NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes (Ex 13.6 – Ex 13.8)

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 cm2

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 πr2h
                                             = (1/3 × π × 5 × 5 × 12) cm3
                                             = 100π cm3

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 πr2h
                                              = (1/3 × π × 12 × 12 × 5) cm3
                                              = 240π cm3
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 πr2h
                                = (1/3 × 22/7 × 5.25 × 5.25 × 3) m3
                                = 86.625 m3
Also,
l2 = h+ r2
⇒ l2 = 3+ (5.25)2
⇒ l2 = 9 + 27.5625
⇒ l= 36.5625
⇒ l = √36.5625 = 6.05 m
Area of canvas = Curve surface area
                         = πrl = (22/7 × 5.25 × 6.05) m2
                         = 99.825 m(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 πr3
                                                     = (4/3 × 22/7 × 7 × 7 × 7) cm3
                                                     = 4312/3 cm3

(ii) Radius of the sphere(r) = 0.63 m
Volume of the sphere = 4/3 πr3
                                   = (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m3
                                   = 1.05 m3

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 πr3
                              = (4/3 × 22/7 × 14 × 14 × 14) cm3
                              = 34496/3 cm3

(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 πr3
                              = (43×227×0.105×0.105×0.105) m3
                              = 0.004851 m3

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 cm3?

Answer

Diameter of the ball = 4.2 cm
Radius = (4.2/2) cm = 2.1 cm
Volume of the ball = 4/3 πr3
                               = (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm3
                               = 38.808 cm3
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× 1/8
⇒ 8v = 4/3 πr— (i)
Volume of the earth = r3 = 4/3 π(2r)3
                                = 4/3 πr3× 8
⇒ V/8 = 4/3 πr3 — (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 πr3
                                = (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm3
                                = 303.1875 cm3
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– 2/3 πr3
                                  = 2/3 π(R– r3)
                                  = 2/3 × 22/7 × [(1.01)3−(1)3] m3
                                  = 44/21 × (1.030301 – 1) m3
                                  = (44/21 × 0.030301) m3
                                  = 0.06348 m3(approx)

7. Find the volume of a sphere whose surface area is 154 cm2.

Answer

Let r cm be the radius of the sphere
So, surface area = 154cm2
⇒ 4πr= 154
⇒ 4 × 22/7 × r= 154
⇒ r= (154×7)/(4×22) = 12.25
⇒ r = 3.5 cm
Volume = 4/3 πr3
             = (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm3
             = 539/3 cm3

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= 249.48 m

 (ii) Let r be the radius of the dome.
Surface area = 2πr2
⇒ 2 × 22/7 × r= 249.48
⇒ r= (249.48×7)/(2×22) = 39.69
⇒ r2=  39.69
⇒ r = 6.3m
Volume of the air inside the dome = Volume of the dome
                                                       = 2/3 πr3
                                                                    = (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m3
                                                       = 523.9 m(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 πr3 — (i)
Volume of the new sphere of radius r′ = 4/3 πr’3 — (ii)
A/q,
4/3 πr’3= 27 × 4/3 πr3
⇒ r’= 27r3
⇒ r’

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) 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 πr3
                                                  = (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm3
                                                  = 22.46 mm(approx.)

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