**Exercise 13.4**

**1. Find the surface area of a sphere of radius:**

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

**Answer**

(i) Radius of the sphere (r) = 10.5 cm

Surface area = 4πr

^{2}

= (4 × 22/7 × 10.5 × 10.5) cm

^{2}

= 1386 cm

^{2}

(ii) Radius of the sphere (r) = 5.6 cm

Surface area = 4πr

^{2}

= (4 × 22/7 × 5.6 × 5.6) cm

^{2}

= 394.24 cm

^{2}

(iii) Radius of the sphere (r) = 14 cm

Surface area = 4πr

^{2}

= (4 × 22/7 × 14 × 14) cm

^{2}

= 2464 cm

^{2}

**2. Find the surface area of a sphere of diameter:**

(i) 14 cm (ii) 21 cm (iii) 3.5 m

(i) 14 cm (ii) 21 cm (iii) 3.5 m

**Answer**

(i) r = 14/2 cm = 7cm

Surface area = 4πr

^{2}

^{ }= (4 × 22/7 × 7 × 7)cm

^{2}

=616cm

^{2}

(ii) r = 21/2 cm = 10.5 cm

Surface area = 4πr

^{2}

= (4 × 22/7 × 10.5 × 10.5) cm

^{2}

= 1386 cm

^{2}

(iii) r = 3.5/2 m = 1.75 m

Surface area = 4πr

^{2}

= (4 × 22/7 × 1.75 × 1.75) m

^{2}

= 38.5 m

^{2}

**3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)**

**Answer**

r = 10 cm

Total surface area of hemisphere = 3πr

^{2}

= (3 × 3.14 × 10 ×10) cm

^{2}

= 942 cm

^{2}

**4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.**

**Answer**

Let r be the initial radius and R be the increased radius of balloons.

r = 7cm and R = 14cm

Ratio of the surface area =4πr

^{2}/4πR

^{2}

= r

^{2}/R

^{2}

= (7×7)/(14×14) = 1/4

Thus, the ratio of surface areas = 1 : 4

**5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm**

^{2}.**Answer**

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm

Curved surface area of the hemispherical bowl = 2πr

^{2}

^{ }= (2 × 22/7 × 5.25 × 5.25) cm

^{2}

= 173.25 cm

^{2}

Rate of tin - plating is = ₹16 per 100 cm

^{2}

^{}

Therefor, cost of 1 cm

^{2 }=

^{ }₹16/100

Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100

= ₹27.72

**6. Find the radius of a sphere whose surface area is 154 cm**

^{2}.**Answer**

Let r be the radius of the sphere.

Surface area = 154 cm

^{2}

⇒ 4πr

^{2 }= 154

⇒ 4 × 22/7 × r

^{2 }= 154

⇒ r

^{2 }= 154/(4 × 22/7)

⇒ r

^{2 }= 49/4

⇒ r = 7/2 = 3.5 cm

**7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.**

**Answer**

Let the diameter of earth be r and that of the moon will be r/4

Radius of the earth = r/2

Radius of the moon = r/8

Ratio of their surface area = 4π(r/8)

^{2}/4π(r/2)

^{2}

^{ }= (1/64)/(1/4)

= 4/64 = 1/16

Thus, the ratio of their surface areas is 1:16

**8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.**

**Answer**

Inner radius of the bowl (r) = 5 cm

Thickness of the steel = 0.25 cm

∴ outer radius (R) = (r + 0.25) cm

= (5 + 0.25) cm = 5.25 cm

Outer curved surface = 2πR

^{2}

= (2 × 22/7 × 5.25 × 5.25) cm

^{2}

= 173.25 cm

^{2}

**9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find**

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii).

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii).

**Answer**

(i) The surface area of the sphere with raius r = 4πr

^{2}

(ii) The right circular cylinder just encloses a sphere of radius r.

∴ the radius of the cylinder = r and its height = 2r

∴ Curved surface of cylinder =2πrh

= 2π × r × 2r

= 4πr

^{2}

(iii) Ratio of the areas = 4πr

^{2}:4πr

^{2}= 1:1

**Exercise 13.5**

**1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?**

**Answer**

Dimension of matchbox = 4cm × 2.5cm × 1.5cm

l = 4 cm, b = 2.5 cm and h = 1.5 cm

Volume of one matchbox = (l × b × h)

^{}

= (4 × 2.5 × 1.5) cm

^{3 }= 15 cm

^{3}

Volume of a packet containing 12 such boxes = (12 × 15) cm

^{3 }= 180 cm

^{3}

**2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m**

^{3}= 1000 l)**Answer**

Dimensions of water tank = 6m × 5m × 4.5m

l = 6m , b = 5m and h = 4.5m

Therefore Volume of the tank =ℓbh m

^{3}

=(6×5×4.5)m3=135 m

^{3}

Therefore , the tank can hold = 135 × 1000 litres [Since 1m3=1000litres]

= 135000 litres of water.

**3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?**

**Answer**

Length = 10 m , Breadth = 8 m and Volume = 380 m

^{3}

Volume of cuboid = Length × Breadth × Height

⇒ Height = Volume of cuboid/(Length × Breadth)

= 380/(10×8) m

= 4.75m

**4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m**

^{3}.**Answer**

l = 8 m, b = 6 m and h = 3 m

Volume of the pit = lbh m

^{3}

= (8×6×3) m

^{3}

^{ }= 144 m

^{3}

Rate of digging = ₹30 per m

^{3}

Total cost of digging the pit = ₹(144 × 30) = ₹4320

**5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.**

**Answer**

length = 2.5 m, depth = 10 m and volume = 50000 litres

1m

^{3}= 1000 litres

^{}

∴ 50000 litres = 50000/1000 m

^{3}= 50 m

^{3}

Breadth = Volume of cuboid/(Length×Depth)

= 50/(2.5×10) m

= 2 m

**6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?**

**Answer**

Dimension of tank = 20m × 15m × 6m

l = 20 m , b = 15 m and h = 6 m

Capacity of the tank = lbh m

^{3}

= (20×15×6) m

^{3}

= 1800 m

^{3}

Water requirement per person per day =150 litres

Water required for 4000 person per day = (4000×150) l

= (4000×150)/1000

= 600 m

^{3}

Number of days the water will last = Capacity of tank Total water required per day

=(1800/600) = 3

The water will last for 3 days.

**7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.**

**Answer**

Dimension of godown = 40 m × 25 m × 15 m

Volume of the godown = (40 × 25 × 15) m

^{3}= 10000 m

^{3}

Dimension of crates = 1.5m × 1.25m × 0.5m

Volume of 1 crates = (1.5 × 1.25 × 0.5) m

^{3}= 0.9375 m

^{3}

Number of crates that can be stored =Volume of the godown/Volume of 1 crate

= 10000/0.9375 = 10666.66 = 10666

**8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.**

**Answer**

Edge of the cube = 12 cm.

Volume of the cube = (edge)

^{3}cm

^{3}

= (12 × 12 × 12) cm

^{3}

= 1728 cm

^{3}

Number of smaller cube = 8

Volume of the 1 smaller cube =1728/8 cm

^{3}= 216 cm

^{3}

Side of the smaller cube = a

a

^{3 }= 216

⇒ a = 6 cm

Surface area of the cube = 6 (side)

^{2}

Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)

= 4/1 = 4:1

**9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?**

**Answer**

Depth of river (h) = 3 m

Width of river (b) = 40 m

Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute

= 100/3 m per minute

Volume of water flowing into the sea in a minute = lbh m

^{3}

= (100/3 × 40 × 3) m

^{3}

= 4000 m

^{3}

**Exercise 13.6**

**1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm 3 = 1l)**

**Answer**

(i) Radius of the sphere (r) = 10.5 cm

**Exercise 13.7**

**1. Find the volume of the right circular cone with**

(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm

(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm

**Answer**

(i) Radius (r) = 6 cm

Height (h) = 7 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 6 × 6 × 7) cm

^{3}

= 264 cm

^{3}

(ii) Radius (r) = 3.5 cm

Height (h) = 12 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 3.5 × 3.5 × 12) cm

^{3}

= 154 cm

^{3}

**2. Find the capacity in litres of a conical vessel with**

(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

**Answer**

(i) Radius (r) = 7 cm

Slant height (l) = 25 cm

Let h be the height of the conical vessel.

∴ h = √l

^{2 }- r

^{2}

⇒ h = √25

^{2 }- 7

^{2}

⇒ h = √576

⇒ h = 24 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 7 × 7 × 24) cm

^{3}

= 1232 cm

^{3}

Capacity of the vessel = (1232/1000) l = 1.232 l

(i) Height (h) = 12 cm

Slant height (l) = 13 cm

Let r be the radius of the conical vessel.

∴ r = √l

^{2 }-

^{h}

^{2}

⇒ r = √13

^{2 }- 12

^{2}

⇒ r = √25

⇒ r = 5 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 5 × 5 × 12) cm

^{3}

= (2200/7) cm

^{3}

Capacity of the vessel = (2200/7000) l = 11/35 l

**3. The height of a cone is 15 cm. If its volume is 1570 cm**

^{3}, find the radius of the base. (Use π = 3.14)**Answer**

Height (h) = 15 cm

Volume = 1570 cm

^{3}

Let the radius of the base of cone be r cm

∴ Volume = 1570 cm

^{3}

⇒ 1/3 πr

^{2}h = 1570

⇒ 13 × 3.14 × r

^{2 }× 15 = 1570

⇒ r

^{2 }= 1570/(3.14×5) = 100

⇒ r = 10

**4. If the volume of a right circular cone of height 9 cm is 48π cm**

^{3}, find the diameter of its base.**Answer**

Height (h) = 9 cm

Volume = 48π cm

^{3}

Let the radius of the base of the cone be r cm

∴ Volume = 48π cm

^{3}

⇒ 1/3 πr

^{2}h = 48π

⇒ 13 × r

^{2 }× 9 = 48

⇒ 3r

^{2 }= 48

⇒ r

^{2 }= 48/3 = 16

⇒ r = 4

**5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?**

**Answer**

Diameter of the top of the conical pit = 3.5 m

Radius (r) = (3.5/2) m = 1.75 m

Depth of the pit (h) = 12 m

Volume = 1/3 πr

^{2}h

= (13 × 22/7 × 1.75 × 1.75 × 12) m

^{3}

= 38.5 m

^{3}

1 m

^{3 }= 1 kilolitre

^{}

Capacity of pit = 38.5 kilolitres.

6. The volume of a right circular cone is 9856 cm

(i) height of the cone (ii) slant height of the cone

(iii) curved surface area of the cone

6. The volume of a right circular cone is 9856 cm

^{3}. If the diameter of the base is 28 cm, find(i) height of the cone (ii) slant height of the cone

(iii) curved surface area of the cone

**Answer**

(i) Diameter of the base of the cone = 28 cm

Radius (r) = 28/2 cm = 14 cm

Let the height of the cone be h cm

Volume of the cone = 13πr

^{2}h = 9856 cm

^{3}

⇒ 1/3 πr

^{2}h = 9856

⇒ 1/3 × 22/7 × 14 × 14 × h = 9856

⇒ h = (9856×3)/(22/7 × 14 × 14)

⇒ h = 48 cm

(ii) Radius (r) = 14 m

Height (h) = 48 cm

Let l be the slant height of the cone

l

^{2}= h

^{2 }+ r

^{2}

⇒ l

^{2}= 48

^{2 }+ 14

^{2}

⇒ l

^{2}= 2304+196

⇒ l

^{2 }= 2500

⇒ ℓ = √2500 = 50 cm

(iii) Radius (r) = 14 m

Slant height (l) = 50 cm

Curved surface area = πrl

= (22/7 × 14 × 50) cm

^{2}

= 2200 cm

^{2}

**7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.**

**Answer**

On revolving the ⃤ ABC along the side 12 cm, a right circular cone of height(h) 12 cm, radius(r) 5 cm and slant height(l) 13 cm will be formed.

Volume of solid so obtained = 1/3 πr

^{2}h

= (1/3 × π × 5 × 5 × 12) cm

^{3}

= 100π cm

^{3}

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

**Answer**

On revolving the ⃤ ABC along the side 12 cm, a cone of radius(r) 12 cm, height(h) 5 cm, and slant height(l) 13 cm will be formed.

Volume of solid so obtained =1/3 πr

^{2}h

= (1/3 × π × 12 × 12 × 5) cm

^{3}

= 240π cm

^{3}

Ratio of the volumes = 100π/240π = 5/12 = 5:12

**9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.**

**Answer**

Diameter of the base of the cone = 10.5 m

Radius (r) = 10.5/2 m = 5.25 m

Height of the cone = 3 m

Volume of the heap = 1/3 πr

^{2}h

= (1/3 × 22/7 × 5.25 × 5.25 × 3) m

^{3}

= 86.625 m

^{3}

Also,

l

^{2}= h

^{2 }+ r

^{2}

⇒ l

^{2}= 3

^{2 }+ (5.25)

^{2}

⇒ l

^{2}= 9 + 27.5625

⇒ l

^{2 }= 36.5625

⇒ l = √36.5625 = 6.05 m

Area of canvas = Curve surface area

= πrl = (22/7 × 5.25 × 6.05) m

^{2}

= 99.825 m

^{2 }(approx)

**Exercise 13.8**

**1. Find the volume of a sphere whose radius is**

(i) 7 cm (ii) 0.63 m

(i) 7 cm (ii) 0.63 m

**Answer**

(i) Radius of the sphere(r) = 7 cm

Therefore, Volume of the sphere = 4/3 πr

^{3}

= (4/3 × 22/7 × 7 × 7 × 7) cm

^{3}

= 4312/3 cm

^{3}

(ii) Radius of the sphere(r) = 0.63 m

Volume of the sphere = 4/3 πr

^{3}

= (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m

^{3}

= 1.05 m

^{3}

**2. Find the amount of water displaced by a solid spherical ball of diameter.**

(i) 28 cm (ii) 0.21 m

(i) 28 cm (ii) 0.21 m

**Answer**

(i) Diameter of the spherical ball = 28 cm

Radius = 28/2 cm = 14 cm

Amount of water displaced by the spherical ball = Volume

= 4/3 πr

^{3}

= (4/3 × 22/7 × 14 × 14 × 14) cm

^{3}

= 34496/3 cm

^{3}

(ii) Diameter of the spherical ball = 0.21 m

Radius (r) = 0.21/2 m = 0.105 m

Amount of water displaced by the spherical ball = Volume

= 4/3 πr

^{3}

= (43×227×0.105×0.105×0.105) m

^{3}

= 0.004851 m

^{3}

**3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm**

^{3}?**Answer**

Diameter of the ball = 4.2 cm

Radius = (4.2/2) cm = 2.1 cm

Volume of the ball = 4/3 πr

^{3}

= (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm

^{3}

= 38.808 cm

^{3}

Density of the metal is 8.9g per cm3

Mass of the ball = (38.808 × 8.9) g = 345.3912 g

**4. The diameter of the moon is approximately one-fourth of the diameter of the earth.What fraction of the volume of the earth is the volume of the moon?**

**Answer**

Let the diameter of the moon be r

Radius of the moon = r/2

A/q,

Diameter of the earth = 4r

Radius(r) = 4r/2 = 2r

Volume of the moon = v = 4/3 π(r/2)

^{3}

= 4/3 πr

^{3 }× 1/8

⇒ 8v = 4/3 πr

^{3 }--- (i)

Volume of the earth = r

^{3}= 4/3 π(2r)

^{3}

= 4/3 πr

^{3}× 8

⇒ V/8 = 4/3 πr

^{3}--- (ii)

From (i) and (ii), we have

8v = V/8

⇒ v = 1/64 V

Thus, the volume of the moon is 1/64 of the volume of the earth.

**5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?**

**Answer**

Diameter of a hemispherical bowl = 10.5 cm

Radius(r) = (10.5/2) cm = 5.25cm

Volume of the bowl = 2/3 πr

^{3}

= (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm

^{3}

= 303.1875 cm

^{3}

Litres of milk bowl can hold = (303.1875/1000) litres

= 0.3031875 litres (approx.)

**6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.**

**Answer**

Internal radius = r = 1m

External radius = R = (1 + 0.1) cm = 1.01 cm

Volume of iron used = External volume – Internal volume

= 2/3 πR

^{3 }- 2/3 πr

^{3}

= 2/3 π(R

^{3 }- r

^{3})

= 2/3 × 22/7 × [(1.01)

^{3}−(1)

^{3}] m

^{3}

= 44/21 × (1.030301 - 1) m

^{3}

= (44/21 × 0.030301) m

^{3}

= 0.06348 m

^{3}(approx)

**7. Find the volume of a sphere whose surface area is 154 cm**

^{2}.**Answer**

Let r cm be the radius of the sphere

So, surface area = 154cm

^{2}

⇒ 4πr

^{2 }= 154

⇒ 4 × 22/7 × r

^{2 }= 154

⇒ r

^{2 }= (154×7)/(4×22) = 12.25

⇒ r = 3.5 cm

Volume = 4/3 πr

^{3}

= (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm

^{3}

= 539/3 cm

^{3}

**8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the**

(i) inside surface area of the dome, (ii) volume of the air inside the dome.

(i) inside surface area of the dome, (ii) volume of the air inside the dome.

**Answer**

(i) Inside surface area of the dome =Total cost of white washing/Rate of white washing

= (498.96/2.00) m

^{2 }= 249.48 m

^{2 }

(ii) Let r be the radius of the dome.

Surface area = 2πr

^{2}

^{}

⇒ 2 × 22/7 × r

^{2 }= 249.48

⇒ r

^{2 }= (249.48×7)/(2×22) = 39.69

⇒ r

^{2}= 39.69

⇒ r = 6.3m

Volume of the air inside the dome = Volume of the dome

= 2/3 πr

^{3}

^{ }= (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m

^{3}

= 523.9 m

^{3 }(approx.)

**9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the**

(i) radius r′ of the new sphere, (ii) ratio of S and S′.

(i) radius r′ of the new sphere, (ii) ratio of S and S′.

**Answer**

(i) Volume of 27 solid sphere of radius r = 27 × 4/3 πr

^{3}--- (i)

Volume of the new sphere of radius r′ = 4/3 πr'

^{3}--- (ii)

A/q,

4/3 πr'

^{3}= 27 × 4/3 πr

^{3}

⇒ r'

^{3 }= 27r

^{3}

⇒ r'

^{3 }= (3r)

^{3}

⇒ r′ = 3r

(ii) Required ratio = S/S′ = 4 πr

^{2}/4πr′

^{2 }= r

^{2}/(3r)

^{2}

= r

^{2}/9r

^{2}= 1/9 = 1:9

**10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm**

^{3}) is needed to fill this capsule?**Answer**

Diameter of the spherical capsule = 3.5 mm

Radius(r) = 3.52mm

= 1.75mm

Medicine needed for its filling = Volume of spherical capsule

= 4/3 πr

^{3}

= (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm

^{3}

= 22.46 mm

^{3 }(approx.)

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