# EXERCISE 13.1

**1. 2 cubes each of volume 64 cm**^{3} are joined end to end. Find the surface area of the resulting cuboid.

^{3}are joined end to end. Find the surface area of the resulting cuboid.

**Answer:**

#### Let l be length of each cube

Volume of each cube = l^{3} = 64 cm^{3}

#### ⇒ l3 = 64 cm^{3}

#### ⇒ l = 4 cm

#### If, we join 2 cubes each having side equal to 4 cm, we get a cuboid

#### Length of cuboid = L = 8 cm

#### Height of cuboid = H = 4 cm

#### Breadth of cuboid = B = 4 cm

We know that Surface Area of Cuboid = 2(L.B + B.H + H.L)

#### = 2((8 x 4) + (4 x 4) + (4 x 8))

#### = 2(32 + 16 + 32)

#### = 2(80)

#### = 160 cm^{2}

**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.**

**Answer:**

#### Height of the cylindrical portion 13 − 7 = 6 cm.

#### Area of a Curved surface of cylindrical portion is,

#### = 2πrh

#### = 2 × 22/7 × 7 × 6

#### = 264 cm^{2}

#### Area of a curved of hemispherical portion is

#### =2πr2

#### =2 × 22/7 × 7 × 7

#### = 308 cm^{2}

#### ∴ total surface area is = 308 + 264 = 572 cm^{2}.

**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.**

**Answer:**

#### Total Surface area of toy = Surface area of hemisphere without base + Surface area of cone without base

#### Total height = 15.5 = r + h

#### h = 15.5 − 3.5 = 12 cm

#### A = 2πr^{2} + πr√(r^{2} + (15.5 − r)^{2})

#### A = 2π × 3.5^{2} + π × 3.5√(3.5^{2} + 12^{2})

#### A = 214.5 cm^{2}

**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.**

**Answer:**

#### (i) A cubical block of side 7 cm is surrounded by a hemisphere.

#### ∴ Diameter of hemisphere is = 7 cm.

#### Radius of hemisphere, r = 3.5 cm

#### ∴ Surface area of hemisphere = Curved surface area – Area of base of hemisphere

#### = 2πr^{2} − πr^{2}

#### = 2 × 22/7 × (7/2)^{2 }− 22/7 × (7/2)^{2}

#### = 38.5

#### (ii) The curved surface area of square,

#### = 6 × I^{2}

#### = 6 × (7)^{2}

#### = 6 × 49

#### = 294 sq.cm.

#### ∴ Total surface area of cubical block = Curved surface area of cubical block + Surface area of hemisphere

#### = 294 + 38.5

#### = 332.5 sq.cm.

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

**Answer:**

**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.**

**Answer:**

#### Surface area of capsule = 2πrh + 2 × 2πr^{2}

#### Daimeter of capsule = 5mm

#### Radius of cylinder = Radius of hemisphere = 5/2 = 2.5mm

#### Height (h) =14 − 2.5 − 2.5 = 9mm

#### TSA = 2πr (h + 2r)

#### = 2 × 22/7 × 2.5 (9 + 5)

#### = 220 mm^{2}

**7. A tent is in the shape of a cylinder surmounted buy 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 m**^{2}. (Note that the base of the tent will not be covered with canvas.)

^{2}. (Note that the base of the tent will not be covered with canvas.)

**Answer:**

#### Radius of the cylindrical base = 2m

#### Height of the cylindrical is = 2.1m

#### Slant Height of conical top is = 2.8

#### Curved Surface Area of cylindrical portion is,

#### = 2πrh

#### = 2π × 2 × 2.1

#### = 8.4πm^{2}

#### Curved Surface Area of conical portion is,

#### = πrl

#### = π × 2 × 2.8

#### = 5.6πm^{2}

#### Total Curved Surface Area

#### = 8.4π + 5.6π

#### = 14 × 22/7

#### = 44m^{2}

#### Therefore,

#### Cost of canvas = Rate\times Surface Area

#### = 500 × 44

#### = 22000

**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 cm**^{2}.

^{2}.

**Answer:**

#### Radius = 0.7 cm and height = 2.4 cm

#### Total surface area of structure = Curved surface area of cylinder + Area of top of cylinder + Curved surface area of cone

#### Curved surface area of cylinder = 2πrh

#### = 2π × 0.7 × 2.4

#### = 3.36 π cm^{2}

#### Area of top:

#### = πr^{2}

#### = π × 0.7^{2}

#### = 0.49π cm^{2}

#### Slant height of cone can be calculated as follows:

#### l = √(h^{2}+r^{2})

#### = √(2.4^{2 }+ 0.7^{2})

#### = √(5.76 + 0.49)

#### = √6.25 = 2.5 cm

#### Curved surface area of cone:

#### = πrl

#### = π × 0.7 × 2.5

#### = 1.75π cm^{2}

#### Hence, remaining surface area of structure

#### = 3.36π + 0.49π + 1.75π

#### = 5.6π = 17.6 cm^{2}

#### = 18 cm^{2 }(approx)

**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.**

**Answer:**

#### Total surface area of the article = Lateral surface area of cylinder + 2 × Curved surface area of hemisphere

#### We know that the lateral surface are of a cylinder = 2πrh.

#### And the curved surface area of a hemisphere = 2πr^{2}.

#### ∴ TSA of article = 2πrh + 2 × 2πr^{2}

#### ⇒ 2 × 22/7 × 7/2 × 10 + 2 × 2 × 22/7 × (7/2)^{2}

#### ⇒ 22 × 10 + 22 × 7

#### ⇒ 22 (10 + 7)

#### ⇒ 22 × 17

#### ∴ TSA = 374 cm^{2}

#### Hence, the total surface area of the article is 374 cm^{2}.

# EXERCISE 13.2

**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.**

**Answer:**

#### Total Volume = Volume of cone + Volume of Hemisphere

#### We know that

#### Volume of cone = 1/3 πr^{2}h

#### Volume of hemisphere = 2/3πr^{3}

#### = 1/3πr^{2}h + 2/3πr^{3}

#### = 1/3 πr^{2}(h+2r)

#### = 1/3 × π × 1 × 1 × (1+2)

**= **π cm^{3}

**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 made. (Assume the outer and inner dimensions of the model to be nearly the same.)**

**Answer:**

#### Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm

#### Volume of cylinder = πr^{2}h

#### = π × 1.5^{2 }× 8

#### = 18π cm^{3}

#### Volume of a cone = 1/3 π × 1.5^{2 }× 2

#### = 1.5π cm^{3}

#### Total volume = Volume of cylinder + 2 Volume of a cone

#### = 1.5π + 1.5π + 18π

#### = 21π = 66 cm^{3}

**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 hemispherical ends, with length 5 cm and diameter 2.8 cm (see figure).**

**Answer:**

#### It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.

#### So, the total height of a gulab jamun = 5 cm.

#### Diameter = 2.8 cm

#### So, radius = 1.4 cm

#### ∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm

#### =2.2 cm

#### Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres

#### = πr^{2}h+(4/3)πr^{3}

#### = 4.312π+(10.976/3) π

#### = 25.05 cm^{3}

#### We know that the volume of sugar syrup = 30% of total volume

#### So, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm^{3})

#### = 45×7.515 = 338.184 cm^{3}

**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 id 1.4 cm. Find the volume of wood in the entire stand (see figure).**

**Answer.**

#### Volume of cuboid = length x width x height

#### We know the cuboid’s dimensions as 15 cmx10 cmx3.5 cm

#### So, the volume of the cuboid = 15x10x3.5 = 525 cm^{3}

#### Here, depressions are like cones and we know,

#### Volume of cone = (⅓)πr^{2}h

#### Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm

#### ∴ Volume of 4 cones = 4x(⅓)πr^{2}h

#### = 1.46 cm^{2}

#### Now, volume of wood = Volume of cuboid – 4 x volume of cone

#### = 525-1.46 = 523.54 cm^{2}

**5. A vessel is in the form of inverted cone. Its height is 8 cm and the radius of the 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.**

**Answer:**

#### It is known that,

Volume of cone = volume of water in the cone

= ⅓πr2h = (200/3)π cm3

Now,

Total volume of water overflown= (¼)×(200/3) π =(50/3)π

Volume of lead shot

= (4/3)πr^{3}

= (1/6) π

Now,

The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot

= (50/3)π/(⅙)π

= (50/3)×6 = 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 cm**^{3} of iron has approximately 8 g mass. ( π = 3.14 ).

^{3}of iron has approximately 8 g mass. ( π = 3.14 ).

**Answer:**

#### Given that, the height of the big cylinder (H) = 220 cm

Radius of the base (R) = 24/2 = 12 cm

So, the volume of the big cylinder = πR2H

= π(12)2 × 220 cm^{3}

= 99565.8 cm^{3}

Now, the height of smaller cylinder (h) = 60 cm

Radius of the base (r) = 8 cm

So, the volume of the smaller cylinder = πr2h

= π(8)2×60 cm^{3}

= 12068.5 cm^{3}

∴ Volume of iron = Volume of the big cylinder+ Volume of the small cylinder

= 99565.8 + 12068.5

=111634.5 cm^{3}

We know,

Mass = Density x volume

So, mass of the pole = 8×111634.5

= 893 Kg (approx.)

**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.**

**Answer:**

#### Here, the volume of water left will be = Volume of cylinder – Volume of solid

Given,

Radius of cone = 60 cm,

Height of cone = 120 cm

Radius of cylinder = 60 cm

Height of cylinder = 180 cm

Radius of hemisphere = 60 cm

Now,

Total volume of solid = Volume of Cone + Volume of hemisphere

Volume of cone = 1/3πr2h = 1/3 × π×602×120cm3 = 144×103π cm^{3}

Volume of hemisphere = (⅔)×π×603 cm3 = 144×103π cm^{3}

So, total volume of solid = 144×103π cm3 + 144×103π cm3 = 288 ×103π cm^{3}

Volume of cylinder = π×602×180 = 648000 = 648×103 π cm^{3}

Now, volume of water left will be = Volume of cylinder – Volume of solid

= (648-288) × 103×π = 1.131 m^{3}

**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 cm**^{2} . Check whether she is correct, taking the above as the inside measurements and π = 3.14.

^{2}. Check whether she is correct, taking the above as the inside measurements and π = 3.14.

**Answer:**

#### =

#### Given,

#### For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm

#### For the spherical part, Radius (r) = (8.5/2) = 4.25 cm

#### Now, volume of this vessel = Volume of cylinder + Volume of sphere

#### = π×(1)^{2}×8+(4/3)π(4.25)^{3}

#### = 346.51 cm^{3}

# EXERCISE 13.3

**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.**

**Answer:**

#### It is given that radius of the sphere (R) = 4.2 cm

#### Also, Radius of cylinder (r) = 6 cm

#### Now, let height of cylinder = h

#### It is given that the sphere is melted into a cylinder.

#### So, Volume of Sphere = Volume of Cylinder

#### ∴ (4/3)×π×R^{3 }= π×r^{2}×h.

#### h = 2.74 cm

**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.**

**Answer:**

**For Sphere 1:**

#### Radius (r_{1}) = 6 cm

#### ∴ Volume (V_{1}) = (4/3)×π×r_{1}^{3}

**For Sphere 2:**

#### Radius (r_{2}) = 8 cm

#### ∴ Volume (V_{2}) = (4/3)×π×r_{2}^{3}

**For Sphere 3:**

#### Radius (r_{3}) = 10 cm

#### ∴ Volume (V_{3}) = (4/3)× π× r_{3}^{3}

#### Also, let the radius of the resulting sphere be “r”

#### Now,

#### Volume of resulting sphere = V_{1}+V_{2}+V_{3}

#### (4/3)×π×r^{3} = (4/3)×π×r_{1}^{3}+(4/3)×π×r_{2}^{3} +(4/3)×π×r_{3}^{3}

#### r^{3 }= 6^{3}+8^{3}+10^{3}

#### r^{3 }= 1728

#### r = 12 cm

**3. A 20 m deep well with diameter 7 m 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.**

**Answer:**

#### Diameter of well = 7 m

#### It is given that the shape of the well is in the shape of a cylinder with a diameter of 7 m

#### So, radius = 7/2 m

#### Also, Depth (h) = 20 m

#### Volume of the earth dug out will be equal to the volume of the cylinder

#### Let the height of the platform = H

#### Volume of soil from well (cylinder) = Volume of soil used to make such platform

#### π×r^{2}×h = Area of platform × Height of the platform

#### We know that the dimension of the platform is = 22×14

#### So, Area of platform = 22×14 m^{2}

#### ∴ π×r^{2}×h = 22×14×H

#### ⇒ H = 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.**

**Answer:**

#### Given, Depth (h_{1}) of well = 14 m

#### Diameter of the circular end of the well =3 m

#### So, Radius (r_{1}) = 3/2 m

#### Width of the embankment = 4 m

#### From the figure, it can be said that the embankment will be a cylinder having an outer radius (r_{2}) as 4+(3/2) = 11/2 m and inner radius (r_{1}) as 3/2m

#### Now, let the height of embankment be h_{2}

#### ∴ Volume of soil dug from well = Volume of earth used to form embankment

#### π×r_{1}^{2}×h_{1} = π×(r_{2}^{2}-r_{1}^{2}) × h_{2}

#### Solving this, we get,

#### The height of the embankment (h_{2}) as 1.125 m.

**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.**

**Answer:**

#### Number of cones will be = Volume of cylinder / Volume of ice cream cone

#### For the cylinder part,

#### Radius = 12/2 = 6 cm

#### Height = 15 cm

#### ∴ Volume of cylinder = π×r^{2}×h = 540π

#### For the ice cone part,

#### Radius of conical part = 6/2 = 3 cm

#### Height = 12 cm

#### Radius of hemispherical part = 6/2 = 3 cm

#### Now,

#### Volume of ice cream cone = Volume of conical part + Volume of hemispherical part

#### = (⅓)×π×r^{2}×h+(⅔)×π×r^{3}

#### = 36π +18π

#### = 54π

#### ∴ Number of cones = (540π/54π)

#### = 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 x 10cm x 3.5?**

**Answer:**

#### For silver coin, Diameter = 1.75 cm

#### Radius (r) of circular end of coins = 1.75/2 = 0.875 cm

#### Now, the number of coins to be melted to form the required cuboids be “n”

#### So, Volume of n coins = Volume of cuboids

#### n × π × r^{2 }× h_{1} = l × b × h

#### n×π×(0.875)^{2}×0.2 = 5.5×10×3.5

#### Or, n = 400

**7. A cylindrical bucket, 32 cm and 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.**

**Answer:**

#### let “r_{2}” be the radius of the circular end of the conical heap.

#### We know that volume of the sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.

#### ∴ Volume of sand in the cylindrical bucket = Volume of sand in conical heap

#### π×r_{1}^{2}×h_{1} = (⅓)×π×r_{2}^{2}×h_{2}

#### π×18^{2}×32 = (⅓)×π ×r_{2}^{2}×24

#### Or, r_{2}= 36 cm

#### And,

#### Slant height (l) = √(36^{2}+24^{2}) = 12√13 cm.

**8. Water in a canal 6 m wide 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?**

**Answer:**

#### Breadth (b) = 6 m and Height (h) = 1.5 m

#### It is also given that

#### The speed of canal = 10 km/hr

#### Length of canal covered in 1 hour = 10 km

#### Length of canal covered in 60 minutes = 10 km

#### Length of canal covered in 1 min = (1/60)x10 km

#### Length of canal covered in 30 min (l) = (30/60)x10 = 5km = 5000 m

#### We know that the canal is cuboidal in shape. So,

#### Volume of canal = lxbxh

#### = 5000x6x1.5 m^{3}

#### = 45000 m^{3}

#### Now,

#### Volume of water in canal = Volume of area irrigated

#### = Area irrigated x Height

#### So, Area irrigated = 56.25 hectares

#### ∴ Volume of canal = lxbxh

#### 45000 = Area irrigatedx8 cm

#### 45000 = Area irrigated x (8/100)m

#### Or, Area irrigated = 562500 m^{2 }= 56.25 hectares.

**9. A farmer connects a pipe of internal 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 km/h, in how much time will the tank be filled?**

**Answer:**

#### Volume of water that flows in t minutes from pipe = t×0.5π m^{3}

#### Radius (r_{2}) of circular end of cylindrical tank =10/2 = 5 m

#### Depth (h_{2}) of cylindrical tank = 2 m

#### Let the tank be filled completely in t minutes.

#### Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.

#### Volume of water that flows in t minutes from pipe = Volume of water in tank

#### t×0.5π = π×r_{2}^{2}×h_{2}

#### Or, t = 100 minutes.

# EXERCISE 13.4

**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.**

**Answer:**

#### Here,

#### Radius (r_{1}) of the upper base = 4/2 = 2 cm

#### Radius (r_{2}) of lower the base = 2/2 = 1 cm

#### Height = 14 cm

#### Now, Capacity of glass = Volume of frustum of cone

#### So, Capacity of glass = (⅓)×π×h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})

#### = (⅓)×π×(14)(2^{2}+1^{2}+ (2)(1))

#### ∴ The capacity of the glass = 102×(⅔) 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.**

**Answer:**

#### Slant height (l) = 4 cm

#### Circumference of upper circular end of the frustum = 18 cm

#### ∴ 2πr_{1} = 18

#### Or, r_{1} = 9/π

#### Similarly, circumference of lower end of the frustum = 6 cm

#### ∴ 2πr_{2} = 6

#### Or, r_{2} = 3/π

#### Now, CSA of frustum = π(r_{1}+r_{2}) × l

#### = π(9/π+3/π) × 4

#### = 12×4 = 48 cm^{2}

**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.**

**Answer:**

#### ,

#### For the lower circular end, radius (r_{1}) = 10 cm

#### For the upper circular end, radius (r_{2}) = 4 cm

#### Slant height (l) of frustum = 15 cm

#### Now,

#### The area of material to be used for making the fez = CSA of frustum + Area of upper circular end

#### CSA of frustum = π(r_{1}+r_{2})×l

#### = 210π

#### And, Area of upper circular end = πr_{2}^{2}

#### = 16π

#### The area of material to be used for making the fez = 210π + 16π = (226 x 22)/7 = 710 2/7

#### ∴ The area of material used = 710 2/7 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 total cost of milk which can completely fill the container at the rate of Rs. 20 per liter. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm**^{2}. ( take π = 3.14)

^{2}. ( take π = 3.14)

**Answer:**

#### Given,

#### r_{1} = 20 cm,

#### r_{2} = 8 cm and

#### h = 16 cm

#### ∴ Volume of the frustum = (⅓)×π×h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})

#### = 1/3 ×3.14 ×16((20)^{2}+(8)^{2}+(20)(8))

#### = 1/3 ×3.14 ×16(400 + 64 + 160) = 10449.92 cm^{3} = 10.45 lit

#### It is given that the rate of milk = Rs. 20/litre

#### So, Cost of milk = 20×volume of the frustum

#### = 20 × 10.45

#### = Rs. 209

#### Now, slant height will be

#### l = 20 cm

#### So, CSA of the container = π(r_{1}+r_{2})×l

#### = 1758.4 cm^{2}

#### Hence, the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)

#### = 1758.4+πr^{2} = 1758.4+π(8)^{2}

#### = 1758.4+201 = 1959.4 cm^{2}

#### ∴ Total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157

**5. A metallic right circular cone 20 cm high and whose vertical angle is 60**^{0} 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 1/16 cm, find the length of the wire.

^{0}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 1/16 cm, find the length of the wire.

**Answer:**

# 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 cm**^{2 }.

^{2 }.

**Answer:**

#### Given that,

#### Diameter of cylinder = 10 cm

#### So, radius of the cylinder (r) = 10/2 cm = 5 cm

#### ∴ Length of wire in completely one round = 2πr = 3.14×5 cm = 31.4 cm

#### It is given that diameter of wire = 3 mm = 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-

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

#### 40 x 31.4 cm = 1256 cm

#### Radius of the wire = 0.3/2 = 0.15 cm

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

#### = π(0.15)^{2}×1257.14

#### = 88.898 cm^{3}

#### 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 π as found appropriate)**

**Answer:**

#### 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’

#### Now, AD/CA = AB/CB

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

#### AD = 2/5 cm

#### 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

#### Solving this we get,

#### V = 30.14 cm^{3}

#### The surface area of the double cone will be

#### = 52.75 cm^{2}

**3. A cistern, internally measuring 150 cm x 120 cm x 110 cm has 129600 cm**^{2} 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 22.5 cm x 7.5 cm x 6.5 cm?

^{2}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 22.5 cm x 7.5 cm x 6.5 cm?

**Answer:**

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

#### So, volume = 1980000 cm^{3}

#### Volume to be filled in cistern = 1980000 – 129600

#### = 1850400 cm^{3}

#### 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

#### = n/(17)×(22.5×7.5×6.5)

#### For the condition given in the question,

#### 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+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 km**^{2} , 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.

^{2}, 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.

**Answer:**

#### 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×(75/1000)] km

#### And,

#### Depth = (3/1000) km

#### Now, volume of 3 rivers = 3×[1072×(75/1000)]×(3/1000)

#### = 0.7236 km^{3}

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

#### = 0.7280 km^{3}

#### 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, 0.7280 km^{3} = 0.7236 km^{3}

#### So, the question statement is true.

**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 figure).**

**Answer:**

#### Given,

#### Diameter of upper circular end of frustum part = 18 cm

#### So, radius (r_{1}) = 9 cm

#### Now, the radius of the lower circular end of frustum (r_{2}) will be equal to the radius of the circular end of the cylinder

#### So, r_{2} = 8/2 = 4 cm

#### Now, height (h_{1}) of the frustum section = 22 – 10 = 12 cm

#### And,

#### Height (h_{2}) of cylindrical section = 10 cm (given)

#### Now, the slant height will be-

#### Or, l = 13 cm

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

#### = π(r_{1}+r_{2})l+2πr_{2}h_{2}

#### Solving this we get,

#### Area of tin sheet required = 782 4/7 cm^{2}

**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.**

**Answer:**

#### Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r_{1} and r_{2} are the radii of the frustum ends of the cone and h be the frustum height.

#### Now, consider the ΔABG and ΔADF,

#### Here, DF||BG

#### So, ΔABG ~ ΔADF

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

#### = π(r_{1}+r_{2})l+πr_{2}^{2}+πr_{1}^{2}

#### ∴ Surface area of frustum = π[(r_{1}+r_{2})l+r_{1}^{2}+r_{2}^{2}]