# NCERT solution for class 10 maths surface area and volume ( Chapter 13)

#### Solution for Exercise - 13.1

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

Let the length of each edge of the cube be a cm.

Then, Volume $$= \ 64$$cm3

$$=> \ a^3 \ = \ 64 \ => \ a \ = \ 4$$ cm

When two cubes of equal volumes (i.e., equal edges are joined end to end, we get a cuboid such that its.

l = Length $$= \ 4 \ + \ 4 \ = \ 8$$ cm

b = Breadth = 4 cm

and h = Height = 4 cm

∴ Surface area of the cuboid

2(lb + bh + hl)

$$= \ 2(8 \ × \ 4 \ + \ 4 \ × \ 4 \ + \ 4 \ × \ 8)$$

$$= 2(32 \ + \ 16 \ + \ 32)$$

$$= (2 \ × \ 80)$$

$$= \ 160$$ cm2

Q.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.

Here r, the radius of hemisphere = 7 cm ,

h , the height of cylinder = (13 - 7) = 6 cm

Clearly, radius of the base of cylindrical part is also r cm

Surface area of the vessel = Curved surface area of the cylindrical part + Curved surface area of hemispherical part

$$= \ 2 \pi rh \ + \ 2 \pi r^2 \ = \ 2 \pi r(h \ + \ r)$$

$$= \ 2 \ × \ \frac{22}{7} \ × \ 7 \ × \ 13 \ = \ 572$$ cm2

Q.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 toy.

We have VO = 15.5 cm, OA = OO' = 3.5 cm

Let r be the radius of the base of cone and h be the height of conical part of the by.

Then r = OA = 3.5 cm

h = VO = VO' - OO = (15.5 - 3.5) cm = 12 cm

Also radius of the hemisphere = OA = r = 3.5 cm

Total surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

$$= \ \pi rl \ + \ 2 \pi r^2$$

where l = Slant height $$= \ \sqrt{OA^2 \ + \ OV^2} \ = \ \sqrt{(3.5)^2 \ + \ 12^2} \ = \ \sqrt{156.25} \ = \ 12.5$$cm

$$= \ \pi r(l \ + \ 2r) \ = \ \frac{22}{7} \ × \ 3.5 \ × \ 12.5 \ + \ 2 \ × \ 3.5$$

$$= \ \frac{22}{7} \ × \ 3.5 \ × \ 19.5 \ = \ 214.5$$ cm2

Q.4 A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter of the hemisphere can have? Find the surface area of the solid?

The greatest diameter that a hemisphere can have = 7 cm .

Surface area of the solid after surmounted hemisphere

$$= \ 6 l^2 \ - \ \pi R^2 \ + \ 2 \pi R^2$$

$$= \ 6l^2 \ + \ \pi R^2$$

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

$$= \ 6 \ × \ 49 \ + \ 11 \ × \ \frac{7}{2}$$

$$= \ 294 \ + \ 38.5 \ = \ 332.5$$ cm2

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

Edge of the cube = l

Diameter of the hemisphere = l

∴ Radius of the hemisphere $$= \frac{l}{2}$$

∴ Area of the remaining solid after cutting out the hemispherical depression.

$$= \ 6 l^2 \ - \ \pi ( \frac{l}{2} )^2 \ + \ 2\pi (\frac{l}{2})^2 \ = \ 6l^2 \ + \ \pi (\frac{l}{2})^2$$

$$= \ 6 l^2 \ + \ \pi \ × \ \frac{l^2}{4} \ = \ \frac{l^2}{4} (24 \ + \ \pi)$$

Q.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.

Let r cm be the radius and h cm be the height of the cylinder. Then.

$$r \ = \ \frac{5}{2} \ = \ 2.5$$mm

and $$h \ = \ (14 \ - \ 2 \ × \ \frac{5}{2} \ = 14 \ - \ 5 \ = \ 9$$mm

Also the radius of hemisphere $$= \ \frac{5}{2}$$ cm

Now, surface area of the capsule = Curved surface of cylinder + Surface area of two hemispheres

$$= \ 2 \pi rh \ + \ 2 \ × \ 2\pi r^2 \ = \ 2\pi r(h \ + \ 2r)$$

$$= \ 2 \ × \ \frac{22}{7} \ × \ \frac{5}{2} \ × \ (9 \ + \ 2 \ × \ \frac{5}{2}) \ = \ 220$$ mm2

Q.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 Rs 500 per m2 . (Note that the base of the tent will not covered with canvas).

We have, Total canvas used = Curved surface area of cylinder + Curved surface area of cone

$$= \ 2 \pi rh \ + \ \pi rl \ = \ \pi r(2h \ + \ l)$$

$$= \ \frac{22}{7} \ × \ 2 \ × \ (2 \ × \ 2.1 \ + \ 2.8) \ = \ 44$$ m2

Now, cost of 1 m2 the canvas for the tent = Rs 500

So, cost of 44 m2 the canvas for the tent = Rs 44 × 500 = Rs 22000

Q.8 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 total surface area of the remaining solid to the nearest cm2.

Radius of the cylinder $$= \ \frac{1.4}{2} \ = \ 7$$cm

Height of the cylinder $$= \ 2.4$$ cm

Radius of the cone $$= \ 0.7$$ cm

Height of the cone $$= \ 2.4$$ cm

Slant height of the cone $$= \ \sqrt{(0.7)^2 \ + \ (2.4)^2} \ = \ \sqrt{0.49 \ + \ 5.76}$$

$$= \ \sqrt{6.25} \ = \ 2.5$$ cm

Surface area of the remaining solid = Curved surface area of cylinder + Curved surface of the cone + Area of upper circular base of cylinder

$$= \ 2 \pi rh \ + \ \pi rl \ + \ \pi r^2 \ = \ \pi r(2h \ + \ l \ + \ r)$$

$$= \ \frac{22}{7} \ × \ 0.7 \ × \ (2 \ × \ 2.4 \ + \ 2.5 \ + \ 0.7)$$

$$= \ 22 \ × \ 0.1 \ × \ (4.8 \ + \ 2.5 \ + \ 0.7)$$

$$= \ 2.2 \ × \ 8.0 \ = \ 17.6 \ = \ 18$$ cm2

Q.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 m, find the total surface area of the article.

Surface area of the article when it is ready = Curved surface area of cylinder + 2 × Curved surface area of hemisphere.

$$= \ 2 \pi rh \ + \ 2 \ × \ 2 \pi r^2 \ = \ 2 \pi r(h \ + \ 2r)$$

where r = 3.5 m and h = 10 cm

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

$$= \ 2 \ × \ \frac{22}{7} \ × \ 3.5 \ × \ 17 \ = \ 374$$ cm2

#### Solution for Exercise - 13.2

Q.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 $$\pi$$.

Volume of the solid = Volume of the cone + Volume of the hemisphere

$$= \ \frac{1}{3} \pi r^2 h \ + \ \frac{2}{3} \pi R^3$$

$$= \ \frac{1}{3} \pi r^2 \ × \ r \ + \ \frac{2}{3} \pi r^3$$        [∵ h = r and R = r]

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

$$= \ \pi(1)^3 \ = \ \pi$$ cm3        [∵ r = 1 cm]

Q.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 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 the Rachel made. (Assume the outer and inner dimensions of the model be nearly the same).

Volume of the air contained in the model = Volume of the cylindrical portion of the model + Volume of its two conical ends.

$$= \ \pi r^2 h_1 \ + \ 2 \ × \ \frac{1}{3} \pi r^2 h_2 \ = \ \pi r^2(h_1 \ + \ \frac{2}{3} h_2)$$

where $$r \ = \ \frac{3}{2}$$ cm , $$h_1 \ = \ 8$$ cm and $$h_2 \ = \ 2$$ cm

$$= \ \frac{22}{7} \ × \ ( \frac{3}{2})^2 \ × \ (8 \ + \ \frac{2}{3} \ × \ 2)$$ $$= \ \frac{22}{7} \ × \ \frac{9}{4} \ × \ \frac{24 \ + \ 4}{3}$$

$$= \ \frac{22}{7} \ × \ \frac{9}{4} \ × \ \frac{28}{3}$$ $$= \ 66$$ cm3

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

Volume of the gulab jamun = Volume of the cylindrical portion + Volume of the hemispherical ends

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

where r = 1.4 cm, h = 2.2 cm

$$= \ \frac{22}{7} \ × \ (1.4)^2 \ × \ (2.2 \ + \ \frac{4}{3} \ × \ 1.4) \ = \ \frac{22}{7} \ × \ 1.96 \ × \ ( \frac{6.6 \ + \ 5.6}{3} )$$

$$= \ \frac{22}{7} \ × \ 1.96 \ × \ \frac{12.2}{3}$$ cm3

Volume of 45 gulab jamuns

$$= \ 45 \ × \ \frac{22}{7} \ × \ 1.96 \ × \ \frac{12.2}{3}$$ cm3

Quantity of syrup in gulab jamuns = 30% of their volume

$$= \ \frac{30}{100} \ × \ 45 \ × \ \frac{22}{7} \ × \ 1.96 \ × \ \frac{12.2}{3}$$

$$= \ \frac{9 \ × \ 11 \ × \ 1.96 \ × \ 12.2}{7} \ = \ 338.184$$

$$= \ 338$$ cm3

Q.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 m and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).

Volume of wood in the entire stand = Volume of the cuboid - 4 × Volume of a depression (i.e., cone)

$$= \ 15 \ × \ 10 \ × \ 3.5 \ - \ 4 \ × \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 0.5 \ × \ 0.5 \ × \ 1.4$$

$$= \ 525 \ - \ 1.47 \ = \ 523.53$$ cm3

Q.5 A vessel 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 up to 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.

Height of the conical vessel, h = 8 cm.

Its radius r = 5 cm

Volume of cone = Volume of water in cone $$= \ \frac{1}{3} \pi r^2 h$$

$$= \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 5 \ × \ 5 \ × \ 8 \ = \ \frac{4400}{21}$$ cm3

Volume of water flows out = Volume of lead shots

$$= \ \frac{1}{4}$$ of the volume of water in the cone

$$= \ \frac{1}{4} \ × \ \frac{4400}{21} \ = \ \frac{1100}{21}$$ cm3

Volume of one spherical lead shot $$= \ \frac{4}{3} \pi r^3 \ = \ \frac{4}{3} \ × \ \frac{22}{7} \ × \ ( \frac{5}{10})^3$$

$$= \ \frac{11}{21}$$ cm3

∴ Number of lead shots dropped into the vessel

$$= \frac{ \ Volume \ of \ water \ flows \ out}{Volume \ of \ one \ lead \ shot} \ = \frac{ \frac{1100}{21}}{ \frac{11}{21}}$$

$$= \ \frac{1100}{21} \ × \ \frac{21}{11} \ = \ 100$$

Q.6 A solid iron pole consists of a cylindrical 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 $$\pi \ = \ 3.14$$ )

Volume of the solid iron pole = Volume of the cylindrical portion + Volume of the other cylindrical portion

$$= \ \pi (r_1)^2 h_1 \ + \ \pi (r_2)^2 h_2$$

$$= \ 3.14 \ × \ (12)^2 \ × \ 220 \ + \ 3.14 \ × \ (8)^2 \ × \ 60$$

$$= \ 3.14 \ × \ 144 \ × \ 220 \ + \ 3.14 \ × \ 64 \ × \ 60$$

$$= \ 99475.2 \ + \ 12057.6$$

$$= \ 111532.8$$ cm3

∴ The mass of the pole $$= \ 111532.8 \ × \ 8$$ grams

$$= \ \frac{111532.8 \ × \ 8}{1000} \ = \ 892.26$$ kg

Q.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 circular 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.

Volume of the cylinder $$= \ \pi r^2 h \ = \ \frac{22}{7} \ × \ (60)^2 \ × \ 180$$

$$= \ \frac{22 \ × \ 3600 \ × \ 180}{7} \ = \ \frac{14256000}{7}$$ cm3

Volume of the solid = Volume of cone + Volume of hemisphere

$$= \ \frac{1}{3} \ × \ \frac{22}{7} \ × \ 60^2 \ × \ 120 \ + \ \frac{2}{3} \ × \ \frac{22}{7} \ × \ 60^3$$

$$= \ \frac{3168000}{7} \ + \ \frac{3168000}{7} \ = \ \frac{6336000}{7}$$ cm3

Volume of water left in the cylinder = Volume of the cylinder - Volume of the solid

$$= \ \frac{14256000}{7} \ - \ \frac{6336000}{7} \ = \frac{7920000}{7}$$

$$= \ 1131428.57142 \ = \ 1.131$$ m3

Q.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 measuring 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 $$\pi \ = \ 3.14$$

Volume of spherical vessel $$= \ \pi (1)^2 (8) \ + \ \frac{4}{3} \pi (4.25)^3$$

$$= \ 3.14(8 \ + \ \frac{4}{3} \ × \ 76.765625) \ = \ 3.14(8 \ + \ 102.353 )$$

$$= \ 3.14 \ × \ 110.353 \ = \ 346.51$$ cm3

∴ She is incorrect. The correct answer is 346.51 cm3

#### Solution for Exercise - 13.3

Q.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.

Volume of the sphere $$= \ \frac{4}{3} \pi r^3 \ = \ \frac{4}{3} \ × \ \pi \ × \ (4.2)^3$$ cm3

If h is the height of a cylinder of radius 6 cm. Then its volume $$= \ \pi (6)^2 h \ = \ 36\pi h$$ cm3

∵ The volume of metal in the form of sphere and cylinder remains the same , we have

$$36 \pi h \ = \ \frac{4}{3} \ × \ \pi \ × \ (4.2)^3$$

$$=> \ h \ = \ \frac{1}{36} \ × \ \frac{4}{3} \ × \ (4.2)^3$$

$$=> \ h \ = \ 2.744$$ cm

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

Sum of the volumes of 3 gives spheres.

$$= \ \frac{4}{3} \pi ((r_1)^3 \ + \ (r_2)^3 \ + \ (r_3)^3)$$

$$= \ \frac{4}{3} \pi (6^3 \ + \ 8^3 \ + \ 10^3)$$

$$= \ \frac{4}{3} \pi (216 \ + \ 512 \ + \ 1000)$$

$$= \ \frac{4}{3} \pi (1728)$$ cm3

Let R be the radius of the new spheres whose volume is the sum of the volumes of 3 given spheres.

$$=> \ \frac{4}{3} \pi R^3 \ = \ \frac{4}{3} \pi (1728)$$

$$=> \ R^3 \ = \ 1728 \ => \ R \ = \ 12$$ cm

∴ The radius of the resulting sphere is 12 cm.

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

Let h m be the required height of the platform.

The shape of the platform will be like the shape of a cuboid $$22 \ m \ × \ 14 \ m \ × \ h$$ with a hole in the shape of cylinder of radius 3.5 m and depth h m.

The volume of the platform will be equal to the volume of the earth dug out from the well.

Now, the volume of the earth = Volume of the
cylindrical well $$= \ \pi r^2 h \ = \ \frac{22}{7} \ × \ 12.25 \ × \ 20 \ = \ 770$$ m3

Also, the volume of the platform $$= \ 22 \ × \ 14 \ × \ h$$ m3

But volume of the platform = Volume of the well

$$=> \ 22 \ × \ 14 \ × \ h \ = \ 770$$

$$=> \ h \ = \ \frac{770}{22 \ × \ 14} \ = \ 2.5$$ m

∴ Height of the platform = 2.5 m

Q.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.

Let h be the required height of the embankment.

The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m and external radius $$4 \ + \ 1.5 \ = \ 5.5$$ m

The volume of the embankment will be equal to the volume of earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

$$= \ \pi \ × \ (1.5)^2 \ × \ 14 \ = \ 31.5\pi$$ m3

Also, the volume of the embankment

$$= \ \pi (5.52 \ - \ 1.52)h \ = \ \pi \ × \ 7 \ × \ 4h$$

$$= \ 28 \pi h$$ m3

∴ $$28 \pi h \ = \ 31.5 \pi$$

$$=> \ h \ = \ \frac{31.5}{28} \ = \ 1.125$$ m

Hence, the required height of the embankment = 1.125 m

Q.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 height12 cm and diameter 6 cm, having hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Volume of the cylinder $$= \ \pi r^2 h \ = \ \pi ( \frac{12}{2})^2 \ × \ 15 \ = \ \pi \ × \ 6^2 \ × \ 15$$ cm3

Volume of a cone having hemispherical shape on the top

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

$$= \ \frac{1}{3} \pi ( \frac{6}{2})^2 (12 \ + \ 2 \ × \ \frac{6}{2} ) \ = \ \frac{1}{3} \pi \ × \ 3^2 \ × \ 18$$ cm3

Let the number of cone that can be filled with ice cream be n.

Then, $$\frac{1}{3} \pi \ × \ 3^2 \ × \ 18 \ × \ n \ = \ \pi \ × \ 6^2 \ × \ 15$$

$$=> \ n \ = \ \pi \ × \ 6 \ × \ 6 \ × \ 15 \pi \ × \ 3 \ × \ 3 \ × \ 18 \ × \ 3 \ = \ 10$$

∴ 10 cones can be filled with ice cream.

Q.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?

The shape of the coin will be like the shape of a cylinder of radius $$\frac{1.75}{2} \ = \ 0.875$$ cm and of height 2 mm = 0.2 cm

Its volume $$= \ \pi r^2 h \ = \ \frac{22}{7} \ × \ 0.875 \ × \ 0.875 \ × \ 0.2$$

$$= \ 0.48125$$ cm3

Volume of the cuboid $$= \ 5.5 \ × \ 10 \ × \ 3.5 \ = \ 192.5$$ cm3

Number of coins required to form the cuboid

$$= \ \frac{Volume \ of \ the \ cuboid}{Volume \ of \ the \ coin} \ = \ \frac{192.5}{0.48125} \ = \ 400$$

∴ 400 coins must be melted to form a cuboid.

Q.7 A cylindrical bucket, 32 m high and with radius of base 18 cm, is filled with sand. The 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.

Volume of the sand = Volume of the cylindrical bucket $$= \ \pi r^2 h \ = \ \pi \ × \ 18 \ × \ 18 \ × \ 32$$ cm3

Volume of the conical heap $$= \ \frac{1}{3} \pi r^2 h$$ where, h = 24 cm

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

The volume of the conical heap will be equal to that of sand.

∴ $$8\pi r^2 \ = \ \pi \ × \ 18 \ × \ 18 \ × \ 32$$

$$=> \ r^2 \ = \ 18 \ × \ 18 \ × \ 4 \ = \ 18^2 \ × \ 2^2$$

$$=> \ r \ = \ 18 \ × \ 2 \ = \ 36$$

Here, slant height, $$l \ = \ \sqrt{r^2 \ + \ h^2}$$

$$=> \ l \ = \ \sqrt{36^2 \ + \ 24^2} \ = \ 12 \sqrt{9 \ + \ 4} \ = \ 12 \sqrt{13}$$

∴ The radius of the conical heap is 36 cm and its slant height is $$12 \sqrt{13}$$ cm

Q.8 Water in a canal 6 m wid and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Width of the canal = 6 m

Depth of the canal = 1.5 m

Length of water column per hour = 10 km

Length of water column in 30 minutes or 12 hour $$= \ \frac{1}{2} \ × \ 10 \ = \ 5000$$ m

Volume of water flown in 30 minutes $$= \ 1.5 \ × \ 6 \ × \ 5000 \ = \ 45000$$ m3

∵ $$8$$ cm $$= \ \frac{8}{100}$$ m

i.e., 0.08 m standing water is desired

∴ Area irrigated in 30 minutes

$$= \ \frac{Volume}{Height} \ = \ \frac{45000}{0.08} \ = \ 562500$$ m2 or $$56.25$$ hectares

Q.9 A farmer connects a pipe of internal diameter 20 cm from a cannal into a cylindrical tank in his field, which is 10 m in diameter and 2m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Diameter of the pipe = 20 cm

=> Radius of the pipe = 10 cm

Length of water column per hour = 3 km = 3 × 1000 × 100 cm

Volume of water flown in one hour $$= \ \pi \ × \ 100 \ × \ 300000$$ cm3

Tank to be filled = Volume of cylinder (with r = 5m = 500 cm and h = 2m = 200 cm) $$= \ \pi \ × \ 500 \ × \ 500 \ × \ 200$$ cm3

Time required to fill the tank $$= \ \frac{Volume \ of \ tank}{Volume \ of \ water \ flown}$$

$$= \ \frac{ \pi \ × \ 500 \ × \ 500 \ × \ 200}{ \pi \ × \ 100 \ × \ 300000}$$

$$= \ \frac{5}{3} \ = \ 100$$ minutes.

Q.9 A farmer connects a pipe of internal diameter 20 cm from a cannal into a cylindrical tank in his field, which is 10 m in diameter and 2m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Diameter of the pipe = 20 cm

=> Radius of the pipe = 10 cm

Length of water column per hour = 3 km = 3 × 1000 × 100 cm

Volume of water flown in one hour $$= \ \pi \ × \ 100 \ × \ 300000$$ cm3

Tank to be filled = Volume of cylinder (with r = 5m = 500 cm and h = 2m = 200 cm) $$= \ \pi \ × \ 500 \ × \ 500 \ × \ 200$$ cm3

Time required to fill the tank $$= \ \frac{Volume \ of \ tank}{Volume \ of \ water \ flown}$$

$$= \ \frac{ \pi \ × \ 500 \ × \ 500 \ × \ 200}{ \pi \ × \ 100 \ × \ 300000}$$

$$= \ \frac{5}{3} \ = \ 100$$ minutes.

#### Solution for Exercise - 13.4

Q.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.

Capacity of the glass, $$V \ = \ { \pi \ × \ h}{3} \ × \ (R^2 \ + \ r^2 \ + \ Rr)$$

Here, R = 2 cm , r = 1 cm , h = 14 cm

∴ $$V \ = \ \frac{ \frac{22}{7}}{3} \ × \ (2^2 \ + \ 1^2 \ + \ 2 \ × \ 1) \ = \ \frac{44}{3} \ × \ (4 \ + \ 1 \ + \ 2)$$

$$= \ \frac{44}{3} \ × \ 7 \ = \ \frac{308}{3}$$ cm3

Q.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.

Slant height, l = 4 cm (Given)

$$2 \pi r_1 \ = \ 6 \ => \ \pi r_1 \ = \ 3$$

and $$2 \pi r_2 \ = \ 18 \ => \ \pi r_2 \ = \ 9$$

Curved surface of the frustum $$= \ ( \pi r_1 \ + \ \pi r_2)l$$

$$= \ (3\ + \ 9) \ × \ 4 \ = \ 48$$ cm2

Q.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 cmand its slant height is 15 cm, find the area of material used for making it.

Here R = 10 cm, r = 4 cm and l = 15 cm

Area of the material used for making the feza = Surface area of frustum + The surface of top circular section

$$= \ \pi (R \ + \ r)l \ + \ \pi r^2$$

$$= \ \frac{22}{7} (10 \ + \ 4) 15 \ + \ \frac{22}{7} \ × \ 4 \ × \ 4$$

$$= \ \frac{22}{7} \ × \ 14 \ × \ 15 \ + \ \frac{352}{7}$$

$$= \ \frac{4620 \ + \ 352}{7} \ = \frac{4972}{7}$$ cm2

Q.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 milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take $$\pi \ = \ 3.14$$ ).

Here R = 20 cm, r = 8 cm and h = 16 cm

Capacity of the container = Volume of the frustum $$= \ \frac{1}{3} \pi h(R^2 \ + \ r^2 \ + \ Rr)$$

$$= \ \frac{1}{3} \ × \ 3.14 \ × \ 16(20^2 \ + \ 8^2 \ + \ 20 \ × \ 8) \ = \ \frac{50.24}{3} \ × \ (400 \ + \ 64 \ + \ 160)$$

$$= \ \frac{50.24}{3} \ × \ 624 \ = \ 50.24 \ × \ 208$$

$$= \ \frac{10449.92}{1000}$$ litres

Cost of milk @ Rs 20 per litre $$= \ 20 \ × \ \frac{10449.92}{1000}$$

$$= \ 208.99$$ $$\approx$$ Rs $$209$$

To find the slant height

$$l \ = \ \sqrt{16^2 \ + \ 12^2} \ = \ \sqrt{256 \ + \ 144}$$

$$= \ \sqrt{400} \ = \ 20$$ cm

$$= \ \pi (R \ + \ r)l \ = \ \frac{22}{7} \ × \ (20 \ + \ 8) \ × \ 20$$

$$= \ \frac{22}{7} \ × \ 28 \ × \ 20 \ = \ 1758.4$$ cm2

and area of the bottom

$$= \ \pi r^2 \ = \ \frac{22}{7} \ × \ (8)^2 \ = \ \frac{22}{7} \ × \ 64 \ = \ 200.96$$ cm2

∴ Total area of metal required $$= 1758.4 \ + \ 200.96 \ = \ 1959.36$$ cm2

Cost of metal sheet used to manufacture the container @ Rs 8 per 100 cm2 $$= \ \frac{8}{100} \ × \ 1959.36 \ =$$ Rs $$156.75$$

Q.5 A metallic right circular cone 20 cm high and whose vertical angle is 60º 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 wire of diameter $$\frac{1}{16}$$ cm, find the length of the wire.

Let ABC be the metallic cone, DECB is the required frustum.

Let the two radii of the frustum be $$DO' \ = \ r_2$$ and $$BO \ = \ r_1$$

From the ∆ ADO' and ABO,

$$r_2 \ = \ h_1 tan30 \ = \ 10 \ × \ \frac{1}{ \sqrt{3}}$$

$$r_1 \ = \ (h_1 \ + \ h_2) tan30º \ = \ 20 \ × \ \frac{1}{ \sqrt{3}}$$

Volume of the frustum DBCE $$= \ \frac{\pi h_2}{3}((r_1)^2 \ + \ (r_1)(r_2) \ + \ (r_2)^2)$$

$$= \ \frac{ \pi h_2}{3} \ × \ [ \frac{400}{3} \ + \ \frac{200}{3} \ + \ \frac{100}{3}] \ = \frac{ \pi \ × \ 10}{3} \ × \ \frac{700}{3}$$

Volume of the wire of length l and diameter D

$$= \ \pi ( \frac{D}{2})^2 \ × \ l \ = \ \frac{ \pi D^2}{4} \ × \ l$$       [ $$V \ = \ \pi r2h$$ ]

∴ Volume of the frustum = Volume of the wire drawn from it

$$=> \ \frac{ \pi \ × \ 10}{3} \ × \ \frac{700}{3} \ = \ \frac{ \pi D^2}{4} \ × \ l$$       [ ∵ D=116]

$$l \ = \ \frac{10 \ × \ 700 \ × \ 4}{3 \ × \ 3 \ × \ 16 \ × \ 16} \ = \ 7964.44$$ m

#### Solution for 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 .

Given that,

Diameter of cylinder = 10 cm

So, radius of the cylinder (r) $$\frac{10}{2}$$ cm = 5 cm

? Length of wire in completely one round $$= \ 2 \pi r \ = 3.14 \ x \ 5 \ = \ 31.4$$ cm

It is given that diameter of wire = 3 mm $$= \ \frac{3}{10}$$ cm

? The thickness of cylinder covered in one round = 3/10 m

Hence, the number of turns(rounds) of the wire to cover 12 cm will be $$= \ \frac{12}{ \frac{3}{10}} \ = \ 12 \ × \ \frac{10}{3}$$

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds

=> $$40 \ x \ 31.4 \ = \ 1256$$ cm

Radius of the wire $$= \ \frac{0.3}{2} \ = \ 0.15$$ cm

Volume of wire = Area of cross-section of wire × Length of wire

$$= \ \pi(0.15)^2 \ × \ 1257.14 \ = \ 88.898$$ cm3

We know,

Mass = Volume × Density

$$= \ 88.898 \ × \ 8.88 \ = \ 789.41$$ gm

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 $$\pi$$ as found appropriate.)

Let us consider the ∆ ;ABA

Here,

AS = 3 cm, AC = 4 cm

So, Hypotenuse BC = 5 cm

We have got 2 cones on the same base AA’ where the radius = DA or DA’

By putting the value of CA, AB and CB we get,

We also know,

DB/AB = AB/CB

So, DB = 9/5 cm

As, CD = BC - DB,

CD = 16/5 cm

Now, volume of double cone will be

$$[ \frac{1}{3} \pi \ × \ ( \frac{12}{5})^2 \frac{9}{5} \ + \ \frac{1}{3} \pi \ × \ ( \frac{12}{5})^2 \ × \ \frac{16}{5} ]$$ cm3

p>Solving this we get,

V = 30.14 cm3

The surface area of the double cone wil be

$$( \pi \ × \ \frac{12}{3} \ × \ 3) \ + \ ( \pi \ × \ \frac{12}{3} \ × \ 4)$$ $$= \ \pi \ × \ \frac{12}{5}[3 \ + \ 4]$$

$$= \ 52.75$$ cm3

3. A cistern, internally measuring 150 cm × 120 cm × 100 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 ocerflowing the water, each being 22.5 cm × 7.5 cm × 6.5 cm?

Given that the dimension of the cistern = 150 × 120 × 110

So, volume = 1980000 cm3

Volume to be filled in cistern = 1980000 - 129600

= 1850400 cm3

Now, let the number of bricks placed be “n”

So, volume of n bricks will be = n × 22.5 × 7.5 × 6.5

Now as each brick absorbs one-seventeenth of its volume, the volume will be

$$= \ \frac{n}{17} \ × \ 22.5 \ × \ 7.5 \ × \ 6.5$$

For the condition given in the question,

The the volume of n bricks has to be equal to volume absorbed by n bricks + Volume to be filled in cistern

Or, n × 22.5 × 7.5 × 6.5 = 1850400 + $$\frac{n}{17}$$ × (22.5 × 7.5 × 6.5)

Solving this we get,

n = 1792.41

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.

From the question, it is clear that

Total volume of 3 rivers = 3 × [(Surface area of a river) × Depth]

Given,

Surface area of a river = [1072 × \frac{75}{1000} \)] km

And,

Depth = $$\frac{3}{1000}$$ km

Now, volume of 3 rivers = 3 × [1072 × $$\frac{75}{1000}$$] × $$\frac{3}{1000}$$

= 0.72 km3

Now, volume of rainfall = total surface area × total height of rain

$$= \ 9780 \ × \ \frac{10}{100 \ × \ 1000}$$

= 9.7 km3

For the total rainfall was approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.

But, 9.7 km3 $$\ne$$ 0.72 km3

So, the question statement is false.

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 (see Fig.).

Given,

Diameter of upper circular end of frustum part = 18 cm

So, radius (r1) = 9 cm

Now, the radius of the lower circular end of frustum (r2) will be equal to the radius of the circular end of the cylinder

So, r2 = 8/2= 4 cm

Now, height (h1) of the frustum section = 22 - 10 = 12 cm

And,

Height (h2) of cylindrical section = 10 cm (given)

Now, the slant height will be, $$l \ = \ \sqrt{(r_1 \ - \ r_2)^2 \ + \ (h_1)^2}$$

Or, l = 13 cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

$$= \ (r_1 \ + \ r_2)l \ + \ 2 \pi r_2 h_2$$

Solving this we get,

Area of tin sheet required = 782 × $$\frac{4}{7}$$ cm2

6. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r1 and r2 are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the ∆ ABG ∆ ADF,

Here, DF||BG

So, $$∆ \ ABG \ \sim \ ∆ \ ADF$$

=> $$\frac{DF}{BG} \ = \ \frac{AF}{AG} \ = \ \frac{AD}{AB} \ => \ \frac{r_2}{r_1} \ = \ \frac{h_1 \ - \ h}{h_1} \ = \ \frac{l_1 \ - \ l}{l_1}$$

$$=> \ l_1 \ = \ \frac{r_1 l}{r_1 \ - \ r_2}$$

The total surface area of frustum will be equal to the total CSA ot frustum + the area ol upper circular end + area of the lower circular end

$$= \pi (r_1 \ + \ r_2)l \ + \ \pi (r_2)^2 \ + \ \pi (r_1)^2$$

∴ Surface area of frustum $$= \ \pi [r_1 \ + \ r_2)l \ + \ (r_1)^2 \ + \ (r_2)^2]$$

7. Derive the formula for the volume of the frustum of a cone.

Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r1 and r2 are the radii of the frustum ends of the cone and h be the frustum height.

Now, consider the ∆ ABG ∆ ADF,

Here, DF||BG

So, $$∆ \ ABG \ \sim \ ∆ \ ADF$$

=> $$\frac{DF}{BG} \ = \ \frac{AF}{AG} \ = \ \frac{AD}{AB} \ => \ \frac{r_2}{r_1} \ = \ \frac{h_1 \ - \ h}{h_1} \ = \ \frac{l_1 \ - \ l}{l_1}$$

$$=> \ \frac{r_2}{r_1} \ = \ 1 \ - \ \frac{h}{h_1} \ = \ 1 \ - \ \frac{l}{l_1}$$

=> $$1 \ - \ \frac{h}{h_1} \ = \ \frac{r_2}{r_1} \ => \ \frac{h}{h_1} \ = \ \frac{r_1 \ - \ r}{r_1}$$

=> $$h_1 \ = \ \frac{r_1 h}{r_1 \ - \ r_2}$$

The total volume of frustum of the cone will be = Volume of cone ABC – Volume of cone ADE

$$= \ \frac{1}{3} \pi (r_1)^2 h_1 \ – \ \frac{1}{3} \pi (r_2)^2 (h_1 \ – \ h)$$

$$= \ ( \frac{ \pi}{3})[(r_1)^2 h_1 \ - \ (r_2)^2 (h_1 \ - \ h)]$$

$$= \ ( \frac{ \pi}{3})[ ( \frac{(r_1)^3 h}{r_1 \ - \ r_2} ) \ - \ (r_2)^2 ( \frac{r_1 h}{r_1 \ - \ r_2} \ - \ h)]$$

$$= \ ( \frac{ \pi}{3})[ ( \frac{(r_1)^3 h}{r_1 \ - \ r_2} ) \ - \ (r_2)^2 \ ( \frac{hr_1 \ - \ hr_1 \ + \ hr_2}{r_1 \ - \ r_2} ) ]$$

$$= \ \frac{ \pi }{3} h [ \frac{(r_1)^3 \ - \ (r_2)^3}{r_1 \ - \ r_2} ]$$

$$= \ \frac{ \pi }{3} h [ \frac{(r_1 \ - \ r_2)(r_1 ^2 \ + \ r_2^2 \ + \ r_1 r_2}{r_1 \ - \ r_2} ]$$

∴ Volume of frustum of the cone $$= \ \frac{1}{3} \pi h(r_1^2 \ + \ r_2^2 \ + \ r_1 r_2)$$