i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm Formula for Surface area of sphere = 4πr2 = 4×(22/7)×72 = 616 Surface area of the sphere is 616 cm2 (ii) Radius (r) of the sphere = 21/2 = 10.5 cm Surface area of a sphere = 4πr2 = 4×(22/7)×10.52 = 1386 Surface area of the sphere is 1386 cm2 Therefore, the surface area of a sphere having diameter 21cm is 1386 cm2 (iii) Radius(r) of sphere = 3.5/2 = 1.75 cm Surface area of a sphere = 4πr2 = 4×(22/7)×1.752 = 38.5 Surface area of the sphere is 38.5 cm2 3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14] Solution: Radius of the hemisphere, r = 10cm Formula: Total surface area of hemisphere = 3πr2 = 3×3.14×102 = 942 The total surface area of the given hemisphere is 942 cm2. 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. Solution: Let r1 and r2 be the radii of the spherical balloon and spherical balloon when air is pumped into it, respectively. So, r1 = 7cm r2 = 14 cm Now, required ratio = (initial surface area)/(Surface area after pumping air into balloon) = 4πr12/4πr22 = (r1/r2)2 = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4. 5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7) Solution: Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm Formula for Surface area of hemispherical bowl = 2πr2 = 2×(22/7)×(5.25)2 = 173.25 Surface area of hemispherical bowl is 173.25 cm2 Cost of tin-plating 100 cm2 area = Rs 16 Cost of tin-plating 1 cm2 area = Rs 16 /100 Cost of tin-plating 173.25 cm2 area = Rs. (16×173.25)/100 = Rs 27.72 Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72. 6. Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7) Solution: Let the radius of the sphere be r. Surface area of sphere = 154 (given) Now, 4πr2 = 154 r2 = (154×7)/(4×22) = (49/4) r = (7/2) = 3.5 The radius of the sphere is 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. Solution: If the diameter of the earth is d, then the diameter of the moon will be d/4 (as per the given statement) Radius of earth = d/2 Radius of moon = ½×d/4 = d/8 Surface area of moon = 4π(d/8)2 Surface area of earth = 4π(d/2)2  The ratio between 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 5cm. Find the outer curved surface of the bowl. (Assume π =22/7) Solution: Given: Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2πr2, where r is the radius of the hemisphere = 2×(22/7)×(5.25)2 = 173.25 cm2 Therefore, the outer curved surface area of the bowl is 173.25 cm2. 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).  Solution: (i) Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder) = 4πr2/4πr2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1. Exercise 13.4 of Class 9 Maths involves application-level real-time problems that help students to think and apply the relevant formula. It helps to apply the total surface area of a sphere and hemisphere. Learn the NCERT Solutions of Class 9 Maths Chapter 13 along with other learning materials and notes provided by BYJU’S. The problems are solved in a detailed way with relevant formulas and figures to score well in the CBSE exams. Key Features of NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volume Exercise 13.4 These NCERT Solutions help you solve and revise all questions of Exercise 13.4.Helps to find the radius of a sphere and the total surface area of a hemisphere.It follows NCERT guidelines which help in preparing the students accordingly.Stepwise solutions given by our subject expert teachers will help you to secure more marks.
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i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm
Formula for Surface area of sphere = 4πr2
= 4×(22/7)×72 = 616
Surface area of the sphere is 616 cm2
(ii) Radius (r) of the sphere = 21/2 = 10.5 cm
Surface area of a sphere = 4πr2
= 4×(22/7)×10.52 = 1386
Surface area of the sphere is 1386 cm2
Therefore, the surface area of a sphere having diameter 21cm is 1386 cm2
(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm
Surface area of a sphere = 4πr2
= 4×(22/7)×1.752 = 38.5
Surface area of the sphere is 38.5 cm2
3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]
Solution:
Radius of the hemisphere, r = 10cm
Formula: Total surface area of hemisphere = 3πr2
= 3×3.14×102 = 942
The total surface area of the given hemisphere is 942 cm2.
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.
Solution:
Let r1 and r2 be the radii of the spherical balloon and spherical balloon when air is pumped into it, respectively. So,
r1 = 7cm
r2 = 14 cm
Now, required ratio = (initial surface area)/(Surface area after pumping air into balloon)
= 4πr12/4πr22
= (r1/r2)2
= (7/14)2 = (1/2)2 = ¼
Therefore, the ratio between the surface areas is 1:4.
5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7)
Solution:
Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm
Formula for Surface area of hemispherical bowl = 2πr2
= 2×(22/7)×(5.25)2 = 173.25
Surface area of hemispherical bowl is 173.25 cm2
Cost of tin-plating 100 cm2 area = Rs 16
Cost of tin-plating 1 cm2 area = Rs 16 /100
Cost of tin-plating 173.25 cm2 area = Rs. (16×173.25)/100 = Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.
6. Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)
Solution:
Let the radius of the sphere be r.
Surface area of sphere = 154 (given)
Now,
4πr2 = 154
r2 = (154×7)/(4×22) = (49/4)
r = (7/2) = 3.5
The radius of the sphere is 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.
Solution:
If the diameter of the earth is d, then the diameter of the moon will be d/4 (as per the given statement)
Radius of earth = d/2
Radius of moon = ½×d/4 = d/8
Surface area of moon = 4π(d/8)2
Surface area of earth = 4π(d/2)2

The ratio between 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 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)
Solution:
Given:
Inner radius of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm
Formula for outer CSA of hemispherical bowl = 2πr2, where r is the radius of the hemisphere
= 2×(22/7)×(5.25)2 = 173.25 cm2
Therefore, the outer curved surface area of the bowl is 173.25 cm2.
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).

Solution:
(i) Surface area of sphere = 4πr2, where r is the radius of sphere
(ii) Height of cylinder, h = r+r =2r
Radius of cylinder = r
CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)
= 4πr2
(iii) Ratio between areas = (Surface area of sphere)/(CSA of Cylinder)
= 4πr2/4πr2 = 1/1
Ratio of the areas obtained in (i) and (ii) is 1:1.
Exercise 13.4 of Class 9 Maths involves application-level real-time problems that help students to think and apply the relevant formula. It helps to apply the total surface area of a sphere and hemisphere.
Learn the NCERT Solutions of Class 9 Maths Chapter 13 along with other learning materials and notes provided by BYJU’S. The problems are solved in a detailed way with relevant formulas and figures to score well in the CBSE exams.
Key Features of NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volume Exercise 13.4
These NCERT Solutions help you solve and revise all questions of Exercise 13.4.Helps to find the radius of a sphere and the total surface area of a hemisphere.It follows NCERT guidelines which help in preparing the students accordingly.Stepwise solutions given by our subject expert teachers will help you to secure more marks.
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