NCERT Solutions for Class 10 Maths Chapter 13 - Surface Areas and Volumes

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    Important Points

    1. Solid Figures : The objects which occupy space (i.e., they have three dimensions) are called solids. The solid figures can be derived from the plane figures.

    e.g., In Fig. (a), we have a paper cut in the form as shown. It is a plane figure but when we fold the paper along the dotted lines, a box can be made as shown in Fig. (b).

    Surface Areas and Volumes_im(1)

    2. Surface Area : The sum of the areas of the plane figures making up the boundary of a solid figure is called its surface area e.g., the area of paper in fig. (a) is the surface area of box.

    3. Volume : The measure of part of space occupied by a solid is called its volume.

    4. Parallelopiped (cuboid) : A cuboid or parallelopiped is a region bounded by its six rectangular faces. A parallelopiped whose faces are rectangles is called a rectangular parallelopiped or a rectangular solid or a cuboid.

    Surface Areas and Volumes_im(4)

    (i) A cuboid has six rectangular plane surfaces called faces.

    (ii) A cuboid has 8 corners called the vertices.

    (iii) Volume of cuboid = l × b × h

    where, l = length, b = breadth and h = height

    (iv) Whole surface of cuboid = 2(lb + bh + lh)

    $$\text{(v) Diagonal of the cuboid =}\sqrt{l^{2}+b^{2}+h^{2}}$$

    (vi) Area of 4 walls = 2(l + b).h

    5. Cube : If the faces of a rectangular parallelopiped be squares, then it is called a cube.

    (i) Edge of cube = Length = Breadth = Height

    (ii) Volume of a cube = (Edge)3

    (iii) Total Surface Area (TSA) of a cube = 6 × (Edge)2

    $$\text{(iv) Diagonal of a cube =}\space\sqrt{3}×\text{Edge}$$

    6. Right Circular Cylinder : A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides which remains fixed. A cylinder shown in figure is described by revolving the rectangle ABCD about the side BC. The examples of right circular cylinder are many such as water pipes, powder box, laboratory beakers etc.

    (i) Volume of the cylinder

    = (Area of base) × Height

    = πr2h cu units

    where, r is the radius of the base and h is the height of the cylinder.

    (ii) Curved surface area
    = Circumference of the base × Height
    = 2 πrh sq. units

    (iii) Total surface area = Curved surface area + Area of two ends

    = 2πry + 2πr2 = 2πr(h + r)

    $$\text{Also, note h}=\frac{\text{V}}{\pi r^{2}}\\\Rarr r=\sqrt{\frac{\text{V}}{\pi h}}$$

    7. Hollow Cylinder : The volume of material in a hollow cylinder is the differences between the volume of a cylinder having the external dimensions and the volume of a cylinder having the internal dimensions.

    Let R and r be the external and internal radii of the hollow cylinder and h be its height. Then,

    (i) Volume of material = π(R2 – r2)h

    (ii) Total surface area = 2π(R + r)(h + R – r)

    (iii) Curved surface area

    = 2πRh + 2πrh = 2π(R + r)h

    (iv) Total outer surface area

    = 2πrh + πR2 + π(R2 – r2)

    8. Right Circular Cone : A right circular cone is a solid generated by the revolution of a right angled triangle about one of its sides containing the right angle is axis. It is a 3-D shape that has one circular base and narrows smoothly from base to a point, called vertex. In the adjoining figure a cone of height h and radius r is generated by revolving the right ΔAOB along AO.

    $$\text{The slant height of the cone is\space}\text{l = AC}\\=\sqrt{r^{2}+h^{2}}\\\text{(i) Volume of cone =}\frac{1}{3}\pi r^{2}h\space\text{cu units}$$

    (ii) Curved surface area of cone = πrl sq units.

    (iii) Total surface area of a cone = πr(l + r) sq. units.

    9. Frustum of a Cone : If a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of a cone.

    Surface Areas and Volumes_im(9)

    Let R and r be the radii of base and top of the frustum of a cone. Let h be the height, then

    (i) Volume of frustum of right circular cone

    $$=\frac{\pi h}{3}[\text{R}^{2}+\text{r}^{2}+\text{Rr}]\space\text{cm}^{3}$$

    (ii) Lateral surface area of frustum of right circular cone

    = π(R+r) l sq.units

    $$\text{where, slant height, l =}\sqrt{h^{2}+(R^{2}-r^{2})}$$

    (iii) Total surface area of frustum of right circular cone = Area of base + Area of top + Lateral surface area

    =  π [R2 + r2 + l(R + r)]

    (iv) Total surface area of bucket

    = π [(R2 + r) l + r2]

    (∵ It is open at the bigger end)

    10. Sphere : A sphere is a solid that is round in shape and the points on its surface are at equidistant from the centre.

    (i) A sphere is the locus of a point which moves in space such that its distance from a fixed point in space remains constant. The fixed point is called the centre of sphere and the constant distance is called the radius of sphere.

    (ii) If figure, O is centre and r is radius, all radii are equal.

    (iii) The section of a sphere cut by any plane is a circle.

    (iv) If the cutting plane passes through the centre of the sphere the section is called a great circle.

    $$\text{(v)\space}\text{Volume of sphere =}\frac{4}{3}\pi r^{3}\space\text{cu units}$$

    (vi) Surface area of sphere = 4πr2 sq. units.

    (vii) Volume of a hollow sphere =

    $$\frac{4}{3}\pi(\text{R}^{3}-\text{r}^{3})\text{cu units.}$$

    where, r = inner radius and R = outer radius.

    11. Hemisphere : A plane passing through the centre cuts the sphere in two equal parts called a hemisphere.

    Surface Areas and Volumes_im(11)

    (i) Volume of hemisphere =

    $$=\frac{2}{3}\pi r^{2}\space\text{cu units}$$

    (ii) Curved surface area of hemisphere = 2πr2 sq. units.

    (iii) Total surface area = 2πr2 + πr2 = 3πr2 sq. units.

    12. Surface Area of a Combination of Solids : The total surface area of the new solid is the sum of the curved surface area of each of the individual parts. e.g.,

    TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere

    where, TSA → Total Surface Area

    and CSA → Curved Surface Area

    Exercise 13.1

    $$•\text{ Unless stated otherwise, take}\space\pi=\frac{22}{7}.$$

    1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

    Sol. Given, the volume of a cube is 64 cm3. Let each side of the cube be a cm.

    Surface Areas and Volumes13.1(1)

    Then, a2 = 64 = 42

    ⇒ a = 4 cm

    When we join cubes end to end, then length of the new cuboid become (4 + 4) = 8 cm, b = 4 cm, h = 4 cm.

    ∴ Surface area of the resulting cuboid

    = 2(lb + bh + hl)

    = 2(8 × 4 + 4 × 4 + 4 × 8)

    = 2(32 + 16 + 32)

    = 2(80) = 160 cm2

    2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

    Sol. Given, vessel is a combination of hollow hemisphere and a hollow cylinder. Also, we have

    AB = DC = 14 cm (Diameter)

    ∴ OB = O′C = O′P

    $$=\frac{\text{AB}}{2}=\frac{\text{14}}{2}=\text{7 cm}\space\text{(Radius)}$$

    and PO = 13 cm

    ∴ OO′ = PO – O′P

    = 13 – 7 = 6 cm

    ∴ Radius of hemisphere = Height of hemisphere

    = 7 cm

    Now, the inner surface area of the vessel

    = Surface area of cylinder
    + Surface area of hemisphere

    = 2πrh + 2πr2

    $$2×\frac{22}{7}×7×6+2×\frac{22}{7}×(7)^{2}$$

    = 2 × 22 × 6 + 2 × 22 × 7

    = 44(6 + 7) = 44 × 13

    = 572 cm2

    3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the boy.

    Sol. Given, toy is a combination of hemisphere and a cone.

    Surface Areas and Volumes_13.1(3)

    Here, AD = 15.5 cm

    OC = OD = OB = 3.5 cm

    (Radius of cone & hemisphere)

    OA = AD – OD

    = 15.5 – 3.5 = 12 cm

    ∴ Height of the cone = 12 cm

    ∴ Total surface area of the toy

    = Surface area of cone + Surface area of hemisphere

    = πrl + 2πr2

    $$=\pi r\sqrt{h^{2}+r^{2}}+2\pi r^{2}\\=\frac{22}{7}×3.5×\sqrt{(12)^{2}+(3.5)^{2}}+2×\frac{22}{7}×(3.5)^{2}\\=11\sqrt{144+12.25}+22×3.5\\=11\sqrt{156.25}+11×7$$

    = 11(12.5) + 77 = 137.5 + 77

    = 214.5 cm2

    4. 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.

    Sol. Given, a cubical block is surmounted by a hemisphere. Therefore, its diameter must be equal to the side of cube, i.e., 7 cm.

    $$\therefore\space\text{Radius of hemisphere (r) =}\frac{7}{2}\text{cm}$$

    Now, total curved surface area of solid

    = Surface area of five faces of the cube + Surface area of hemisphere except circular face PQRS + [Area of face ABCD – Area of circular space PQRS]

    = 5 × (side)2 + 2π(radius)2 + [(side)2 – π(radius)2]

    $$= 5×(7)^{2}+2×\frac{22}{7}×\bigg(\frac{7}{2}\bigg)^{2} + \begin{Bmatrix}(7)^{2}-\bigg(\frac{7}{2}\bigg)^{2}\end{Bmatrix}\\=5×49+11×7+\bigg(49-\frac{77}{2}\bigg)\\=245+77+49-\frac{77}{2}\\=294+\frac{77}{2}\\=\frac{665}{2}=332.5\space\text{cm}^{2}$$

    Hence, required area of the solid is 332.5 cm2.

    5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

    Sol. Given, side of the cube = diameter of the hemisphere = l

    Surface Areas and Volumes_13.1(5)

    $$\therefore\space\text{Radius of a hemisphere =}\frac{l}{2}$$

    ∴ Now,
    Required surface area of the remaining solid

    = [Area of one face ABCD of the cube – Area of circle PQRS] + [Area of remaining five faces of the cube + Area of hemisphere]

    $$=\bigg(l^{2}-\pi\frac{l^{2}}{4}\bigg)+5l^{2}+2\pi\frac{l^{2}}{4}\\6l^{2}+\frac{\pi l^{2}}{4}=\frac{l^{2}}{4}(\pi+24)\text{sq. units}$$

    Chapter 13 Surface Areas and Volumes
    Play Video about Chapter 13 Surface Areas and Volumes

    6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

    Surface Areas and Volumes13.1(6)

    Sol. In the given medicine capsule, two hemisphere and one cylinder are included.

    and diameter of the capsule = 5 mm

    $$\text{Radius}=\frac{5}{2}=2.5\space\text{mm}$$

    Length of the capsule = 14 mm

    ∴ Length of the cylinder = 14 – (2.5 + 2.5) = 9 mm

    Now, surface area of a hemisphere = 2πr2

    $$=2×\frac{22}{7}×2.5×2.5\\=\frac{275}{7}\text{mm}^{2}$$

    Surface Areas and Volumes_13.1(6)

    Surface area of the cylinder = 2πrh

    $$=2×\frac{22}{7}×2.5×9\\=\frac{22}{7}×45\\=\frac{990}{7}\text{mm}^{2}$$

    ∴ Required surface area of medicine capsule

    = 2 × Surface area of a hemisphere + Surface area of the cylinder

    $$= 2 ×\frac{275}{7}+\frac{900}{7}=\frac{550}{7}+\frac{990}{7}\\=\frac{1540}{7}=220\space\text{mm}^{2}$$

    7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ` 500 per m2. (Note that the base of the tent will not be covered with canvas)

    Sol. A tent which is combination of a cylinder and a cone.

    Surface Areas and Volumes_13.1(7)

    Slant height (l) of the cone = 2.8 m

    Radius of the cone, r = Radius of cylinder

    $$=\frac{\text{Diameter}}{2}=\frac{\text{4}}{2}=2m$$

    and height of the cylinder, h = 2.1 m

    ∴ Required surface area of the tent

    = Surface area of cone + Surface area of cylinder

    = πrl + 2πrh

    = πr(l + 2h)

    $$=\frac{22}{7}×2×(2.8+2.1)\\=\frac{44}{7}(2.8+4.2)\\\frac{44}{7}×7=44\space\text{m}^{2}$$

    Surface Areas and Volumes13.1(7_1)

    Hence, the cost of the canvas of the tent at the rate of ₹ 500 per m2

    = Surface area × Cost per m2

    = 44 × 500 = ₹ 22000

    8. From a solid cylinder whose height is 2.4 cm and diameter 14 cm, a conical cavity of the same height and same diameter is followed out. Find the total surface area of the remaining solid to the nearest cm2.

    Sol. Given, diameter of cylinder = diameter of conical cavity = 1.4 cm

    Surface Areas and Volumes13.1(8)
    ∴ Radius of cylinder = Radius of conical cavity $$=\frac{\text{Diameter}}{2}\\=\frac{1.4}{2}=0.7\space\text{cm}$$ Height of the cylinder = Height of the conical cavity = 2.4 cm ∴ Slant height of the conical cavity (l) $$=\sqrt{h^{2}+r^{2}}\\=\sqrt{(2.4)^{2}+(0.7)^{2}}\\=\sqrt{5.76+0.49}=\sqrt{6.25}$$ Total surface area of remaining solid = Surface area of conical cavity + Total surface area of cylinder = πrl + (2πrh + πr2) = πr(l + 2h + r) 2.2 × 8 = 17.6 = 18 cm2

    9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

    Sol. Given,

    Height of the cylinder, h = 10 cm

    and radius of base of cylinder = Radius of hemisphere (r) = 3.5 cm

    Now, required total surface area of the article

    = 2 × Surface area of hemisphere
    + Curved surface area of cylinder

    = 2 × (2πr2) + 2πrh

    = 2πr(2r + h)

    $$2×\frac{22}{7}×3.5×(2×3.5×10)\\=\frac{22}{7}×7×(7+10)$$

    =22×17=374 cm2

    Exercise 13.2

    • Unless stated otherwise, take π = $$\frac{22}{7}.$$

    1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

    Sol. Given, solid is a combination of a cone and a hemisphere.

    Surface Areas and Volumes_13.2(1)

    According to the figure, Radius of the cone (r) = Radius of the hemisphere = 1 cm

    Height of the cone (h) = 1 cm

    ∴ Required volume of the solid

    = Volume of the cone + Volume of the hemisphere

    $$=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^{3}\\=\frac{1}{3}\pi(1)^{2}(1)+\frac{2}{3}\pi(1)^{3}\\=\frac{\pi}{4}+\frac{2}{3}\pi\\=\frac{(\pi+2\pi)}{3}=\frac{3\pi}{3}$$

    = π cm3

    2. Rachel, an engineering, student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel mode. (Assume the outer and inner dimensions of the model to be nearly the same.)

    Sol. Given, model is a combination of a cylinder and two cones. Also, we have, diameter of the model, BC = ED = 3 cm

    $$\therefore\space r=\frac{3}{2}=1.5\space\text{cm}$$

    Height of cone, y1 = 2 cm and length of the model, AF = 12 cm

    ∴ OO′ = AF – (AO + O′F)

    = 12 – (2 + 2)

    = 8 cm

    ∴ Height of cylinder, h2 = 8 cm

    Now, volume of the air inside the model

    = Volume of air inside (cone + cylinder + cone)

    $$=\bigg(\frac{1}{3}\pi r^{2}h_1+\pi r^{2}h_2+\frac{1}{3}\pi r^{2}h_1\bigg)\\=\frac{1}{3}\pi r^{2}(h_1+3h_2+h_1)\space\text{...(i)}\\=\frac{1}{3}×\frac{22}{7}×1.5×1.5(2+3×8+2)\\=\frac{22}{7}×\frac{2.25}{3}×(2+24+2)\\=\frac{22}{7}×0.75×28$$

    = 22 × 3

    = 66 cm3

    3. A gulab jamun, contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

    Surface Areas and Volumes13.2(3)

    Sol. Let r be the radius of the hemispheres and cylinder both h1 be the height of the hemisphere which is equal to its radius and h2 be the height of the cylinder. Given, length = 5 cm, diameter = 2.8 cm.

    $$\therefore\space r=h_1=\frac{2.8}{2}=1.4\space\text{cm}$$

    and h2 = PQ – (PR + SQ)

    = 5 – (1.4 + 1.4)

    = 5 – 2.8 = 2.2 cm

    ∴ Volume of one gulab jamun

    = 2 × (Volume of hemisphere) + Volume of cylinder

    $$= 2×\begin{Bmatrix}\frac{2}{3}\pi r^{3}\end{Bmatrix}+\pi r^{2}h_2\\=\frac{4}{3}\pi r^{3}+\pi r^{2}h_2\\=\pi r^{2}\begin{Bmatrix}\frac{4\pi r}{3}+h_2\end{Bmatrix}\\=\frac{22}{7}×1.4×1.4\begin{Bmatrix}\frac{4}{3}×r+2.2\end{Bmatrix}\\=\frac{22}{7}×\frac{14}{10}×\frac{14}{10}\begin{Bmatrix}\frac{4}{3}×\frac{14}{10}+\frac{22}{10}\end{Bmatrix}\\=22×\frac{1}{5}×\frac{7}{5}\begin{Bmatrix}\frac{28}{15}+\frac{11}{5}\end{Bmatrix}\\=\frac{154}{25}×\frac{61}{15}=\frac{9394}{375}\text{cm}^{3}$$

    ∴ Volume of 45 gulab jamuns

    $$= 45×\frac{9394}{375}=\frac{1}{75}×84546$$

    = 1127.28 cm3

    ∴ One gulab jamun, contains sugar syrup upto about 30% of its volume

    $$\text{i.e.}\space\bigg(\frac{9394}{375}\bigg)×30\%$$

    Hence, quantity of syrup found in 45 gulab jamun

    $$=1127.28×\frac{30}{100}\\=1127.28×\frac{3}{10}=338.184$$

    = 338 cm3

    4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

    Sol. Given, length of cuboid (l) = 15 cm

    Breadh of cuboid (b) = 10 cm

    and height of cuboid (h) = 3.5 cm

    ∴ Volume of cuboid = l × b × h

    = 15 × 10 × 3.5

    = 525 cm3

    Also, given that

    Radius of conical depression r = 0.5 cm

    and height of conical depression, h = 1.4

    ∴ Volume of a conical depression

    $$=\frac{1}{3}\pi×r^{2}×h\\=\frac{1}{3}×\frac{22}{7}×0.5×0.5×1.4\\=\frac{22}{3}×\frac{1}{2}×\frac{1}{2}×\frac{2}{10}\\=\frac{11}{30}\space\text{cm}^{3}$$

    13.2(4_2)Surface Areas and Volumes13.2(4)

    ∴ Volume of 4 conical depressions

    = 4 × Volume of a conical depression

    $$=4×\frac{11}{30}\\=\frac{22}{15}\space\text{cm}^{3}$$

    Hence, the volume of wood in the entire wood

    = Volume of cuboid – Volume of 4 conical depressions

    $$= 525-\frac{22}{15}$$

    = 525 – 1.46

    = 523.54 cm3

    5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

    Sol. Given,

    Height of the vessel (h) = 8 cm

    Radius of the vessel (h) = 5 cm

    ∴ Volume of water filled in a vessel

    $$=\frac{1}{3}×\pi×r×r×h\\=\frac{1}{3}×\frac{22}{7}×5×5×8\\=\frac{4400}{21}\space\text{cm}^{3}$$

    Radius of a lead shot (sphere) = 0.5 cm

    ∴ Volume of a lead shot (sphere)

    $$=\frac{4}{3}×\pi r^{3}\\=\frac{4}{3}×\frac{22}{7}×\frac{1}{2}×\frac{1}{2}×\frac{1}{2}=\frac{11}{21}\space\text{cm}^{3}$$

    Let ‘n’ be the number of lead shots dropped in the vessel.

    ∴ n × Volume of a lead shot (sphere)

    $$=\frac{1}{4}×\text{Volume of water filled in a vessel}$$

    (∵ Each lead shot flow out the water from the vessel equal to its volume)

    $$\Rarr\space n×\frac{11}{21}=\frac{4400}{21}×\frac{1}{4}$$

    ⇒ n × 11 = 1100

    ⇒ n = 100

    ∴ Required number of lead shots (n) = 100.

    6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)

    Surface Areas and Volumes_13.2(6)

    Sol. Given,

    Height of the first cylinder, h1 = 220 cm

    $$\text{Radius of the first cylinder, r}_1=\frac{24}{2}=12\space\text{cm}$$

    Height of the second cylinder, h2 = 60 cm

    $$\text{and radius of the second cylinder, r}_2 =\frac{16}{2}=8\space\text{cm}$$

    ∴ Volume of iron pole = Volume of first cylinder + Volume of second cylinder

    = πr12h1 + πr22h2

    = π(r12h1 + r22h2)

    = 3.14 {144 × 220 + 64 × 60}

    = 3.14 (31680 + 3840)

    = 3.14 × 35520 cm3

    Also, given that

    1 cm3 of iron has approximately mass = 8 g

    $$=\frac{8}{1000}\space\text{kg}$$

    ∴ (3.14 × 25520) cm3 of iron has approximately mass

    $$= 3.14×35520 ×\frac{8}{1000}$$

    = 3.14 × 284.160

    = 892.26 kg

    7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm 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.

    Sol. Given,

    Height of the cylinder (h) = 180 cm = 1.8 m

    $$(\because 1m=100\space \text{cm})\Rarr 1\space\text{cm}=\frac{1}{100}\text{m}$$

    and radius of the cylinder (r) = 60 cm = 0.6 m

    ∴ Volume of water in a right circular cylinder

    = πr2h

    $$=\frac{22}{7}×0.6××0.6×8$$

    $$=\frac{14.256}{7}\space\text{m}^{3}$$

    The solid immersed in the cylindrical tank is a combination of a cone and a hemisphere.

    Height of the cone (h1) = 120 cm = 1.2 m

    Radius of the cone (r1) = 60 cm = 0.6 m

    and radius of the hemisphere (r2) = 60 cm = 0.6 m

    ∴ Volume of the solid immersed

    = Volume of the cone + Volume of the hemisphere

    $$=\frac{1}{3}×\pi r_1^{2}h_1+\frac{2}{3}\pi r_2^{3}\\=\frac{1}{3}×\frac{22}{7}×(0.6)^{2}×(1.2)+\frac{2}{3}×\frac{22}{7}×(0.6)^{3}\\=\frac{22}{21}×(0.6)^{2}[1.2+2×0.6]\\=\frac{22}{21}×0.36(1.2+1.2)\\=\frac{22}{21}×0.36×2.4\\=\frac{19.008}{21}m^{3}=\frac{6.336}{7}m^{3}$$

    The volume of water left in the cylinder

    = Volume of water filled in a right circular cylinder – Volume of the solid

    $$=\frac{14.256}{7}-\frac{6.336}{7}=\frac{7.92}{7}$$

    = 1.131428 m3 = 1.131 m3

    8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuing the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and π = 3.14.

    Sol.

    Given, spherical glass vessel is a combination of sphere as its base and a cylinder as its neck.

    Also, we have height of the cylinder (h1) = 8 cm

    $$\text{Radius of the cylinder (r)}_1=\frac{2}{2}=1\space\text{cm}\\\text{and radius of the sphere (r}_2) =\frac{8.5}{2}\space\text{cm}$$

    ∴ Volume of water filled in a spherical glass vessel

    = Volume of cylinder + Volume of sphere

    $$=\pi r_1^{2}h_1+\frac{4}{3}\pi r_2^{3}$$

    $$=\pi r_1^{2}h_1+\frac{4}{3}\pi r_2^{3}\\=3.14×1×1×8+\frac{4}{3}\\×3.14×\frac{8.5}{2}×\frac{8.5}{2}×\frac{8.5}{2}\\=25.12+\frac{1928.3525}{6}$$

    = 25.12 + 321.39

    = 346.51

    So, the correct answer is 346.51 cm3.

    Exercise 13.3

    • Unless stated otherwise take

    $$\pi=\frac{22}{7},$$

    1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius
    6 cm. Find the height of the cylinder.

    Sol. Given, the radius of the sphere (r) = 4.2 cm and radius of the cylinder (r1) = 6 cm

    Then, the volume of the sphere

    $$=\frac{4}{3}×\pi r^{3}\\=\frac{4}{3}×\frac{22}{7}×4.2×4.2×4.2$$

    = 4 × 22 × 1.4 × 0.6 × 4.2

    = 310.464 cm3

    Let h be the height of the cylinder formed

    As per question

    Volume of the sphere = Volume of the cylinder

    $$\Rarr\space 310.464=\frac{22}{7}×6×6×h\\\Rarr\space h=\frac{310.464×7}{22×6×6}\\=\frac{2173.248}{792}$$

    ⇒ h = 2.744 cm

    2. Metallic spheres or radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

    Sol. Let r1, r2 and r3 be the radius of given three spheres respectively and R be the radius of a single solid sphere.

    Given, r1 = 6 cm, r2 = 8 cm and r3 = 10 cm

    Volume of first metallic sphere (V1) =

    $$\frac{4}{3}\pi r_1^{3}\\=\frac{4}{3}\pi(6)^{3}\\=\frac{864}{3}\pi\space\text{cm}^{3}$$

    Volume of second metallic sphere (V2) =

    $$\frac{4}{3}\pi r_2^{3}\\=\frac{4}{3}\pi(8)^{3}\\=\frac{2048}{3}\pi\space\text{cm}^{3}$$

    Volume of third metallic sphere (V3) =

    $$=\frac{4}{3}\pi r_3^{2}\\=\frac{4}{3}\pi(10)^{3}\\=\frac{4000}{3}\pi\space\text{cm}^{3}$$

    Volume of a resulting solid sphere formed (V) =

    $$=\frac{4}{3}\pi R^{3}$$

    According to the question

    Volume of three metallic sphere

    = Volume of a single solid sphere

    ⇒ V1 + V2 + V3 = V

    $$\Rarr\space\frac{864}{3}\pi+\frac{2048}{3}\pi+\frac{4000}{3}\pi\\=\frac{4}{3}\pi R^{3}\\\Rarr\space\frac{6912}{3}=\frac{4}{3}\text{R}^{3}$$

    ⇒ R3 = 1728 = (12)3

    ⇒ R = 12 cm

    3. A 20 m deep well with diameter 7 m is dug and the Earth from digging is evently spread out to form a platform 22 m by 14 m. Find the height of the platform.

    Sol. Given, the height of deep well which form a cylinder (h) = 20 m

    13.3(3)Surface Areas and Volumes

    $$\text{Radius of deep well (r}_1) =\frac{7}{2}m$$

    Length of the platform, l = 22 m

    and breadh of the platform b = 14 m

    Let height of the platform be h′
    A.T.Q.,

    Volume of deep well (cylinder) = Volume of platform (cuboid)

    πr12h1 = l × b × h′

    $$\Rarr\space\frac{22}{7}×\frac{7}{2}×\frac{7}{2}×20\\= 22 × 14 × h^′\\\Rarr\space\frac{11×7}{2}×20\\22 × 14 × h^′\\\Rarr\space h'=\frac{11×7×20}{2×22×14}=\frac{5}{2}$$

    ∴ Height of the platform (h2) = 2.5 m.

    4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

    Sol. Given, the height of deep well which form a cylinder (h1) = 14 m

    $$\text{Radius of deep well (r}1) =\frac{3}{2}\text{m}$$

    Let the width and height of the embarkment form a circular ring are d and h2.

    Here, d = 4 cm (Given)

    Now, the volume of deep well (cylinder) formed

    $$=\pi r_1^{2}h_1\\=\frac{22}{7}×\frac{3}{2}×\frac{3}{2}×14=99\space m^{3}$$

    Here, we observe that the embankment form a hollow cylinder.

    ∴ Volume of hollow cylinder (embankment)

    $$=\pi(r_1+d)^{2}.h_2-\pi r_1^{2}h_2\\=\pi\bigg(\frac{3}{2}+4\bigg)^{2}h_2-\pi\bigg(\frac{3}{2}\bigg)^{2}h_2\\=\pi\bigg[\bigg(\frac{11}{2}\bigg)^{2}h_2-\frac{9}{4}h_2\bigg]$$

    According to the question

    Volume of deep well (cylinder)

    = Volume of embankment (hollow cylinder)

    $$\Rarr\space 99=\pi\bigg[\bigg(\frac{11}{2}\bigg)^{2}×h_2-\frac{9}{4}h_2\bigg]\\\Rarr\space\frac{7×99}{22}=\frac{121}{4}×h_2-\frac{9}{4}h_{2}-\frac{112h_2}{4}\\\Rarr\space h_2=\frac{7×99×4}{22×112}\\=\frac{9}{8}=1.125\space\text{m}$$

    ∴ Height of the embankment is 1.125 cm.

    5. A container shaped like a right circular cylinder having diameter 12 cm and height
    15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

    Sol. Let the height and radius of ice cream container (cylinder) be h1 and r1.

    Given, h1 = 15 cm and r1 = 6 cm

    ∴ Volume of the ice cream in the container

    = Volume of a cylinder

    $$=\pi r_1^{2}h_1\\=\frac{22}{7}×6×6×15\\=\frac{11880}{7}\text{cm}^{3}$$

    Now, the cone full with ice-cream have top surface like a hemisphere.

    Also, we have

    Height of the cone, h2 = 12 cm

    Diameter of the cone, r1 = 6 cm

    ∴ Radius of the cone r2 = 3 cm = Radius of the hemisphere

    ∴ Volume of cone full with ice-cream

    = Volume of cone + Volume of hemisphere (upper surface level)

    $$=\frac{1}{3}\pi r_2^{2}h_2+\frac{2}{3}\pi r_2^{3}\\=\frac{1}{3}\pi(r_2^{2}h_2+2r_2^{3})\\=\frac{1}{3}×\frac{22}{7}[(3)^{2}×12+2(3)^{3}]\\=\frac{22}{3×7}×9×(12+6)\\=\frac{22×3×18}{7}=\frac{1188}{7}\text{cm}^{3}$$

    Let ‘n’ be number of cones in which ice cream is filled then

    Volume of ice cream container

    = n × volume of one cone full with ice cream

    $$\Rarr\space\frac{11880}{7}=n×\frac{1188}{7}\\\Rarr n=10$$

    Hence, the number of cones can be filled are 10.

    6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

    Sol. We know that, every coin has a shape of cylinder. Let radius and height of the coin are r1 and h1.

    13.3(6)Surface Areas and Volumes

    Given, diameter of coin = 1.75 cm

    $$\therefore\space r_1=\frac{1.75}{2}\text{cm}\\\space\text{and}\space h_1=2\space\text{mm}=0.2\space\text{cm}\\\therefore\space\text{Volume of a coin}\\=\pi r_1^{2}h_1\\=\frac{22}{7}×\frac{1.75}{2}×\frac{1.75}{2}×0.2\\=\frac{13.475}{28}\space\text{cm}^{3}$$

    The length, breadth and height of the cuboid formed are 5.5 cm, 10 cm and 3.5 cm, respectively.

    ∴ Volume of the cuboid

    = Length × Breadth × Height

    = 5.5 × 10 × 3.5 = 192.5 cm3

    Let ‘n’ be the number of coins required.

    Then, n × Volume of a coin = Volume of the cuboid

    $$\Rarr\space n×\frac{13.475}{28}=192.5\\\Rarr\space n=\frac{192.5×28}{13.475}=400$$

    Hence, 400 silver coins must be melted to form a cuboid.

    7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

    Sol. The height and radius of cylindrical bucket are 32 cm and 18 cm, respectively.

    ∴ Volume of cylindrical bucket = π(Radius)2 × (Height) = π(18)2 × 32

    According to the question

    Let the radius and slant height of the heap of sand are r and l.

    Given, the height of the heap of sand h = 24 cm.

    ∴ Volume of the heap of sand

    $$=\frac{1}{3}\pi r^{2}(24)=8\pi r^{2}$$

    According to the question

    Volume of the heap of sand

    = Volume of the cylindrical bucket

    ⇒ 8πr2 = π × 18 × 18 × 32

    ⇒ r2 = 18 × 18 × 4

    ⇒ r = 36 cm

    Now, the slant height of the conical heap of sand

    $$l=\sqrt{h^{2}+r^{2}}\\=\sqrt{(24)^{2}+(36)^{2}}\\=\sqrt{576+1296}=\sqrt{1872}\\\text{I=}\sqrt{144×13}\\=12\sqrt{13}\space\text{cm}$$

    8. Water in a canal, 6 m wide and 1.5 m deep is flowing with a speed of 10 kmh–1. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed ?

    Sol. Given, speed of flow of water (l) = 10 kmh–1

    Canal is in the shape of a cuboid, where breadth = 6 m, height = 1.5 m

    Speed of canal = 10 km/hr

    Surface Areas and Volumes13.3(8)

    Length of canal in 1 hour = 10 km

    Length of canal in 60 minutes = 10 km

    Length of canal in 1 minute

    $$=\frac{1}{60}×10\space\text{km}$$

    Length of canal in 30 minute =

    $$\frac{30}{60}×10$$

    = 5 km = 5000 m

    Volume of canal = l × r × h = 5000 × 6 × 1.5 m

    Volume of water in canal = Volume area integrated

    Volume of water in canal = Area × Height

    $$5000×6×1.5=\text{Area}×\frac{8}{100}\\\text{Area}=\frac{5000×6×1.5×100}{8}$$

    = 562500 m2

    $$=\frac{562500}{10000}\text{hectares}$$

    = 56.25 hectares

    9. A farmer connects a pipe of intrenal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 kmh–1, in how much time will the tank be filled?

    Sol. Given, speed of flow of water = 3 kmh–1

    = 3 × 1000 mh–1

    ∴ Length of water in 1 h = 3000 m

    Now, area of face of pipe which form a circle

    $$=\pi\bigg(\frac{20}{2}\bigg)^{2}\space(\because\space\text{A}=\pi r^{2})\\=100\pi\space\text{cm}^{2}=\frac{\pi}{100}m^{2}$$

    Volume of cylindrical tank

    $$=\pi×\bigg(\frac{10}{2}\bigg)^{2}×2=50\pi\space m^{3}$$

    (∵ V = πr2h)

    ∴ Time required to fill the tank

    $$=\frac{\text{Volume of cylindrical tank}}{\text{tankArea of face of pipe× Length of water}}\\=\frac{50\pi}{\frac{\pi}{100}×3000}\text{h}\\=\frac{50×60×100}{3000}\text{min}$$

    = 100 min

    Exercise 13.4

    • Unless stated otherwise take

    $$\pi=\frac{22}{7},$$

    1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

    Sol. Let the height of the frustum of a cone be h.

    and radius of both ends of the frustum are r1 and r2, respectively.

    Given, h = 14 cm, r1 = 1 cm and r2 = 2 cm.

    ∴ The volume (capacity) of drinking glass (frustum of a cone)

    $$=\frac{\pi}{3}h(r_1^{2}+r_2^{2}+r_1r_2)\\=\frac{1}{3}×\frac{22}{7}×(14)[(1)^{2}+(2)^{2}+(1)(2)]\\=\frac{44}{3}(1+4+2)=\frac{44×7}{3}\\=\frac{308}{3}=102\frac{2}{3}\text{cm}^{3}$$

    2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

    Sol. Let the slant height of the frustum be l and radius of the both ends of the frustum be r1 and r2 respectively.

    Given that, l = 4 cm Perimeter of one end of frustum which is circular in shape = 2πr1 = 18 $$\Rarr\space r_1=\frac{9}{\pi}\space\text{cm}$$ and perimeter of other end of frustum which is also circular in shape = 2πr2 = 6 $$\Rarr\space r_2=\frac{3}{\pi}\space\text{cm}$$ ∴ Required curved surface area of the frustum π(r1+r2)l $$=\pi\begin{Bmatrix}\frac{9}{\pi}+\frac{3}{\pi}\end{Bmatrix}×4$$ = 12 × 4 = 48 cm2

    3. A fez, the cap used by the turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

    13.4(3)Surface Areas and Volumes
    Sol. Let the slant height of fez be l and the radius of upper end which is closed be r1 and the other end which is open be r2.

    Given, l = 15 cm, r1 = 4 cm and r2 = 10 cm

    ∴ Area of upper end of fez which is circular and closed = πr12

    = π(4)2 = 16π cm2

    Hence, total curved surface area of the fez

    = Lateral surface area of frustum + Area of upper close end

    = π(r1 + r2)l + πr12

    = π{(4 + 10) 15 + 16}

    $$=\frac{22}{7}(210+16)\\=\frac{22×226}{7}=\frac{4972}{7}\\710\frac{2}{7}\text{cm}^{2}$$

    4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container. If it costs ₹ 8 per 100 cm2. (Take π = 3.14)

    Sol. Let h be the height of the container. Also, consider the radius of its lower end as r1 and upper end as r2.
    Given, h = 16 cm, r1 = 8 cm and r2 = 20 cm

    ∴ Volume of the container

    $$=\frac{\pi}{3}×h(r_1^{2}+r_2^{2}+r_1r_2)\\=\frac{3.14}{3}×16[(8)^{2}+(20)^{2}+8×20]\\=\frac{3.14×16×(64+400+160)}{3}\\=\frac{3.14×16×624}{3}\text{cm}^{3}\\=\frac{3.14×16×624}{3×1000}\text{L}\space\bigg(\because 1\space \text{cm}^{3}=\frac{1}{1000}\text{L}\bigg)$$

    ∵ Cost of 1 L milk = ₹ 20

    $$\therefore\space\text{Cost of}\space\bigg(\frac{3.14×16×624}{3×1000}\bigg)\text{L milk}$$

    $$= ₹\frac{3.14×16×624×20}{3×1000}\\=\frac{626995.2}{3000}$$

    = 208.99

    = ₹ 209

    Now, the area of its lower end (which is closed) = πr12 = 64 π cm2

    13.4(4)Surface Areas and Volumes

    $$\text{From figure, PC = DP}^′\\ =\frac{\text{CD-PP'}}{2}\\=\frac{40-16}{2}(\because\space\text{PP}^{'}=\text{AB})\\=\frac{24}{2}=12\space\text{cm}$$

    ∴ In right angled BPC,

    BC2 = PC2 + BP2

    (By Pythagoras theorem)

    ⇒ BC2 = (12)2 + (16)2

    = 144 + 256

    = 400 = (20)2

    ⇒ BC = 20 cm

    which is the slant height (l) of the container.

    Now, the curved surface area of frustum

    = π(r1 + r2) l + πr12

    = 3.14(8 + 20) × 20 + 3.14 × 64

    = 3.14(28 × 20 + 64)

    = 3.14 × 624

    = 1959.36 cm2

    ∴ Cost of metal sheet per 100 cm2 = ₹ 8

    ∴ Cost of metal sheet per 1959.36 cm2

    $$= ₹\frac{8×1959.36}{100}$$

    = ₹ 156.7488

    = ₹ 156.75

    5. A metallic right circular cone 20 m high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $$\frac{1}{16}\text{cm},$$ find the length of the wire.

    Sol. Height of right circular cone = h = 20 cm

    Right circular cone is cut into two parts at the middle.

    We get right circular cone of height 10 cm and frustum of cone of height 10 cm.

    Height of the frustum of the cone = h1 = 10 cm

    Let, the radius of the upper end of the frustum of the cone = r1 cm

    Let, the radius of the lower end of the frustum of the cone = r2 cm

    Now, from ΔAPQ,

    $$\frac{r_1}{\text{AQ}}=\text{tan\space 30\degree}\\\Rarr\space\frac{r_1}{10}=\frac{r}{\sqrt{3}}\\\Rarr\space r_1=\frac{10}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\=\frac{10\sqrt{3}}{\sqrt{3}}\text{cm}$$

    Now, from ΔABC

    $$\frac{\text{r}_2}{\text{AC}}=\text{tan}\space30\degree$$

    $$\Rarr\space\frac{r_2}{20}=\frac{1}{\sqrt{3}}\\\Rarr\space r_2=\frac{20}{\sqrt{3}}\\=\frac{20\sqrt{3}}{3}\text{cm}$$

    Let, length of the wire = l cm

    Radius of the wire = r=

    $$\frac{1}{16×2}=\frac{1}{32}\space\text{cm}$$

    A.T.Q.,

    Volume of the frustum of the cone = Volume of wire

    $$\Rarr\space\frac{1}{3}\pi h[(r_1)^{2}+(r_2)^{2}+(r_1)(r_2)]\\=\pi r^{2}l\\\Rarr\space\frac{1}{3}×10\bigg(\frac{100}{3}+\frac{400}{3}+\frac{200}{3}\bigg)\\=\frac{1}{32}×\frac{1}{32}×l\\\Rarr\space l=\frac{l}{3}×10×\frac{700}{3}×32×32$$

    = 796444.44 cm

    Exercise 13.5

    1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

    Sol. Since, the diameter of the wire is 3 mm.

    When a wire is one round wound about a cylinder, it covers a 3 mm of length of the cylinder.

    Given, length of the cylinder = 12 cm = 120 mm

    ∴ Number of rounds to cover 120 mm

    $$=\frac{120}{3}=40$$

    Given, diameter of a cylinder is d= 10 cm

    $$\therefore\space\text{Radius, r =}\frac{10}{2}=5\space\text{cm}$$

    ∴ Length of wire required to complete one round = 2πr = 2π(5) = 10 π cm

    ∴ Length of the wire in covering the whole surface

    = Length of the wire in completing 40 rounds

    = 10π × 40 = 400π cm

    = 400 × 3.14 = 1256 cm

    Now, radius of copper wire =

    $$\frac{3}{2}\space\text{mm}=\frac{3}{20}\space\text{cm}$$

    ∴ Volume of wire =

    $$\pi\bigg(\frac{3}{20}\bigg)^{2}(400\pi)$$

    = 9π2 cm3

    ∴ Mass of wire = 9π2 × density

    = 9(3.14)2 × 8.88

    = 88.74 × 8.88

    = 788 g (approx)

    2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).

    Sol. Here, ABC is a right angled triangle, angled right at A and BC is the hypotenuse.

    $$\therefore\space\text{BC}=\sqrt{3^{2}+4^{2}}\\=\sqrt{25}=5\space\text{cm}$$

    As ΔABC revolves about the hypotenuse BC. It forms two cones ABD and ACD.

    Since, ΔAEB and ΔABC are similar.

    $$\therefore\space\frac{\text{AE}}{\text{CA}}=\frac{\text{AB}}{\text{BC}}\\\Rarr\frac{\text{AE}}{4}=\frac{3}{5}\\\Rarr\space\text{AE}=\frac{12}{5}=2.4\space\text{cm}$$

    So, radius of the base of cone = AE = 2.4

    Now, in right angled ΔABE,

    $$\text{BE}=\sqrt{\text{AB}^{2}-\text{AE}^{2}}\\=\sqrt{9-(2.4)^{2}}\\=\sqrt{3.24}$$

    = 1.8 cm

    ∴ CE = BC – BE

    = 5 – 1.8 = 3.2 cm

    Volume of the cone ABD

    $$=\frac{1}{3}×\frac{22}{7}×(2.4)^{2}×3.2$$

    $$\bigg(\because\text{V}=\frac{1}{3}\pi r^{2}h\bigg)\\=\frac{405.504}{21}$$

    =19.31 cm3

    ∴ Required volume of double cone

    = 10.86 + 19.31 = 30.17 cm3

    Now, surface area of cone ABD

    $$=\pi rl=\frac{22}{7}×2.4×3\\=\frac{158.4}{7}$$

    =22.63 cm2

    Surface area of cone ACD

    $$=\frac{22}{7}×2.4×4\\=\frac{211.2}{7}\\=30.17\space\text{cm}^{2}$$

    ∴ Required surface area of double cone

    = 22.63 + 30.17 = 52.8 cm2

    3. A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water, how many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

    Sol. Given, internal dimensions of cistern

    = 150 cm × 120 cm × 110 cm

    ∴ Volume of cistern = 150 × 120 × 110 cm3

    = 1980000 cm3

    Volume of water in cistern = 129600 m3

    ∴ Volume of cistern to be filled

    = 1980000 – 129600

    = 1850400 cm3

    Let required number of bricks be n

    Volume of each brick = 22.5 × 7.5 × 6.5 = 1096.875

    Then, water absorbed by n bricks

    $$=n\bigg(\frac{1096.875}{17}\bigg)\text{cm}^{3}$$

    According to question

    $$\text{Now, 1850400}+n\bigg(\frac{1096.875}{17}\bigg)\\=n(1096.875)\\\Rarr\space n×1096.875×\frac{16}{17}\\=1850400 \\\Rarr\space n=\frac{1850400×17}{1096.875×16}$$

    = 1792.4102 = 1792 (approx)

    Hence, 1792 bricks can be put in cistern

    4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

    Sol. Given, area of the valley = 97280 km2

    $$\text{and\space rainfall = 10 cm =}\frac{10}{100×1000}\text{km}\\\therefore\space\text{Volume of rainfall =}\frac{97280×10}{100×1000}$$

    = 9.728 km3 ...(i)

    Also, given length of the river = 1072 km

    $$\text{Breadth of the river = 75 m}\\ =\frac{75}{1000}\text{km}$$

    Height (deep) of the river = 3 m

    $$=\frac{3}{1000}\text{km}\\\therefore\space\text{Volume of one river}=\\1072×\frac{75}{1000}×\frac{3}{1000}$$

    = 0.2412 km3

    So, volume of three rivers = 3 × 0.2412

    = 0.7236 km3 ...(ii)

    From eqs. (i) and (ii), it is clear that total rainfall is not approximately equivalent to normal water of three rivers

    5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

    Sol. Given, oil funnel is a combination of a cylinder and a frustum of a cone.

    Height of cylindrical portion h = 10 cm
    and radius of lower end of frustum,

    $$r_2=\frac{8}{2}=4\space\text{cm}$$

    Diameter of top of the funnel = 18 cm

    $$\therefore\space\text{Radius of top of the funnel r}_1\\=\frac{8}{2}=9\text{cm}$$

    Now, height of the frustum of cone

    = 22 – 10 = 12 cm

    ∴ Slant height of the frustum, l

    $$=\sqrt{12^{2}+(9-4)^{2}}\\\sqrt{144+25}\\=\sqrt{169}$$

    l = 13 cm

    Required area of the tin sheet
    = Curved surface area of cylindrical portion + Curved surface area of the frustum

    = 2πr2h + π(r1 + r2)l

    $$=\frac{22}{7}[2×4×10+(9+4)13]\\=\frac{22}{7}(80+169)\\=\frac{22}{7}×249\\=\frac{5478}{7}\\=782\frac{4}{7}\text{cm}^{2}$$

    6. Derive the formula for the curve surface area and total surface area of the frustum of a cone. Using the symbols as explained.

    Sol. Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base.

    Let r1 and r2 be the radii of the ends of the frustum of the cone.

    13.5(6)opSurface Areas and Volumes

    In ΔABG and ΔADF, DF || BG.

    ∴ ΔABG ~ ΔADF

    $$\frac{\text{DF}}{\text{BG}}=\frac{\text{AF}}{\text{AG}}=\frac{\text{AD}}{\text{AB}}\\\frac{r_2}{r_1}=\frac{h_1-h}{h_1}=\frac{l_1-l}{l_1}\\\Rarr\space\frac{r_2}{r_1}=1-\frac{h}{h_1}\\=1-\frac{l}{l_1}\\\Rarr\space 1-\frac{l}{l_1}=\frac{r_2}{r_1}\\\Rarr\space\frac{l}{l_1}=1-\frac{r_2}{r_1}=\frac{r_1-r_2}{r_1}$$

    $$\Rarr\space\frac{l_1}{l}=\frac{r_2}{r_1-r_2}\\\Rarr\space l_1=\frac{r_1l}{r_1-r_2}$$

    CAS of frustum DECB

    = CSA of cone ABC – CSA of cone ADE

    = πr1l1 – πr2(l1 – l)

    $$=\pi r_1\bigg(\frac{r_1l}{r_1-r_2}\bigg)-\pi r_1\bigg(\frac{r_1l}{r_1-r_2}-l\bigg)\\=\frac{\pi r_1^{2}l}{r_1-r_2}-\pi r_2\bigg(\frac{r_1l-r_1l+r_2l}{r_1-r_2}\bigg)\\=\frac{\pi r_1^{2}l}{r_1-r_2}-\frac{\pi r_2^{2}l}{r_1-r_2}\\=\pi rl\bigg[\frac{r_1^{2}-r_2^{2}}{r_1-r_2}\bigg]\\=\pi(r_1+r_2)l$$

    Total surface area of frustum

    = CSA of frustum + Area of upper circular end + Area of lower circular end

    = π(r1 + r2)l + πr22 + πr12

    = π[(r1 + r2)l + r12 + r22]

    7. Derive the formula for the volume of the frustum of a cone. Using the symbols as explained.

    Sol. Let h be the height, l be the slant height and r1 and r2 be the radii of the bases (r1 > r2) of the frustum of a cone. We complete the conical part OCD.

    The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let the height of the cone OAB be h1 and its slant height be l1.

    i.e., OP = h1 and OA = OB = l1

    Then, height of the cone OCD = h1 – h

    ∵ ΔOQD ~ ΔOPB

    (AA similarity criterion)

    $$\frac{\text{OQ}}{\text{OP}}=\frac{\text{QD}}{\text{PB}}\\\Rarr\space\frac{h_1-h}{h_1}=\frac{r_2}{r_1}\\\Rarr\space 1-\frac{h}{h_1}=\frac{r_2}{r_1}\\\Rarr\space\frac{h}{h_1}\\=1-\frac{r_2}{r_1}$$

    $$=\frac{r_1-r_2}{r_1}\\\Rarr\space h_1=\frac{hr_1}{r_1-r_2}\space\text{...(i)}$$

    Now, height of the cone OCD

    $$h_1=h_1-h=\\\Rarr\frac{hr_1}{r_1-r_2}-h\space\space\text{[From Eq. (i)]}\\=\frac{hr_2}{r_1-r_2}\space\text{..(ii)}$$

    ∴ Volume of the frustum of cone

    = Volume of the cone OAB – Volume of the cone OCD

    $$=\frac{1}{3}\pi r_1^{2}h_1-\frac{1}{3}\pi r_2^{2}(h_1-h)\\=\frac{\pi}{3}\bigg[r_1^{2}\frac{hr_1}{r_1-r_2}-r_2^{2}\frac{hr_2}{r_1-r_2}\bigg]\\\text{[From eq. (i) and (ii)]}\\=\frac{\pi h}{3}\bigg[\frac{r_1^{3}-r_2^{3}}{r_1-r_2}\bigg]\\=\frac{1}{3}\pi h(r_1^{2}+r_2^{2}+r_1r_2).\\\lbrack\text{∵ a}^{3}-\text{b}^{3}=(a-b)(a^{2}+b^{2}+ab)\rbrack\\\text{Therefore, volume of the frustum of cone is}$$

    $$\frac{1}{3}\pi h(r_1^{2}+r_2^{2}+r_1r_2).$$

    If A1 and A2 are the surface areas (A1 > A2) the two circular bases, then

    $$\text{A}_1=\pi r_1^{2}\space\text{and}\space\text{A}_2=\pi r_2^{2}$$

    Then, volume of the frustum of the cone

    $$=\frac{h}{3}[\pi r_1^{2}+\pi r_2^{2}+\sqrt{\pi r_1^{2}}\sqrt{\pi r_2^{2}}]\\=\frac{h}{3}(\text{A}_1+\text{A}_2+\sqrt{\text{A}_1\text{A}_2})$$

    Selected NCERT Exemplar Problems

    Exercise 13.1

    • Choose the correct answer from the given four options.

    1. A cylindrical pencil sharpened at one edge is the combination of

    (a) a cone and a cylinder

    (b) frustum of a cone and a cylinder

    (c) a hemisphere and a cylinder

    (d) two cylinders

    Sol. (a) a cone and a cylinder

    13.1(1)Selected NCERT Exemplar Problems

    2. A surahi is the combination of

    (a) a sphere and a cylinder

    (b) a hemisphere and a cylinder

    (c) two hemispheres

    (d) a cylinder and a cone

    Sol. (a) a sphere and a cylinder

    13.1(2)Selected NCERT Exemplar Problems

    3. A plumbline (sahul) is the combination of (see figure).

    13.1(3)Selected NCERT Exemplar Problems

    (a) a cone and a cylinder

    (b) a hemisphere and a cone

    (c) frustum of a cone and a cylinder

    (d) sphere and cylinder

    Sol. (b) a hemisphere and a cone

    4. The shape of a glass (tumbler) (see figure) is usually in the form of

    (a) a cone

    (b) frustum of a cone

    (c) a cylinder

    (d) a sphere

    Sol. (b) frustum of a cone

    5. The shape of a gilli, in the gilli-danda game (see figure) is a combination of

    13.1(5)Selected NCERT Exemplar Surface Areas and Volumes

    (a) two cylinders

    (b) a cone and a cylinder

    (c) two cones and a cylinder

    (d) two cylinders and a cone

    Sol. (c) two cones and a cylinder

    6. A shutle cock used for playing badminton has the shape of the combination of

    (a) a cylinder and a sphere

    (b) a cylinder and a hemisphere

    (c) a sphere and a cone

    (d) frustum of a cone and a hemisphere

    Sol. (d) frustum of a cone and a hemisphere

    7. A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is calle

    (a) a frustum of a cone

    (b) cone

    (c) cylinder

    (d) sphere

    Sol. (a) a frustum of a cone

    13.1(7)Surface Areas and Volumes

    8. A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is moulded to form a solid sphere. The radius of the sphere is

    (a) 21 cm

    (b) 23 cm

    (c) 25 cm

    (d) 19 cm

    Sol. (a) 21 cm

    Explanation:

    Given, dimensions of the cuboid

    = 49 cm × 33 cm × 24 cm

    ∴ Volume of the cuboid = 49 × 33 × 24

    = 38808 cm3

    Let the radius of the sphere is r, then

    $$\text{Volume of the sphere =}\frac{4}{3}\pi r^{3}$$

    According to the question,

    $$\frac{4}{3}\pi r^{3} = 38808\\\Rarr\space4×\frac{22}{7}×r^{3}=38808×3\\r^{3}=\frac{38808×3×7}{4×22}$$

    = 441 × 21

    ⇒ r3 = 21 × 21 × 21

    ⇒ r = 21 cm

    9. Twelve solid spheres of the same size are made are melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

    (a) 4 cm

    (b) 3 cm

    (c) 2 cm

    (d) 6 cm

    Sol. (c) 2 cm

    Explanation:

    Given, diameter of the cylinder = 2 cm

    ∴ Radius of the cylinder = 1 cm

    and height of the cylinder = 16 cm

    ∴ Volume of the cylinder = π × (1)2 × 16

    = 16 π cm3

    Now, let the radius of solid sphere = r

    $$\text{Then, its volume =}\frac{4}{3}\pi r^{3}\space\text{cm}^{3}$$

    According to the question,

    Volume of the twelve solid sphere

    = Volume of cylinder

    $$\Rarr\space 12 ×\frac{4}{3}\pi r^{3}=16\pi$$

    ⇒ r3 = 1 ⇒ 1 = 1 cm

    ∴ Diameter of each sphere,

    d = 2r = 2 × 1 = 2 cm

    10. The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is

    (a) 4950 cm2

    (b) 4951 cm2

    (c) 4952 cm2

    (d) 4953 cm2

    Sol. (a) 4950 cm2

    Explanation:

    Given, the radius of the top of the bucket, R = 28 cm

    Radius or the bottom of the bucket, r = 7 cm

    and Slant height of the bucket, l = 45 cm

    ∴ Curved surface area of the bucket

    = πl(R + r)

    = π × 45 (28 + 7)

    = π × 45 × 35π

    $$=\frac{22}{7}×45×45$$

    = 4950 cm2

    11. A right circular cylinder of radius r cm and height h cm (h > 2r) just encloses a sphere of diameter

    (a) r cm

    (b) 2r cm

    (c) h cm

    (d) 2h cm

    Sol. (b) 2r cm

    Explanation:

    Because the diameter of the sphere encloses in the cylinder is equal to diameter of cylinder which is 2r cm.

    12. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

    (a) 32.7 L

    (b) 33.7 L

    (c) 34.7 L

    (d) 31.7 L

    Sol. (a) 32.7 L

    Explanation:

    Given, diameter of one end of the bucket,

    2R = 44 ⇒ R = 22 cm

    and diameter of the other end,

    2r = 24 ⇒ r = 12 cm

    Height of the bucket,

    h = 35 cm

    Capacity of the bucket

    =  Volume of the frustum of the cone

    $$=\frac{1}{3}\pi h[\text{R}^{2}+\text{r}^{2}+\text{Rr}]\\=\frac{1}{3}×\pi×35[(22)^{2}+(12)^{2}+22×12]\\=\frac{35\pi}{3}[484+144+264]\\=\frac{35\pi×892}{3}\\=\frac{35×22×892}{3}$$

    = 32706.6 cm3

    = 32.7 L

    (∵ 1000 cm3 = 1L)

    13. In a right circular cone, the cross-section made by a plane parallel to the base is a

    (a) circle

    (b) frustum of a cone

    (c) sphere

    (d) hemisphere

    Sol. (b) frustum of a cone

    We know that if a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the core.

    14. Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface area is

    (a) 3 : 4

    (b) 4 : 3

    (c) 9 : 16

    (d) 16 : 9

    Sol. (d) 16 : 9

    Explanation:

    Let the radii of the two spheres are r1 and r2 respectively.

    Volume of the sphere of radii,

    $$r_1=\frac{r}{3}\pi r_1^{3}\\\therefore\space\text{...(i)}$$

    and Volume of the sphere of radii,

    $$r_2=\frac{4}{3}\pi r_2^{3}\\\text{...(ii)}$$

    Given, ratio of volumes

    $$\lambda=64:27\Rarr\frac{\frac{4}{3}\pi r_1^{3}}{\frac{4}{3}\pi r_2^{3}}=\frac{64}{27}\\\Rarr\space\frac{r_1^{3}}{r_2^{3}}=\frac{64}{27}\\\Rarr\space\frac{r_1}{r_1}=\frac{4}{3}\space\text{...(iii)}$$

    Now,

    ratio of the surface area

    $$=\frac{4\pi r_1^{2}}{4\pi r_2^{2}}=\frac{r_1^{3}}{r_2^{3}}=\frac{r_1^{2}}{\bigg(\frac{3r_1}{4}\bigg)^{2}}\\\bigg[\text{From eq. (iii),}r_2=\frac{3}{4}r_1\bigg]\\=\frac{r_1^{2}}{\frac{9r_1^{2}}{16}}=\frac{16}{9}\space\text{or 16:9}$$

    15. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and $$\frac{1}{8}$$

    $$\text{it is assumed that}\space\frac{1}{8}\space\text{space}\space\text{of the cube remains unfilled.}$$

    Then the number of marbles that the cube can accomodate is

    (a) 142244

    (b) 142344

    (c) 142444

    (d) 142544

    Sol. (a) 142244

    Explanation:

    Given, edge of the cube = 22 cm

    ∴ Volume of the cone = (22)3 = 10648 cm3

    Also, given diameter of marble = 0.5 cm

    ∴ Radius of marble,

    $$r=\frac{0.5}{2}=0.25\space\text{cm}$$

    Volume of one marble =

    $$\frac{4}{3}\pi r^{3}\\=\frac{4}{3}×\frac{22}{7}×(0.25)^{3}\\=\frac{1.375}{21}=0.0655\space\text{cm}^{3}$$

    $$\frac{4}{3}\pi r^{3}\\=\frac{4}{3}×\frac{22}{7}×(0.25)^{3}\\=\frac{1.375}{21}=0.0655\space\text{cm}^{3}\\\text{Filled space of cube = 10648 – 10648}×\frac{1}{8}\\=10648×\frac{7}{8}$$

    = 9317 cm3

    Now,

    Required number of marbles

    $$=\frac{\text{Total space filled by marbles}}{\text{Volume of one marble}}\\=\frac{9317}{0.0655}=142244\text{(approx.)}$$

    16. A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form a cone of base diameter 8 cm. The height of the cone is

    (a) 12 cm

    (b) 14 cm

    (c) 15 cm

    (d) 18 cm

    Sol. (b) 14 cm

    Given, internal diameter of spherical shell = 4 cm

    ∴ Internal radius of spherical shell,

    $$r_1=\frac{4}{2}\space\text{cm}=2\space\text{cm}$$

    and external diameter of shell = 8 cm

    ∴ External radius of shell,

    $$r_2=\frac{8}{2}=4\space\text{cm}$$

    13.1(16)Selected NCERT Surface Areas and Volumes

    Now,

    Volume of the spherical shell

    $$=\frac{4}{3}\pi[r_2^{3}-r_1^{3}]\\=\frac{4}{3}\pi[4^{3}-2^{3}]\\=\frac{4}{3}\pi[64-8]\\=\frac{224}{3}\pi\space\text{cm}^{3}$$

    Let, height of the cone = h cm

    Diameter of the base of cone = 8 cm

    $$\therefore\space\text{Radius of the base of cone =}\frac{8}{2}=4\text{cm}$$

    Now,

    Volume of cone = Volume of spherical shell

    $$\therefore\space\frac{1}{3}\pi(4)^{2}h=\frac{224}{3}\pi\\\Rarr\space h=\frac{224}{16}=14\space\text{cm}$$

    17. A mason constructs a wall of dimensions 270 cm × 300 cm × 250 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that

    $$\frac{1}{8}\space\text{space}$$

    is covered by the mortar.

    Then, the number of bricks used to construct the wall is

    (a) 11100

    (b) 11200

    (c) 11000

    (d) 11300

    Sol. (b) 11200

    Explanation:

    Volume of the wall = 270 × 300 × 350

    = 2,83,50,000 cm3

    $$\frac{1}{8}\space\text{space of wall is covered by mortar}$$

    So,

    remain space of wall = 28350000 – 28350000  × $$\frac{1}{8}$$

    = 28350000 – 3543750

    = 24806250 cm3

    Now,

    volume of one brick = 22.5 × 11.25 × 8.75

    = 2214.844 cm3

    ∴ Required number of bricks

    $$=\frac{24806250}{2214.844}$$

    = 11200 (approx.)

    18. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck of each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

    (a) 0.36 cm3

    (b) 0.35 cm3

    (c) 0.34 cm3

    (d) 0.33 cm3

    Sol. (a) 0.36 cm3

    Explanation:

    Given diameter of cylinder = diameter of hemisphere = 0.5 cm

    ∴ Radius of cylinder (r) = radius of hemisphere

    $$(r)=\frac{0.5}{2}=0.25\space\text{cm}$$

    and total length of capsule = 2 cm

    ∴ Length of cylindrical part of capsule,

    h = 2 – (0.25 + 0.25) = 1.5 cm

    Now, capacity of capsule

    = Volume of cylindrical part + 2 × Volume of hemisphere

    $$=\pi r^{2}h+2×\frac{2}{3}\pi r^{3}\\=\frac{22}{7}\bigg[(0.25)^{2}×1.5+\frac{4}{3}×(0.25)^{3}\bigg]\\=\frac{22}{7}[0.09375+0.0208]\\=\frac{22}{7}×0.11455$$

    = 0.36 cm3

    19. During conversion of a solid from one shape of another, the volume of the new shape will

    (a) increase

    (b) decrease

    (c) remain unaltered

    (d) be doubled

    Sol. (c) remain unaltered

    Explanation:

    During conversion of a solid from one shape to another, the volume of the shape will remain unaltered.

    20. If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

    (a) 4πr2

    (b) 6πr2

    (c) 3πr2

    (d) 8πr2

    Sol. (a) 4πr2

    Explanation:

    Because curved surface area of a hemisphere is 2πr2. Here, we join two solid hemispheres of same base radius r and for curved surface area we not consider the area of base.

    Exercise 13.2

    • State whether the following statements are true or false. Justify your answer.

    1. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination of 6πr2.

    Sol. False.

    Total surface area of hemisphere

    = Curved surface area + Area of base

    = 2πr2 + πr2

    Here, two identical solid hemisphere of equal radius are stuck together. So, base of both hemisphere is common.

    ∴ Total curved surface area of the combination

    = 2πr2 + πr2 + 2πr2 = 5πr2

    2. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4πrh + 4πr2.

    Sol. False.

    Since, the total surface area of cylinder of radius r and height h = 2πrh + 2πr2.

    When one cylinder is placed over the other cylinder of same height and radius, then height of the new cylinder = 2h

    and radius of the new cylinder = r

    ∴ Total curved surface area of the new cylinder = 2πr(2h) + 2πr2 = 4πrh + 2πr2.

    3. A solid ball is exactly fitted inside the cubical box of side the volume of the ball is

    $$\frac{4}{3}\pi a^{3}.$$

    Sol. False.
    Because solid ball is exactly fitted inside the cubical box of side a. So, a be the diameter for the solid ball.

    $$\therefore\space\text{Radius of the ball =}\frac{a}{2}.$$

    $$\therefore\space\text{Radius of the ball =}\frac{a}{2}.\\\text{So, volume of the ball}\\=\frac{4}{3}\pi\bigg(\frac{a}{2}\bigg)^{3}=\frac{1}{6}\pi a^{3}.$$

    4. The volume of the frustum of a cone is$$\frac{\textbf{1}}{\textbf{3}}\pi \textbf{h[r}_\textbf{1}^{\textbf{2}}+\textbf{r}_\textbf{2}^{\textbf{2}}-\textbf{r}_\textbf{1}\textbf{r}_\textbf{2}]$$, where, h is vertical height of the frustum and r1, r2 are the radii of the ends.

    Sol. False.

    Since, the volume of the frustum of a cone is

    $$\frac{1}{3}\pi h[r_1^{2}+r_2^{2}-r_1r_2],$$

    where h is vertical height of the frustum and r1, r2 are the radii of the ends.

    5. The curved surface area of a frustum of a cone is

    $$\pi l(r_1+r_2,\text{where I = }\sqrt{h^{2}+(r_1-r_2)^{2}},$$

    r1 and r2 are the radii of the two ends of the frustum and h is the vertical height.

    Sol. True.

    6. An open metallic bucket is in the shape of frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to the curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.

    Sol. True.

    Because the resulting figure is

    Here, ABCD is a frustum of a cone and CDEF is a hollow cylinder.

    7. A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone.

    The total surface area of the combined solid is

    $$\pi\text{rl}\sqrt{r^{2}+h^{2}}+2r+2h].$$

    Sol. False.

    We know that total curved surface area of a cone of radius r and height h = curved surface area + area of base = πrl + πr2.

    $$\text{where, l=}\sqrt{h^{2}+r^{2}}$$

    and total curved surface area of a cylinder of base radius r and height h = curved surface area + area of both base = 2πrh + 2πr2.

    Here, when we placed a cone over a cylinder one base is common for both

    So, total curved surface area of the combined solid

    = πrl + 2πrh + 2πr2

    = πr[l + 2h + 2r]

    $$=\pi r[\sqrt{r^{2}+h^{2}}+2h+2r]$$

    8. The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the figure is

    $$\frac{\pi r^{2}}{3}[3h-2r].$$

    Sol. True. We know that capacity of cylindrical vessel = πr2h cm3

    and capacity of hemisphere

    $$=\frac{2}{3}\pi r^{3}\text{cm}$$

    From the figure, capacity of the cylindrical vessel

    $$=\pi r^{2}h-\frac{2}{3}\pi r^{3}\\=\frac{1}{3}\pi r^{2}[3h-2r]$$

    Exercise 13.3

    1. Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed.

    Sol. Given, edges of three solid cubes are 3 cm, 4 cm and 5 cm respectively. So,

    Volume of first cube = (3)3 = 27 cm3

    Volume of second cube = (4)3 = 64 cm3

    and Volume of third cube = (5)3 = 125 cm3

    ∴ Sum of volume of three cubes

    = (27 + 64 + 125) cm3

    = 216 cm2

    Let the edge of the resulting cube = a cm.

    Then volume of the resulting cube

    a3 = 216

    ⇒ a = 6 cm

    2. A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm, respectively. Find the height of the bucket.

    Sol. Given, volume of the frustum

    = 28.490 L

    = 28.49 × 1000 cm3 (1L = 1000 cm3)

    and radius of the top (r1) = 28 cm

    radius of the bottom (r2) = 21 cm

    Let height of the bucket = h cm

    Now, volume of the bucket

    $$=\frac{1}{3}\pi h(r_1^{2}+r_2^{2}+r_1r_2)$$

    = 28490   (Given)

    $$\Rarr\frac{1}{3}×\frac{22}{7}×h(28^{2}+21^{2}+28×21)\\=28490\\\Rarr\space h(784+441+588)\\=\frac{28490×3×7}{22}$$

    ⇒ 1813 h = 1295 × 21

    $$\Rarr\space h=\frac{1295×21}{1813}\\\Rarr\space\frac{27195}{1813}=15\space\text{cm}$$

    3. A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

    Sol. Let ORN be the cone then given, radius of the base of the cone, r1 = 8 cm

    and height of the cone, h

    OM = 12 cm

    Let P be the mid-point of OM, then

    $$\text{OP = PM =}\frac{12}{2}=6\text{cm}$$

    Now, ΔOPD ~ ΔOMN

    $$\therefore\space\frac{\text{OP}}{\text{OM}}=\frac{\text{PD}}{\text{MN}}\\\Rarr\space\frac{6}{12}=\frac{\text{PD}}{8}$$

    $$\therefore\space\frac{\text{OP}}{\text{OM}}=\frac{\text{PD}}{\text{MN}}\\\Rarr\space\frac{6}{12}=\frac{\text{PD}}{8}\\=\frac{1}{2}=\frac{\text{PD}}{8}\\\Rarr\text{PD}=4\space\text{cm}$$

    The plane CD divides the cone into two parts, namely

    (i) a smaller cone of radius 4 cm and height 6 cm and (ii) frustum of a cone for which

    Radius of the top of the frustum, r1 = 8 cm

    Radius of the bottom, r2 = 4 cm

    and height of the frustum, h = 6 cm

    ∴ Volume of smaller cone

    $$=\bigg(\frac{1}{3}\pi×4×4×6\bigg)\space\text{cm}^{3}$$

    = 32π cm3

    and volume of the frustum of cone

    $$=\frac{1}{3}×\pi×6(8)^{2}+(4)^{2}+8×4]\space\text{cm}^{3}$$

    = 2π (64 + 16 + 32) = 224 π cm3

    ∴ Required ratio = volume of cone : volume of frustum = 32 π : 224 π = 1 : 7.

    4. Two identical cubes each of volume 64 cm3 are joined together end to end. What is the surface area of the resulting cuboid?

    Sol. Let the length of a side of a cube = a cm

    13.3(4)_SELECTNECERTSurface Areas and Volumes

    Given, volume of the cube,

    a3 = 64 cm3 ⇒ a = 4 cm

    On joining two cubes, we get a cuboid whose

    length, l = 2a cm

    breadth, b = a cm

    and height, h = a cm

    Now, surface area of the resulting cuboid

    = 2(lb + bh + hl)

    = 2(2a.a + a.a + a.2a)

    = 2(2a2 + a2 + 2a2)

    = 2(5a2)

    = 10a2 = 10(4)2

    = 160 cm2

    5. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

    Sol. Given, diameter of a marble = 1.4 cm

    $$\therefore\space\text{Radius of marble =}\frac{1.4}{2}=0.7\space\text{cm}$$

    So, volume of one marble = volume of sphere

    $$=\frac{4}{3}\pi(0.7)^{3}\\=\frac{4}{3}×\pi×0.343\\=\frac{1.372}{3}\pi\text{cm}^{3}$$

    Also, given diameter of beakter = 7 cm

    $$\therefore\space\text{Radius of beaker =}\frac{7}{2}=3.5\space\text{cm}$$

    Height of water level raised = 5.6 cm

    ∴ Volume of the raised water in beaker

    = Volume of cylinder

    = π(3.5)2 × 5.6

    = 68.6 π cm3

    Now, required number of marbles

    $$=\frac{\text{Volume of the raised water in beaker}}{\text{Volume of one sphericall marble}}\\=\frac{68.6\pi}{1.372\pi}×3$$

    = 150 marbles

    6. How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm2.

    Sol. Given, dimension of cuboidal solid = 9 cm × 11 cm × 12 cm

    ∴ Volume of cuboidal solid = 9 × 11 × 12

    = 1188 cm3

    Diameteter of lead shot = 3 cm

    $$\therefore\space\text{Radius of shot, r =}\frac{3}{2}×1.5\space\text{cm}\\\text{Volume of shot =}\frac{4}{3}\pi r^{3}\\=\frac{4}{3}×\frac{22}{7}×(1.5)^{3}\\=\frac{297}{21}=14.143\space\text{cm}^{3}\\\therefore\space\text{Required number of shots}\\=\frac{1188}{14.143}=84\space\text{(approx)}$$

    7. Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities are 2 : 1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

    13.3(7)_SELECTNECERTSurface Areas and Volumes

    Sol. Let volume of cone A is 2V and of cone B is V. Again, let height of the cone A = h1 cm, then height of cone B = (21 – h1) cm.

    Given, diameter of the cone = 6 cm

    $$\therefore\space\text{Radius of the cone =}\frac{6}{2}=3\text{cm}$$

    Now, volume of the cone, A = 2V

    $$=\frac{1}{3}\pi r^{2}h\\=\frac{1}{3}\pi(3)^{2}h_1\\\Rarr\space\text{V}=\frac{1}{6}\pi×9h_1\\=\frac{3}{2}h_1\pi\space\text{...(i)}$$

    and volume of the cone, B

    $$=\text{V}=\frac{1}{3}\pi(3)^{2}-(21-h_1)$$

    = 3π(21 – h1) ...(ii)

    From eqs. (i) and (ii), we get

    $$\frac{3}{2}h_1\pi=3\pi(21-h_1)$$

    ⇒ h1 = 2(21 – h1)

    ⇒ 3h1 = 42

    $$\Rarr\space h_1=\frac{42}{3}=14\space\text{cm}$$

    ∴ Height of cone, B = 21 – h1 = 21 – 14

    = 7 cm

    Now, volume of the cone, A

    $$=\frac{1}{3}×\frac{22}{7}×3×3×14\\=3×14×\frac{22}{7}$$

    = 132 cm3

    and Volume of the cone B,

    $$=\frac{1}{3}×\frac{22}{7}×3×3×7$$

    = 66 cm3

    Now, volume of the cylinder

    = πr2h

    $$=\frac{22}{7}×(3)^{2}×21$$

    = 594 cm3

    ∴ Required volume of the remaining portion

    = Volume of the cylinder
    – (Volume of cone A + Volume of cone B)

    = 594 – (132 + 66)

    = 396 cm3

    8. An ice cream cone full of ice cream having radius 5 cm and height 10 cm as shown in figure.

    Calculate the volume of ice cream, provided that its $$\frac{\textbf{1}}{\textbf{6}}$$ part is left unfilled with ice-cream.

    13.3(8)_sel ncert Surface Areas and Volumes

    Sol. Given, ice cream cone is the combination of a hemisphere and a cone.

    Also given,

    Radius of hemisphere = 5 cm

    $$\therefore\space\text{Volume of hemisphere =}\frac{2}{3}\pi r^{3}\\=\frac{2}{3}×\frac{22}{7}×(5)^{3}\\=\frac{5500}{21}$$

    = 261.90 cm3

    Now, radius of the cone = 5 cm

    and height of the cone = 10 – 5 = 5 cm

    $$\therefore\space\text{Volume of the cone =}\frac{1}{3}\pi r^{2}h\\=\frac{1}{3}×\frac{22}{7}×(5)^{2}×5\\=\frac{2750}{21}$$

    = 130.95 cm3

    Now, total volume of ice cream cone

    = 261.90 + 130.95

    = 392.85 cm3

    $$\text{Since}\space\frac{1}{6}\space\text{part is left unfilled with ice cream.}\\\therefore\space\text{Required volume of ice cream}\\=392.85 – 392.85 ×\frac{1}{6}$$

    = 392.85 – 65.475

    = 327.4 cm3

    Exercise 13.4

    1. A solid metallic hemisphere of radius 8 cm is melted and recasted into a right circular cone of base radius 6 cm. Determine the height of the cone.

    Sol. Let height of the cone be h.

    Given, radius of the base of the cone = 6 cm

    ∴ Volume of circular cone

    $$=\frac{1}{3}\pi r^{2}h\\=\frac{1}{3}\pi(6)^{2}h\\=\frac{36\pi h}{3}$$

    = 12πh cm3

    Also, given radius of the hemisphere = 8 cm

    $$\therefore\space\text{Volume of the hemisphere}\\=\frac{2}{3}\pi r^{3}\\=\frac{2}{3}\pi(8)^{3}\\=\frac{512×2\pi}{3}\text{cm}^{3}$$

    According to the question,

    Volume of the cone = Volume of the hemisphere

    $$\Rarr\space12\pi h=\frac{512×2\pi}{3}\\\Rarr\space h=\frac{512×2\pi}{12×3\pi}\\=\frac{256}{9}\text{cm}$$

    = 28.44 cm

    2. A rectangular water tank of base 11 m × 6 m contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

    Sol. Given, dimension of base of rectangular tank = 11 m × 6 m

    Height of water in the tank = 5 m

    Volume of the water in rectangular tank
    = 11 × 6 × 5 = 330 m3

    Also, given radius of the cylindrical tank
    = 3.5 m

    Let height of water level in cylindrical tank is h.

    Then, volume of the water in cylindrical tank

    = πr2h = π(3.5)2 × h

    $$=\frac{22}{7}×3.5×3.5×h$$

    = 11.0 × 3.5 × h

    = 38.5 h m2

    According to the question,

    Volume of water is same in both tank.

    ∴ 330 = 38.5 h

    $$\Rarr\space h=\frac{330}{38.5}=\frac{3300}{385}$$

    ⇒ h = 8.57 m or 8.6 m

    Hence, height of the water level in cylindrical tank = 8.6 m.

    3. Water is flowing at the rate of 15 kmh–1 through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm ?

    Sol. Given, length of the tank = 50 m and width of the tank = 44 m

    13.4(3)_Select ncertSurface Areas and Volumes

    $$\text{Depth required of water = 21 cm} =\frac{21}{100}\text{m}\\\therefore \text{Volume of water in the tank}\\=\bigg(50×44×\frac{21}{100}\bigg)m^{3}$$

    = 462 m3

    Also, given radius of the pipe = 7 cm

    $$=\frac{7}{100}\text{m}$$

    and speed of water flowing through the pipe

    = (15 × 1000) mh–1

    = 15000 mh–1

    Now, volume of water flow in 1 h = πR2H

    $$=\bigg(\frac{22}{7}×\frac{7}{100}×\frac{7}{100}×15000\bigg)\text{m}^{3}$$

    = 231 m3

    ∵  231 m3 of water falls in the tank in 1 h.

    $$\therefore\space\text{1 m}^3\space\text{water falls in the tank in}\space\frac{1}{231}\space\text{hr.}$$

    ∴ 462 m3 of water falls in the tank in

    $$\bigg(\frac{1}{231}×462\bigg)\text{h = 2hr}$$

    Hence, the required time = 2 hr.

    4. A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

    Sol. Given, radius of the base of the bucket = 18 cm

    Height of the bucket = 32 cm

    So, volume of the sand in the bucket

    = πr2h = π(18)2 × 32

    = 10368 π

    Also, given height of the conical heap (h) = 24 cm

    Let radius of heap be r cm.

    Then, volume of the sand in the heap

    $$=\frac{1}{3}\pi r^{2}h\\=\frac{1}{3}\pi r^{2}×24=8\pi r^{2}$$

    According to the question,

    Volume of the sand in cylindrical bucket

    = Volume of the sand in conical heap

    ⇒ 10368π = 8πr2

    ⇒ 10368 = 8r2

    $$\Rarr\space r^{2}=\frac{10368}{8}=1296 $$

    ⇒ r = 36 cm

    Again, let the slant height of the conical heap = l

    Now, l2 = h2 + r2

    = (24)2 + (36)2

    = 576 + 1296 = 1872

    l = 43.267 cm

    Hence, radius of conical heap of sand = 36 cm

    and slant height of conical heap = 43.267 cm

    5. A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains

    $$41\frac{19}{21}\text{m}^{3}$$

    of air. If the internal diameter of dome is equal to its total height above the floor, find the height of building?

    Sol. Let, the total height of the building be ‘H’ m.

    Let, the radius of the base be ‘r’ m.

    Therefore, the radius of the hemispherical dome is ‘r’ m.

    13.4(5)_SELECTNCERTSurface Areas and Volumes

    Now, given that internal diameter = Total height

    ⇒ 2r = H

    Total height of the building = Height of the cylinder + Radius of the dome

    ⇒ H = h + r

    ⇒ 2r = h + r

    ⇒ r = h

    Volume of air inside the building = Volume of the cylinder + Volume of the hemisphere

    $$\Rarr\space 41\frac{19}{21}=\pi r^{2}h+\frac{2}{3}\pi r^{3}\\\Rarr\space\frac{880}{21}=\pi h^{2}h+\frac{2}{3}\pi h^{3}\\\Rarr\space\frac{880}{21}=\pi h^{3}\bigg(1+\frac{2}{3}\bigg)=\pi h^{3}\bigg(\frac{5}{3}\bigg)$$

    ⇒ h = 2 m

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

    6. A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl?

    Sol. Given, radius of hemispherical bowl, r = 9 cm

    and radius of cylindrical bottles, R = 1.5 cm and height, h = 4 cm

    ∴ Number of required cylindrical bottles

    $$=\frac{\text{Volume of hemispherical bowl}}{\text{Volume of one cylindrical bottlle}}\\=\frac{\frac{2}{3}\pi r^{3}}{\pi R^{2}h}\\=\frac{\frac{2}{3}×\pi×9×9×9}{\pi×1.5×1.5×4}$$

    = 54 bottles

    7. Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cms–1 in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?

    Sol. Given, radius of tank, r1 = 40 cm
    Let height of water level in tank = h1
    Also, given internal radius of circular pipe, r2 =1 cm
    and length of water flow in 1 s = 80 cm

    ∴ Length of water flow in 30 min (h2)

    = 80 × 60 × 30

    = 144000 cm

    According to question

    Volume of water in cylindrical tank
    = Volume of water flow from the circular pipe in half an hour

    ∴ πr12h1 = πr22h2

    ⇒ 40 × 40 × h1 = 1 × 1 × 144000

    $$\Rarr\space h_1=\frac{144000}{40×40}=90\space\text{cm}$$

    Hence, level of water of cylindrical tank rises by 90 cm in half an hour.

    8. The rain water from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

    Sol. Given, length and breadth of roof are 22 m and 20 m respectively

    Let the rainfall be a cm.

    ∴ Volume of water on the roof

    $$=22×20×\frac{a}{100}\\=\frac{22a}{5}\text{m}^{3}$$

    Also, we have radius of base of the cylindrical vessel = 1 m

    and height of the cylindrical vessel = 3.5 m

    ∴ Volume of water in the cylindrical vessel when it is just full

    $$=\bigg(\frac{22}{7}×1×1×\frac{7}{2}\bigg)\space m^{3}$$

    = 11 m3 (V = πR2h)

    Now, volume of water on the roof

    = Volume of water in the vessel

    $$\Rarr\space\frac{22a}{5}=11\\\Rarr\space a=\frac{11×5}{22}=2.5$$

    Hence, the rainfall is 2.5 cm.

    9. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pin is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth of a litre?

    Sol. Given, length of the barrel of a fountain pen = 7 cm

    Diameter = 5 mm

    $$=\frac{5}{10}\space\text{cm}=\frac{1}{2}\space\text{m}$$

    $$\therefore\space\text{Radius of the barrel =}\frac{1}{2×2}=0.25\space\text{cm}$$

    Volume of the barrel = πr2h

    (∵ It is cylindrical in shape)

    $$=\frac{22}{7}\pi(0.25)^{2}×7$$

    = 22 × 0.0625

    = 1.375 cm3

    Also, given volume of ink in the bottle

    $$=\frac{1}{5}\space\text{of litre}\\=\frac{1}{5}×1000\space\text{cm}^{3}$$

    = 200 cm3

    Now, 1.375 cm3 ink is used for writing number of words = 3300

    ∴ 1 cm3 ink is used for writing number of words

    $$=\frac{3300}{1.375}=2400$$

    ∴ 200 cm3 ink is used for writing number of words

    $$=\frac{3300}{1.375}×200$$

    = 2400 × 200

    = 480000 words

    10. Water flows at the rate of 10 m min–1 through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

    Sol. Diameter of the pipe = 5 mm

    $$=\frac{5}{10}\space\text{cm}=\frac{1}{2}\space\text{cm}\\\therefore\space\text{Radius of the pipe}\\=\frac{1}{2}×\frac{1}{2}\text{cm}=\frac{1}{4}\text{cm}$$

    In 1 minute, the length of the water column in the cylindrical pipe = 10 m = 1000 cm

    ∴ Volume of water that flows out of pipe in 1 minute

    $$=\pi×\frac{1}{4}×\frac{1}{4}×1000\space\text{cm}^{3}$$

    Also, volume of the cone

    $$=\bigg(\frac{1}{3}×\pi×20×20×24\bigg)\space\text{cm}^{3}$$

    Hence, the time needed to fill up this conical vessel

    $$=\bigg(\frac{20×20×24}{3}×\frac{4×4}{1000}\bigg)\\=\bigg(\frac{4×24×16}{30}\bigg)\space\text{minutes}\\=\frac{256}{5}\text{minutes}$$

    = 51.2 minutes

    Hence, the required time is 51.2 minutes.

    11. A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at ₹ 0.05 per dm2.

    Sol. Given, pencils are cylindrical in shape.

    Length of one pencil = 25 cm

    and circumference of base = 1.5 cm

    ∴ 2pr = 1.5, where r is the radius of the base.

    $$\Rarr\space r=\frac{1.5×7}{22×2}=0.2386\space\text{cm}$$

    Now, curved surface area of one pencil = 2πrh

    $$= 2×\frac{22}{7}×0.2386×25\\=\frac{262.46}{7}$$

    = 37.49 cm2

    $$=\frac{37.49}{100}\space\text{dm}^{2}\space\bigg(\because1 \text{cm}=\frac{1}{10}\text{dm}\bigg)$$

    = 0.375 dm2

    ∴ Curved surface area of 120000 pencils = 0.375 × 120000 = 45000 dm2

    Now, cost of colouring 1 dm2 curved surface of the pencils manufactured in one day = ₹ 0.05 per dm2

    ∴ Cost of colouring 45000 dm2 curved surface = ₹ 2250.

    12. 16 glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions
    16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of water filled in the box.

    Sol. Given, dimensions of the cuboidal = 16 cm × 8 cm × 8 cm

    ∴ Volume of the cuboidal = 16 × 8 × 8

    = 1024 cm3

    Also, given radius of one glass sphere = 2 cm

    $$\therefore\space\text{Volume of one glass sphere =}\frac{4}{3}\pi r^{3}\\=\frac{4}{3}×\frac{22}{7}×(2)^{3}\\=\frac{704}{21}$$

    = 33.523 cm3

    Now, volume of 16 glass sphere

    = 16 × 33.523

    = 536.37 cm3

    ∴ Required volume of water

    = Volume of cuboidal – Volume of 16 glass spheres

    = 1024 – 536.37 = 487.6 cm3

    13. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket. (Use π = 3.14)

    Sol. Given, rocket is the combination of a right circular cylinder and a cone.

    Given, diameter of the cylinder = 6 cm

    $$\therefore\space\text{Radius of the cylinder =}\frac{6}{2}=3\space\text{cm}$$

    and height of the cylinder = 12 cm

    ∴ Volume of the cylinder = πr2h

    = 3.14 × (3)2 × 12

    = 399.12 cm3

    and curved surface area of cylinder

    = 2πrl

    = 2 × 3.14 × 3 × 12

    = 226.08

    Now, in right angled ΔAOC,

    $$h=\sqrt{5^{2}-3^{2}}\\=\sqrt{25-9}\\=\sqrt{16}=4$$

    ∴ Height of the cone, h = 4 cm

    Radius of the cone, r = 3 cm

    $$\therefore\space\text{Volume of the cone =}\frac{1}{3}\pi r^{2}h\\=\frac{1}{3}×3.14×(3)^{2}×4\\=\frac{113.04}{3}$$

    = 37.68 cm3

    and total surface area = πr(l + r)

    = 3.14 × 3(5 + 3)

    = 75.36

    Total volume of the rocket

    = 339.12 + 37.68

    = 376.8 cm3

    and total surface area of the rocket

    = 226.08 + 75.36

    = 301.44 cm2

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