These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4
Question 1.
Find the surface area of a sphere of radius.
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
(i) We have given that
Radius of sphere = 10.5 cm
Surface area of sphere = 4πr2
= 4 × \(\frac {22}{7}\) × 5.6 × 5.6
= 394.24 cm2
(ii) We have given that
Radius of sphere = 5.6 cm
Surface area of sphere = 4πr2
= 4 × \(\frac {22}{7}\) × 5.6 × 5.6
= 394.24 cm2
(iii) We have given that
Radius of sphere = 14 cm
Surface area of sphere = 4πr2
= 4 × \(\frac {22}{7}\) × 14 × 14
= 2464 cm2
Question 2.
Find the surface area of a sphere of diameter.
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
Solution:
(i) We have given that
Diameter of sphere = 14 cm
Radius of sphere = 7 cm
Surface area of sphere = 4πr2
= 4 × \(\frac {22}{7}\) × 7 × 7
= 616 cm2
(ii) We have given that
Diameter of sphere = 21 cm
Radius of sphere = 10.5 cm
Therefore,
Surface area of sphere = 4πr2
= 4 × \(\frac {22}{7}\) × 10.5 × 10.5
= 1386 cm2
(iii) We have given that
Diameter of sphere = 3.5 m
Radius os sphere = 1.75 m
Therefore,
Surface area of sphere = 4πr2
= 4 × \(\frac {22}{7}\) × 1.75 × 1.75
= 38.5 m2
Question 3.
Find the total surface area of a hemisphere of radius 10 cm.
Solution:
We have given that
Radius of hemisphere = 10 cm
Therefore, Total surface area of hemisphere = 3πr2
= 3 × 3.14 × 10 × 10 (∴ π = 3.14)
= 942 cm2
Question 4.
The radius of the spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
In first case,
Radius of spherical balloon = 7 cm
Surface area of spherical balloon = 4πr2
= 4 × \(\frac {22}{7}\) × 7 × 7
= 616 cm2
In second case,
radius of spherical balloon = 14 cm
surface area of spherical balloon = 4πr2
= 4 × \(\frac {22}{7}\) × 14 × 14
= 2464 cm2
Therefore the ratio of the surface area of balloons in two cases is 616 : 2464 = 1 : 4
Question 5.
A hemispherical bowl made of brass has an inner diameter of 10.2. Find the cost of tin plating it on the inside at the rate of Rs. 16 per 100 cm2.
Solution:
We have given that
Diameter of hemispherical bowl = 10.5 cm
The radius of hemispherical bowl = 5.25 cm
Curved surface area of hemispherical bowl = 2πr2
= 2 × \(\frac {22}{7}\) × 5.25 × 5.25
= 173.25 cm2
Rate of tin plating = Rs. 16 per 100 cm2
Total cost of tin plating the hemispherical bowl is = 173.25 × \(\frac{16}{100}\) = Rs. 27.72
Question 6.
Find the radius of a sphere whose surface area is 154 cm2.
Solution:
We have given that
Surface area of sphere = 154 cm2
But, Surface area of sphere = 2πr2
⇒ 154 = 4 × \(\frac {22}{7}\) × r2
⇒ r2 = \(\frac{154 \times 7}{22 \times 4}\) = 12.25
⇒ r = √12.25 = 3.5 cm
Therefore, the radius of the sphere whose surface area is 154 cm2 is 3.5 cm.
Question 7.
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Let the diameter of earth = r cm
Diameter of moon = \(\frac{r}{4}\) cm
Again, radius earth = \(\frac{1}{2}\) cm
and radius of moon = \(\frac{r}{8}\) cm
Surface area of earth = 4πr2
= 4 × π × \(\left(\frac{r}{2}\right)^{2}\)
= πr2
and surface area of moon = 4π \(\left(\frac{r}{8}\right)^{2}\) = \(\frac{\pi r^{2}}{16}\)
Ratio of their surface areas = \(\frac{\pi r^{2}}{\frac{16}{\pi r^{2}}}\) = \(\frac{1}{16}\) = 1 : 16
Ratio of moon and earth surface area = 1 : 16
Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner, radius of the bowl is 5 cm. Find the outer curves surface area of the bowl.
Solution:
We have given tha
Inner radius of hemispherical bowl = 5 cm
and thickness of steel = 0.25 cm.
Outer curved surface area of hemispherical bowl = 5.25 cm
Therefore, Outer curved surface area of hemisperical bowl = 2πr2
= 2 × \(\frac {22}{7}\) × 5.25 × 5.25
= 173.25 cm2
Question 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 areas obtained in (i) and (ii)
Solution:
(i) We have given that
Radius of sphere = r
Surface area of sphere = 4πr2
(ii) Surface area of cylinder = 4πrh
= 2 × π × r × 2r (∵ h = 2r)
= 4πr2
(iii) Ratio of the areas obtaining in (i) and (ii) is
\(\frac{4 \pi r^{2}}{4 \pi r^{2}}\) = 1 : 1