
Thus level of water rise in vessel is 1 cm.

Question. A tent is in the shape of cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs.500 per square meter. Use \( \pi = \frac{22}{7} \)
Answer: Height of cylinder \( = 2.1 \text{ m} \)
Radius of cylinder = radius of cone \( = \frac{3}{2} \text{ m} \)
Slant height of cone \( = 2.8 \text{ m} \)
Surface area of tent \( = C.S.A. \text{ of cone} + C.S.A. \text{ of cylinder} \)
\( = \pi rl + 2\pi rh = \pi r(l + 2h) \)
Area of canvas required will be surface area of tent.
Thus \( \pi r(l + 2h) = \frac{22}{7} \times \frac{3}{2} (2.8 + 2 \times 2.1) \)
\( = \frac{33}{7} \times 7 = 33 \text{ m}^2 \)
Total Cost \( = 33 \times 500 \text{ Rs} = 16,500 \text{ Rs} \)

Question. A hemispherical bowl of internal diameter 36 cm contains liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer.
Answer: Volume of bowl \( = \frac{2}{3} \pi r^3 \)
Volume of liquid in bowl \( = \frac{2}{3} \pi \times (18)^3 \text{ cm}^3 \)
Volume of one after wastage \( = \frac{2}{3} \pi (18)^3 \times \frac{90}{100} \text{ cm}^3 \)
Volume of one bottle \( = \pi r^2 h \)
Volume of liquid in 72 bottles \( = \pi \times (3)^2 \times h \times 72 \text{ cm}^2 \)
Volume of bottles = volume in liquid after wastage
\( \pi \times (3)^2 \times h \times 72 = \frac{2}{3} \pi \times (18)^3 \times \frac{90}{100} \)
\( h = \frac{\frac{2}{3} \pi \times (18)^2 \times \frac{90}{100}}{\pi \times (3)^2 \times 72} \)
Hence, the height of bottle = 5.4 cm

Question. A solid right-circular cone of height 60 cm and radius 30 cm is dropped in a right-circular cylinder full of water of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder in cubic metre. Use \(\pi = \frac{22}{7}\)
Answer: Volume of water in cylinder = Volume of cylinder
\(\pi r^2 h = \pi \times (60)^2 \times 180\)
\(= 648000\pi \, \text{cm}^3\)
Water displaced on dropping cone is equal to the volume of solid cone, which is
\(\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times (30)^2 \times 60\)
\(= 18000\pi \, \text{cm}^3\)
Volume of water left in cylinder
= Volume of cylinder - Volume of cone
\(= 648000\pi - 18000\pi = 630000\pi \, \text{cm}^3\)
\(= \frac{630000 \times 22}{1000000 \times 7} \, \text{m}^3 = 1.98 \, \text{m}^3\)

Question. The rain water from \(22 \, \text{m} \times 20 \, \text{m}\) roof drains into cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collect from the roof the root fills \(\frac{4}{5}\)th of cylindrical vessel then find the rainfall in cm.
Answer: Volume of water collected in cylindrical vessel
\(= \frac{4}{5} \times \pi \times (1)^2 \times \left(\frac{7}{2}\right) \, \text{m}^3 = \frac{44}{5} \, \text{m}^3\)
Let the rainfall be \(h \, \text{m}\).
Rain water from roof \(= 22 \times 20 \times h \, \text{m}^3\)
\(22 \times 20 \times h = \frac{44}{5}\)
\(h = \frac{44}{5} \times \frac{1}{22 \times 20} = \frac{1}{50} \, \text{m}^3\)
\(= \frac{1}{50} \times 100 = 2 \, \text{cm}\)

Question. A hollow cylindrical pipe is made up of copper. It is 21 dm long. The outer and inner diameters of the pipe are 10 cm and 6 cm respectively. Find the volume of copper used in making the pipe.
Answer: Height of cylindrical pipe \(h = 21 \, \text{dm} = 210 \, \text{cm}\)
External Radius \(R = \frac{10}{2} = 5 \, \text{cm}\)
Internal Radius \(r = \frac{6}{2} = 3 \, \text{cm}\)
Volume of copper used in making the pipe
= (Volume of External Cylinder) - (Volume of Internal Cylinder)
\(= \pi R^2 h - \pi r^2 h\)
\(= \pi h (R^2 - r^2)\)
\(= \frac{22}{7} \times 210 (5^2 - 3^2) = \frac{22}{7} \times 210 \times 8 \times 2\)
\(= 10560 \, \text{cm}^3\).

Question. A glass is in the shape of a cylinder of radius 7 cm and height 10 cm. Find the volume of juice in litre required to fill 6 such glasses. Use \(\pi = \frac{22}{7}\)
Answer: Radius of the glass = 7 cm
Height of the glass = 10 cm
Volume of 1 glass \(= \pi r^2 h\)
\(= \frac{22}{7} \times 7 \times 7 \times 10\)
\(= 1540 \, \text{cm}^3\)
Volume of juice to fill 6 glasses
\(= 6 \times 1540 = 9240 \, \text{cm}^3\)
Volume in litre \(= \frac{9240}{1000} = 9.240 \, \text{litre}\).

Question. The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. Use \(\pi = \frac{22}{7}\)
Answer: Side of cube \(a = 7 \, \text{cm}\)
The diameter of the largest possible sphere is the side of the cube.
Thus radius of sphere \(= \frac{7}{2} \, \text{cm}\).
Volume of the wood left
= volume of cube - volume of sphere
\(= a^3 - \frac{4}{3} \pi r^3\)
\(= 7 \times 7 \times 7 - \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\)
\(= 7 \times 7 \times 7 (1 - \frac{11}{21})\)
\(= 7 \times 7 \times 7 \times \frac{10}{21} = \frac{490}{3}\)
Hence, Volume of wood \(= 163.3 \, \text{cm}^3\).

Question. In the given figure a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq. metre. Use \(\pi = \frac{22}{7}\)
Answer: Radius of cylinder as well as conical part \(= \frac{3}{2} \, \text{m}\)
Height of cylinder \(h = 2.1 \, \text{m}\)
Slant height of cone \(l = 2.8 \, \text{m}\)
Total canvas required \(= 2\pi rh + \pi rl = \frac{22}{7} \times \frac{3}{2} [2 \times 2.1 + 2.8] \, \text{m}^2\)
\(= \frac{22}{7} \times \frac{3}{2} \times 7.0 = 33 \, \text{m}^2\)
Total cost \(= 33 \times 500 = 16,500\)

Question. A girl empties a cylindrical bucket, full of sand, of radius 18 cm and height 32 cm, on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct upto one place of decimal.
Answer: Volume of cone = Volume of Cylinder
\(\frac{1}{3}\pi r_2^2 h = \pi r_1^2 h_1\)
\(\frac{1}{3} \times r_2^2 \times 24 = 18 \times 18 \times 32\)
\(r_2^2 = 1296\)
Radius of cone \(r_2 = 36 \, \text{cm}\)
Now, slant height of cone
\(l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 36^2}\)
\(= \sqrt{576 + 1296} = \sqrt{1872} = 43.2 \, \text{cm}\).

Question. A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy. Use \(\pi = \frac{22}{7}\)
Answer: As per question the figure is shown below.
Radius of toy = radius of cylinder = 3.5 cm
Vol. of toy = Vol. of cylinder - \(2 \times\) Vol. of hemisphere
\(= \pi r^2 h - 2 \times \frac{2}{3} \pi r^3\)
\(= \pi r^2 [h - \frac{4}{3}r]\)
\(= \frac{22}{7} \times (3.5)^2 [10 - \frac{4}{3} \times 3.5]\)
\(= 22 \times 0.5 \times 3.5 \times 5.33\)
\(= 205.205 \, \text{cm}^3\).

Question. A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area of the vessel. Use \(\pi = \frac{22}{7}\)
Answer: As per question the figure is shown below.
Radius of hemisphere \(= \frac{14}{2} = 7 \, \text{cm}\)
Height of cylinder \(= 13 - 7 = 6 \, \text{cm}\)
Total slanted area of vessel
= S.A of hemisphere + S.A. of cylinder
\(= 2\pi r^2 + 2\pi rh\)
\(= 2\pi r(r + h)\)
\(= 2 \times \frac{22}{7} \times 7(7 + 6)\)
\(= 44 \times 13 = 572 \, \text{cm}^2\)

Question. A toy is in the form of a cone surmounted on a hemisphere of common base of diameter 7 cm. If the height of the toy is 15.5 cm, find the total surface area of the toy. Use \(\pi = \frac{22}{7}\)
Answer: As per question the figure is shown below.
Radius \(r = 3.5 \, \text{cm}\)
and height \(h = 15.5 - 3.5 = 12 \, \text{cm}\)
Slant height of cone,
\(l = \sqrt{h^2 + r^2} = \sqrt{12^2 + 3.5^2} = 12.5\)
Total surface area of the toy
= Surface area of hemisphere + Curved surface area of cone
\(= 2\pi r^2 + \pi rl\)
\(= \pi r(2r + l)\)
\(= \frac{22}{7} \times \frac{7}{2} (2 \times \frac{7}{2} + 12.5)\)
\(= 11 \times 19.5 = 214.5 \, \text{cm}^2\)

Question. The slant height of a frustum of a cone is 4 m and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Answer: Let the radii of frustum be \( r_1 \) and \( r_2 \).
\( 2\pi r_1 = 18 \text{ cm} \) and \( 2\pi r_2 = 6 \text{ cm} \)
\( \pi r_1 = \frac{18}{2} = 9 \text{ cm} \)
\( \pi r_2 = \frac{6}{2} = 3 \text{ cm} \)
and slant height \( l = 4 \text{ cm} \)
Curved surface area of frustum \( = \pi(r_1 + r_2) \times l \)
\( = (\pi r_1 + \pi r_2) \times l \)
\( = (9 + 3) \times 4 \)
\( = 12 \times 4 = 48 \text{ cm}^2 \)
Hence, curved surface area \( = 48 \text{ cm}^2 \)

Question. A metallic solid sphere of radius 10.5 cm melted and recasted into smaller solid cones each of radius 3.5 cm and height 3 cm. How many cones will be made?
Answer: Radius of given sphere \( = 10.5 \text{ cm} \)
Volume of sphere \( = \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \pi \times 10.5 \times 10.5 \times 10.5 \)
\( = 4\pi \times 3.5 \times 10.5 \times 10.5 \text{ cm}^3 \)
Radius of one recasted cone \( = 3.5 \text{ cm} \)
Height \( = 3 \text{ cm} \)
Volume \( = \frac{1}{3} \pi \times 3.5 \times 3.5 \times 3 \)
\( = \pi \times 3.5 \times 3.5 \text{ cm}^3 \)
Let the number of recasted cones be \( n \).
\( n \times \pi \times 3.5 \times 3.5 = 4\pi \times 3.5 \times 10.5 \times 10.5 \)
\( n = \frac{4 \times 3.5 \times 10.5 \times 10.5}{3.5 \times 3.5} = 126 \)
Hence, number of recasted cones \( = 126 \).

Question. A solid metallic sphere of diameter 16 cm is melted and recasted into smaller solid cones, each of radius 4 cm and height 8 cm. Find the number of cones so formed.
Answer: Diameter of sphere \( = 16 \text{ cm} \)
radius \( = \frac{16}{2} = 8 \text{ cm} \)
Volume \( = \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \pi \times 8 \times 8 \times 8 \text{ cm}^3 \)
Radius and height of recasted cones \( = 4 \text{ cm} \) and \( 8 \text{ cm} \) respectively.
Volume of each cone \( = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 4 \times 4 \times 8 \text{ cm}^3 \)
Let number of cones recasted be \( n \)
\( n = \frac{\text{Volume of Sphere}}{\text{Volume of One Cone}} \)
\( = \frac{\frac{4}{3} \times \pi \times 8 \times 8 \times 8}{\frac{1}{3} \times \pi \times 4 \times 4 \times 8} = 16 \)
Hence number of recasted cones \( = 16 \text{ cm} \)

Question. A solid sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged into water, by how much will the level of water rise in the cylindrical vessel?
Answer: Let the rise in level of water be \( h \text{ cm} \).
Radius of sphere \( = 3 \text{ cm} \). radius of cylinder \( = \frac{12}{2} = 6 \text{ cm} \)
Volume of water displaced in cylinder will be equal to the volume of sphere.
Thus \( \pi r^2 h = \frac{4}{3}\pi r^3 \)
\( \pi \times 6 \times 6 \times h = \frac{4}{3} \times \pi \times 3 \times 3 \times 3 \)
\( h = \frac{4 \times 3 \times 3 \times 3}{3 \times 6 \times 6} = 1 \text{ cm} \)
Hence the water level rises \( = 1 \text{ cm} \).

Question. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Answer: As per question the figure is shown below.
Radius of hemisphere \( r = 3.5 \text{ cm} \)
Height of cone \( h = 15.5 - 3.5 = 12 \text{ cm} \)
Slant height of cone \( = \sqrt{r^2 + h^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm} \)
TSA of toy = CSA of cone + CSA of hemisphere
\( \pi r l + 2\pi r^2 = \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times 3.5 \times 3.5 \)
\( = 22 \times 0.5 \times 12.5 + 22 \times 0.5 \times 2 \times 3.5 \)
\( = 11 \times 12.5 + 11 \times 7 \)
\( = 11(12.5 + 7) = 11(19.5) = 214.5 \text{ cm}^2 \)
Thus total surface area of toy is \( 214.5 \text{ cm}^2 \)

Question. A conical vessel, with base radius 5 cm height 24 cm, is full of water. This water emptied into a cylindrical vessel, of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. Use \( \pi = \frac{22}{7} \)
Answer: Here radius and height of conical vessel are 5 cm and 24 cm.
Volume of cone \( = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 5 \times 5 \times 24 \text{ cm}^3 \)
When water is emptied into cylindrical vessel, water will rise in cylindrical vessel. Let rise in height be \( h \).
Volume of water raised \( = \pi r^2 h \)
\( \frac{1}{3} \times \pi \times 25 \times 24 = \pi \times (10)^2 \times h \)
\( 25 \times 8 = 100h \)
\( h = 2 \text{ cm} \)

Question. Water is flowing at the rate of 0.7 m/sec through a circular pipe whose internal diameter is 2 cm into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the level of water in half hour.
Answer: Length of water that flows in 1 sec. \( = 0.7 \text{ m} \)
Length of water that flows out in 30 minutes \( = 0.7 \times 100 \times 60 \times 30 \text{ cm} = 126000 \text{ cm} \)
Volume of water that flows out in 30 minutes \( = \pi \times (1)^2 \times 126000 \text{ cm}^3 = 126000\pi \text{ cm}^3 \)
Let the depth of water in the tank be \( x \text{ cm} \).
Volume of water tank \( = \pi(40)^2 \times x \text{ cm}^3 \)
Volume of tank = Volume of water flows
\( \pi(40)^2 \times x = 126000\pi \)
\( x = 78.75 \text{ cm} \)

Question. A well of diameter 4 m dug 21 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 m to form an embankment. Find the height of the embankment.
Answer: Diameter of earth dug out \( = 4 \text{ m} \)
radius of earth dug out \( = 2 \text{ m} \)
Depth of the earth \( = 21 \text{ m} \)
Volume of earth \( = \pi r^2 d = \frac{22}{7} \times 2 \times 2 \times 21 = 264 \text{ m}^3 \)
Width of embankment \( = 3 \text{ m} \)
Outer radius of ring \( = 2 + 3 = 5 \text{ m} \)
Let the height of embankment be \( h \).
Volume of embankment \( = \pi(R^2 - r^2)h = 264 \)
\( \frac{22}{7} \times (25 - 4) \times h = 264 \)
\( \frac{22}{7} \times 21 \times h = 264 \)
\( 66h = 264 \)
\( h = \frac{264}{66} = 4 \text{ m} \)
Height of embankment is 4 m.

Question. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level into the cylindrical vessel rises by \( 3 \frac{5}{9} \) cm. Find the diameter of the cylindrical vessel.
Answer: Diameter of sphere = 12 cm
Its radius = 6 cm
Volume \( = \frac{4}{3} \pi \times 6^3 \text{ cm}^3 \)
It is submerged into water, in cylindrical vessel, then water turn rise by \( 3 \frac{5}{9} = \frac{32}{9} \) cm
Volume submerged = Volume rise
Let radius of cylinder be \( r \) cm
\( \frac{4}{3} \pi \times 6^3 = \pi \times r^2 \times \frac{32}{9} \) cm
\( \frac{216 \times 3 \times 4}{32} = r^2 \)
\( \frac{4 \times 27 \times 3}{4} = r^2 \Rightarrow 4 \times \frac{81}{4} \text{ cm}^3 \Rightarrow r^2 = 81 \)
\( r = 9 \) cm
Diameter \( 2r = 2 \times 9 = 18 \) cm.

Question. The \( \frac{3}{4} \)th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.
Answer: Radius of conical vessel = 5 cm
and its height = 24 cm
Volume of this vessel \( = \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \pi \times 5 \times 5 \times 24 \)
\( = 200 \pi \text{ cm}^3 \)
Internal radius of cylindrical vessel = 10
Let the height of emptied water be \( h \).
Volume of water in cylinder,
\( \pi r^2 h = \frac{3}{4} \times \text{Volume of cone} \)
\( \pi \times 10 \times 10 \times h = \frac{3}{4} \times 200 \pi \)
\( 100 \pi h = 150 \pi \)
\( h = 1.5 \) cm
Hence the height of water = 1.5 cm

Question. Rampal decided to donate canvas for 10 tents conical in shape with base diameter 14 m and height 24 m to a centre for handicapped person's welfare. If the cost of 2 m wide canvas is Rs. 40 per meter, find the amount by which Rampal helped the money.
Answer: Diameter of tent = 14m and height = 24 m
radius of tent = 7m
Slant height \( = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} \)
\( = \sqrt{576 + 49} = 25 \) m
Surface area of the tent
\( \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ m}^2 \)
Surface area of 10 tents
\( = 550 \times 10 = 5500 \)
Total cost \( = 5500 \times \frac{40}{2} = 110000 \)
Hence, Rampal helped the centre
\( = \text{Rs. } 110000 \)

Question. A cone of maximum size is curved out from a cube edge 14 cm. Find the surface area of remaining solid after the cone is curved out.
Answer: Side of cube = 14 cm.
Cone of maximum size is curved out
Diameter of cone = 14 cm
Radius of cone = 7 cm
Slant height \( l = \sqrt{h^2 + r^2} = \sqrt{14^2 + 7^2} \)
\( = \sqrt{196 + 49} = \sqrt{245} \)
\( = 15.65 \) cm.
Total surface area = Surface area cube + curved Surface area of cone - Circular area of base of cone
\( = 6a^2 + \pi rl - \pi r^2 \)
\( = 6 \times 14 \times 14 + \frac{22}{7} \times 7 \times 15.65 - \frac{22}{7} \times 7 \times 7 \)
\( = 1176 + [22(15.65 - 7)] \)
\( = 1176 + 22 \times 8.65 \)
\( = 223792.8 \text{ cm}^3 \)

Question. Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?
Answer: Water flow in 1 hr
= Area of cross-section \(\times\) Speed of water
\( = 5.4 \times 1.8 \times 25000 \text{ m}^3 \)
\( = 54 \times 18 \times 250 \text{ m}^3 \)
Water flow in 40 minutes
\( = 54 \times 18 \times 250 \times \frac{40}{60} \text{ m}^3 \)
\( = 54 \times 6 \times 500 \text{ m}^3 \)
Irrigated area \( \times \frac{10}{100} = 54 \times 6 \times 500 \)
Irrigated area \( = 54 \times 6 \times 500 \times 10 \)
\( = 1620000 \text{ m}^3 \)

Question. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of same height and same base radius is hollowed out. Find the total surface area of the remaining solid. (Take \( \pi = 3.14 \))
Answer: Height of cylinder = height of cone = 8 cm
radius of cylinder = radius of cone = 6 cm
Slant height of cone \( = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10 \) cm
Total surface area of remaining solid
= Surface area of cylinder + Surface area of cone + area of top
\( = 2 \pi rh + \pi rl + \pi r^2 \)
\( = \pi r(2h + l + r) \)
\( = \frac{22}{7} \times 6(2 \times 8 + 10 + 6) \)
\( = \frac{22}{7} \times 6 \times 32 \)
\( = 603.43 \)
Hence total surface area \( = 603.43 \text{ cm}^2 \)

Long Answer Type Questions

Question. From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. Use \( \pi = \frac{22}{7} \)
Answer: As per question the figure is shown below.
Volume of cylinder,
\( \pi r^2h = \frac{22}{7} \times \frac{42}{10} \times \frac{42}{10} \times 10 \text{ cm}^3 \)
\( = 554.40 \text{ cm}^3 \)
Volume of metal scooped out
\( = 2 \times \text{Volume of hemisphere} \)
\( = 2 \times \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{42}{10}\right)^3 \)
\( = 310.46 \text{ cm}^3 \)
Volume of rest of cylinder
\( = 554.40 - 310.46 = 243.94 \text{ cm}^3 \)
Now from rest volume a wire of thickness 1.4 cm i.e radius 0.7 cm is formed. Let length of wire be \( l \). Thus volume of wire and rest cylinder will be equal.
Volume of wire, \( \pi r^2l = 243.94 \text{ cm}^3 \)
\( \frac{22}{7} \times \frac{7}{10} \times \frac{7}{10} \times l = 243.94 \text{ cm}^3 \)
\( l = \frac{243.94 \times 10 \times 10}{22 \times 7} \)
\( l = 158.4 \) cm

Question. A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is \(166 \frac{5}{6} \text{ cm}^3\). Find the height of the toy. Also find the cost of painting the hemisphere part of the toy at the rate of Rs. 10 per \(\text{cm}^2\). Use \(\pi = \frac{22}{7}\)
Answer: As per question the figure is shown below.
\( \text{Radius of cone} = \text{Radius of hemisphere} \)
\( r = 3.5 \text{ cm} \)
\( \text{Total volume, } V = 166 \frac{5}{6} \text{ cm}^3 = \frac{1001}{6} \text{ cm}^3 \)
Let the height of cone be \( h \).
\( \text{Total volume} = \text{Volume of cone} + \text{Volume of hemisphere} \)
\( \frac{1001}{6} = \frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3 \)
\( \frac{1001}{6} = \frac{1}{3}\pi (3.5)^2h + \frac{2}{3}\pi (3.5)^3 \)
\( \frac{1001}{6} = \frac{1}{3}\pi [12.25h + 2 \times 42.875] \)
\( \frac{1001 \times 3 \times 7}{6 \times 22} = 12.25h + 85.75 \)
\( \frac{21021}{132} = 12.25h + 85.75 \)
\( 159.25 = 12.25h + 85.75 \)
\( 12.25h = 159.25 - 85.75 \)
\( h = \frac{73.5}{12.25} = 6 \)
Height of the toy \( = 6 + 3.5 = 9.5 \text{ cm.} \)
Surface area of hemisphere \( 2\pi r^2 = 2 \times \frac{22}{7} \times 3.5 \times 3.5 = 77 \text{ cm}^2 \)
Cost of painting \( = 10 \times 77 = 770 \text{ Rs} \)

Question. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volume of the cylinder and toy. (Use \(\pi = 3.14\))
Answer: As per question the figure is shown below.
Let \(BPC\) is a hemisphere and \(ABC\) is a cone.
Radius of hemisphere = Radius of cone
\(= \frac{4}{2} = 2 \text{ cm}\)
\(h = \text{Height of cone} = 2 \text{ cm}\)
Volume of toy \(= \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\)
\(= \frac{1}{3}\pi r^2(2r + h) = \frac{1}{3} \times 3.14 \times 2 \times 2(2 \times 2 + 2)\)
\(= \frac{1}{3} \times 3.14 \times 4 \times 6\)
\(= 25.12 \text{ cm}^3\)
Let right circular cylinder \(EFGH\) circumscribe the given solid toy.
Radius of cylinder \(= 2 \text{ cm}\)
Height of cylinder \(= 4 \text{ cm}\)
Volume of right circular cylinder
\(\pi r^2 h = 3.14 \times (2)^2 \times 4 \text{ cm}^3 \quad ...(ii)\)
\(= 50.24 \text{ cm}^3\)
Difference of two volume
\(= \text{Volume of cylinder} - \text{Volume of toy}\)
\(= 50.24 - 25.12 = 25.12 \text{ cm}^3\).

Question. A solid is consisting of a right circular cone of height 120 cm and radius 60 cm standing on hemisphere of radius 60 cm. It is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Answer: As per question the figure is shown below.
Height of cone, \( h = 120 \text{ cm,} \)
Radius of cone \( r = 60 \text{ cm} \)
Radius of hemisphere \( = 60 \text{ cm.} \)
Volume of cone \( = \frac{1}{3}\pi r^2h \)
\( = \frac{1}{3} \times 3.14 \times 60 \times 60 \times 120 \)
\( = 3.14 \times 60 \times 60 \times 40 \)
\( = 452160 \text{ cm}^3 \)
Volume of hemisphere \( = \frac{2}{3}\pi r^3 \)
\( = \frac{2}{3} \times 3.14 \times 60 \times 60 \times 60 \)
\( = 452160 \text{ cm}^3 \)
Total volume \( = \) Volume of cone + Volume of hemisphere
\( = 452160 + 452160 \)
\( = 904320 \text{ cm}^3 \)
Height of cylinder \( = 180 \text{ cm,} \)
radius \( = 60 \text{ cm.} \)
Volume of water in the cylinder
\( = \) Volume of cylinder
\( = \pi r^2h \)
\( = 3.14 \times 60 \times 60 \times 180 \)
\( = 2034720 \text{ cm}^3 \)
Water left in the cylinder \( = \) Volume of water \( - \) Volume of (cone + sphere)
\( = 2034720 - 904320 \)
\( = 1130400 \text{ cm}^3 \)

Question. A cubical block of side 10 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 Rs.5 per 100 sq. cm. [Use \( \pi = 3.14 \)]
Answer: As per question the figure is shown below.
Side of given cube \( a = 10 \text{ cm} \)
Area of cube (excluding base)
\( A_1 = \text{area of 4 walls} + \text{area of Top} \)
\( = 4a^2 + a^2 = 5a^2 = 5(10)^2 = 500 \text{ cm}^2 \)
Let \( r \) be the largest radius of hemisphere. From fig. (ii) we have \( \square ABCD \), in the square of side 10 cm.
In \( \Delta ABC, \angle B = 90^\circ \)
From Pythagoras theorem we have
\( AC^2 = AB^2 + BC^2 \)
\( (2r)^2 = (10)^2 + (10)^2 \)
\( 4r^2 = 200 \text{ cm}^2 \)
\( r = \sqrt{\frac{200}{4}} = 5\sqrt{2} \text{ cm} \)
Hence, the required diameter of hemisphere \( d = 2r = 2 \times 5\sqrt{2} = 10\sqrt{2} \text{ cm} \)
Now, area of unshaded part in fig (ii)
\( A_2 = \text{area of circle} - \text{area of square } ABCD \)
\( = \pi r^2 - (a)^2 = [\pi \times 50 - (10)^2] \)
\( = (157 - 100) = 57 \text{ cm}^2 \)
Now, Total surface area of solid
\( A = A_1 + A_2 + 2\pi r^2 \)
\( = [500 + 57 + 2 \times 3.14 \times 50] \)
\( = 871 \text{ cm}^2 \)
The cost of painting of solid \( = (871 \times \frac{5}{100}) = 43.55 \text{ Rs} \)

Question. A cylindrical tub, whose diameter is 12 cm and height 15 cm is full of ice-cream. The whole ice-cream is to be divided into 10 children in equal ice-cream cones, with conical base surmounted by hemispherical top. If the height of conical portion is twice the diameter of base, find the diameter of conical part of ice-cream cones.
Answer: [Foreign Set I, II, III, 2016]
For cylindrical tub,
Diameter \( D = 12 \text{ cm} \)
Radius \( R = 6 \text{ cm} \)
Height \( H = 15 \text{ cm} \).
Volume \( = \pi R^2 H = \pi(6)^2 \times 15 = 540\pi \text{ cm}^3 \)
Each child will get the ice-cream \( = \frac{540\pi}{10} = 54\pi \text{ cm}^3 \)
For cone, height \( h = 2 \times d = 2(2r) = 4r \)
Volume of cone \( = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 \times 4r = \frac{4}{3}\pi r^3 \)
Volume of hemisphere \( = \frac{2}{3}\pi r^3 \)
Total volume of cone and hemisphere \( = \frac{4}{3}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{6}{3}\pi r^3 = 2\pi r^3 \)
According to question, \( 2\pi r^3 = 54\pi \)
\( r^3 = 27 \)
\( r = 3 \)
Hence, Diameter \( = 2r = 2 \times 3 = 6 \text{ cm} \).