Here we provide NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Class 9 Maths solution Chapter 13 pdf. Now you will get step by step solution to each question. = 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.

**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)

Page No: 236

**Exercise 13.8**

1. Find the volume of a sphere whose radius is

(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

**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.

**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′.

**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 }

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?

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.)

**All Chapter NCERT Solutions For Class 9 Maths**

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