RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Question 1.
Find the surface area of a sphere of radius :
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Answer;
(i) We have : r = radius of the sphere = 10.5 cm
∴ Surface area = 4πr² = (4 × \(\frac{22}{7}\) × 10.5 × 10.5) cm² = 1386 cm²
(ii) We have : r = radius of the sphere = 5.6 cm
∴ Surface area = 4πr² = (4 × \(\frac{22}{7}\) × 5.6 × 5.6) cm² = 394.24 cm²
(iii) We have : r = radius of the sphere = 14 cm
∴ Surface area = 14 × \(\frac{22}{7}\) × 14 × 14) = 2464 cm².

Question 2.
Find the surface area of a sphere of diameter : .
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Answer:
(i) Here r = \(\left(\frac{14}{2}\right)\)cm = 7 cm².
Surface area = 4πr² = (4 × \(\frac{22}{7}\) × 7 × 7) cm² = 616 cm²

(ii) Here r = \(\left(\frac{21}{2}\right)\) cm = 10.5 cm
Surface area = 4πr² = (4 × \(\frac{22}{7}\) × 10.5 × 10.5)cm² = 1386 cm²

(iii) Here r = cm = 1.75 cm
Surface area = 4πr² = (4 × \(\frac{22}{7}\) × 1.75 × 1.75)cm² = 38.5 cm².

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Answer:
Here r = 10 cm
Total surface area of hemisphere = 3πr²
= (3 × 3.14 × 10 × 10) cm² = 942 cm².

Question 4.
The radius of a 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.
Answer:
Let r₁ and r₂ be the radii of balloons in the two cases.
Here r₁ = 7 cm and r₂ = 14 cm
∴ Ratio of their surface areas = \(\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{7 \times 7}{14 \times 14}=\frac{1}{4}\)
Thus, the required ratio of their surface areas is 1 : 4.

Question 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 cm².
Answer:
Here r = \(\left(\frac{10.5}{2}\right)\) cm = 5.25 cm
Curved surface area of the hemisphere = 2πr² = (2 × \(\frac{22}{7}\) × 5.25 × 5.25)
= 173.25 cm²
Rate of tin-plating is ₹ 16 per 100 cm².
∴ Cost of tin-plating the inside of the hemisphere
Thus, the radius of the sphere is 3.5 cm.

Question 6.
Fiüd the radius of a sphere whose surface area is 154 cm².
Answer:
Let r be the radius of the sphere.
Surface area = 154 cm² ⇒ 4πr² = 154
So, 4 × \(\frac{22}{7}\) × r² = 154 ⇒ r² = \(\frac{154 \times 7}{4 \times 22}=\frac{49}{4}\)
r = \(\frac{7}{2}\) =3.5
Thus, the radius of the sphere 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.
Answer:
Let the diameter of earth be R and that of the moon will be \(\frac{R}{2}\)
∴ Ratio of their Surface areas = \(\frac{4 \pi\left(\frac{R}{8}\right)^{2}}{4 \pi\left(\frac{R}{2}\right)^{2}}=\frac{\frac{1}{64}}{\frac{1}{4}}\)
= \(\frac{1}{64} \times \frac{4}{1}=\frac{1}{16}\), i.e 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 curved surface area of the bowl.
Answer:
Inner radius r = 5 cm
Thickness of steel = 0.25 cm
∴ Outer radius R = (r + 0.25) cm = (5 + 0.25) cm = 5.25 cm
∴ Outer curved surface area = 2πR²
= (2 ×\( \frac{22}{7}\) × 5.25 × 5.25)cm² = 173.25 cm²

Question 9.
A right circular cylinder just encloses a sphere of radius r (see figure). Find :

(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 radius of the sphere is r. So, its surface area = 4πr².
(ii) Since, the right circular cylinder just encloses a sphere of radius r, therefore the radius of cylinder = r and its height = 2r.
Curved surface area of cylinder = 2πr (2r) (∵ r = r, h = r)
= 4πr²
(iii) Ratio of areas= 4πr² : 4πr² =1:1.