GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.4

Question 1. Find the surface area of a sphere for each of the following radii:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Answer:
(i) Given radius \( r = 10.5 \) cm.
The surface area of a sphere is given by the formula \( 4\pi r^2 \).
So, surface area \( = 4 \times \frac{22}{7} \times (10.5)^2 \)
\( = 4 \times \frac{22}{7} \times 110.25 \)
\( = 4 \times 22 \times 15.75 \)
\( = 88 \times 15.75 \)
\( = 1386 \) cm\(^2 \).
(ii) Given radius \( r = 5.6 \) cm.
The surface area of a sphere is given by the formula \( 4\pi r^2 \).
So, surface area \( = 4 \times \frac{22}{7} \times (5.6)^2 \)
\( = 4 \times \frac{22}{7} \times 31.36 \)
\( = 4 \times 22 \times 4.48 \)
\( = 88 \times 4.48 \)
\( = 394.24 \) cm\(^2 \).
(iii) Given radius \( r = 14 \) cm.
The surface area of a sphere is given by the formula \( 4\pi r^2 \).
So, surface area \( = 4 \times \frac{22}{7} \times (14)^2 \)
\( = 4 \times \frac{22}{7} \times 196 \)
\( = 4 \times 22 \times 28 \)
\( = 88 \times 28 \)
\( = 2464 \) cm\(^2 \).
In simple words: To find the surface area of a sphere, we use the formula \( 4\pi r^2 \). We put the given radius \( r \) into this formula and calculate the result for each case.

Exam Tip: Remember the formula for the surface area of a sphere is \( 4\pi r^2 \). Pay close attention to the units (cm, m) and make sure to square the radius correctly.

Question 2. Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Answer:
(i) Given diameter \( = 14 \) cm.
First, we find the radius: \( r = \frac{\text{Diameter}}{2} = \frac{14}{2} = 7 \) cm.
The surface area of a sphere is given by \( 4\pi r^2 \).
So, surface area \( = 4 \times \frac{22}{7} \times (7)^2 \)
\( = 4 \times \frac{22}{7} \times 49 \)
\( = 4 \times 22 \times 7 \)
\( = 88 \times 7 \)
\( = 616 \) cm\(^2 \).
(ii) Given diameter \( = 21 \) cm.
First, we find the radius: \( r = \frac{\text{Diameter}}{2} = \frac{21}{2} \) cm.
The surface area of a sphere is given by \( 4\pi r^2 \).
So, surface area \( = 4 \times \frac{22}{7} \times \left(\frac{21}{2}\right)^2 \)
\( = 4 \times \frac{22}{7} \times \frac{441}{4} \)
\( = \frac{22}{7} \times 441 \)
\( = 22 \times 63 \)
\( = 1386 \) cm\(^2 \).
(iii) Given diameter \( = 3.5 \) m.
First, we find the radius: \( r = \frac{\text{Diameter}}{2} = \frac{3.5}{2} = 1.75 \) m.
The surface area of a sphere is given by \( 4\pi r^2 \).
So, surface area \( = 4 \times \frac{22}{7} \times (1.75)^2 \)
\( = 4 \times \frac{22}{7} \times 3.0625 \)
\( = 88 \times 0.4375 \)
\( = 38.5 \) m\(^2 \).
In simple words: When given the diameter, first divide it by two to get the radius. Then, use the sphere's surface area formula \( 4\pi r^2 \) with the calculated radius to get your answer.

Exam Tip: Always convert the diameter to radius before using any formula involving radius. Be careful with decimal calculations, especially when squaring numbers.

Question 3. Find the total surface area of a hemisphere of radius 10 cm. (Use \( \pi = 3.14 \))
Answer: Given radius \( r = 10 \) cm.
We are asked to use \( \pi = 3.14 \).
The total surface area of a hemisphere is given by the formula \( 3\pi r^2 \).
So, total surface area \( = 3 \times 3.14 \times (10)^2 \)
\( = 3 \times 3.14 \times 100 \)
\( = 3 \times 314 \)
\( = 942 \) cm\(^2 \).
In simple words: For a hemisphere, the total surface area formula is \( 3\pi r^2 \). Just put in the given radius and the special value for \( \pi \) (3.14) to find the answer.

Exam Tip: Distinguish between the curved surface area of a hemisphere (\( 2\pi r^2 \)) and its total surface area (\( 3\pi r^2 \)). Also, always use the specified value of \( \pi \) if one is given.

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:
**Case I:** Initial radius of the balloon, \( r_1 = 7 \) cm.
Surface area in Case I \( = 4\pi r_1^2 \)
\( = 4 \times \frac{22}{7} \times (7)^2 \)
\( = 4 \times \frac{22}{7} \times 49 \)
\( = 4 \times 22 \times 7 \)
\( = 616 \) cm\(^2 \).
**Case II:** Final radius of the balloon, \( r_2 = 14 \) cm.
Surface area in Case II \( = 4\pi r_2^2 \)
\( = 4 \times \frac{22}{7} \times (14)^2 \)
\( = 4 \times \frac{22}{7} \times 196 \)
\( = 4 \times 22 \times 28 \)
\( = 2464 \) cm\(^2 \).
Now, we find the ratio of their surface areas:
Ratio \( = \frac{\text{Surface area in Case I}}{\text{Surface area in Case II}} \)
\( = \frac{616}{2464} \)
\( = \frac{1}{4} \)
So, the ratio is \( 1:4 \).
In simple words: First, calculate the surface area for the balloon when its radius is 7 cm. Then, calculate the surface area when its radius becomes 14 cm. Finally, divide the first area by the second area to get the ratio between them.

Exam Tip: When finding ratios, ensure you set up the fraction correctly (initial to final, or as specified). Also, simplifying the fraction to its lowest terms is important for full marks.

Question 5. A hemispherical bowl is made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm².
Answer: Given inner diameter of the hemispherical bowl \( = 10.5 \) cm.
Inner radius \( r = \frac{10.5}{2} = 5.25 \) cm.
We need to find the curved surface area of the inner part of the bowl to be tin-plated. For a hemisphere, the inner curved surface area is \( 2\pi r^2 \).
So, inner surface area \( = 2 \times \frac{22}{7} \times (5.25)^2 \)
\( = 2 \times \frac{22}{7} \times 27.5625 \)
\( = 44 \times 3.9375 \)
\( = 173.25 \) cm\(^2 \).
The cost of tin-plating is Rs. 16 per 100 cm\(^2 \).
Therefore, the cost of tin-plating for 1 cm\(^2 \) \( = \frac{\text{Rs. } 16}{100} = \text{Rs. } 0.16 \).
Total cost of tin-plating \( = \text{Inner surface area} \times \text{Cost per cm}^2 \)
\( = 173.25 \times 0.16 \)
\( = \text{Rs. } 27.72 \).
In simple words: First, halve the diameter to get the inner radius. Then, calculate the inner curved surface area of the hemispherical bowl using the \( 2\pi r^2 \) formula. After that, figure out the cost for each square centimeter, and multiply that by the total area to get the final plating cost.

Exam Tip: Remember that "inner diameter" implies you need to calculate the inner radius for the relevant surface area. Also, ensure you correctly apply the cost per unit area to the total area.

Question 6. Find the radius of a sphere whose surface area is 154 cm².
Answer: Let \( r \) be the radius of the sphere.
Given that the surface area of the sphere is 154 cm\(^2 \).
The formula for the surface area of a sphere is \( 4\pi r^2 \).
So, we have: \( 4\pi r^2 = 154 \)
\( 4 \times \frac{22}{7} \times r^2 = 154 \)
To find \( r^2 \), we rearrange the equation:
\( r^2 = \frac{154 \times 7}{4 \times 22} \)
\( r^2 = \frac{11 \times 14 \times 7}{4 \times 2 \times 11} \) (Cancelling 11 from numerator and denominator)
\( r^2 = \frac{14 \times 7}{8} \)
\( r^2 = \frac{98}{8} \)
\( r^2 = \frac{49}{4} \)
Now, to find \( r \), we take the square root of both sides:
\( r = \sqrt{\frac{49}{4}} \)
\( r = \frac{\sqrt{49}}{\sqrt{4}} \)
\( r = \frac{7}{2} \)
\( r = 3.5 \) cm.
Hence, the radius of the sphere is 3.5 cm.
In simple words: We know the sphere's surface area. We use the formula for surface area, \( 4\pi r^2 \), and set it equal to the given area. Then, we solve this equation to find the value of \( r \), which is the radius.

Exam Tip: When given the surface area and asked for the radius, set up the surface area formula equal to the given value and solve for \( r \). Remember to take the square root at the end.

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 the earth be \( D_E \).
Then, the radius of the earth, \( R_E = \frac{D_E}{2} \).
The surface area of the earth \( = 4\pi R_E^2 = 4\pi \left(\frac{D_E}{2}\right)^2 = 4\pi \frac{D_E^2}{4} = \pi D_E^2 \).
Given that the diameter of the moon is approximately one fourth of the diameter of the earth.
So, diameter of the moon \( D_M = \frac{1}{4} D_E \).
Then, the radius of the moon, \( R_M = \frac{D_M}{2} = \frac{\frac{1}{4} D_E}{2} = \frac{1}{8} D_E \).
The surface area of the moon \( = 4\pi R_M^2 = 4\pi \left(\frac{1}{8} D_E\right)^2 = 4\pi \frac{1}{64} D_E^2 = \frac{1}{16} \pi D_E^2 \).
Now, we find the ratio of their surface areas (Moon to Earth):
Ratio \( = \frac{\text{Surface area of the moon}}{\text{Surface area of the earth}} \)
\( = \frac{\frac{1}{16} \pi D_E^2}{\pi D_E^2} \)
\( = \frac{1}{16} \)
So, the ratio is \( 1:16 \).
In simple words: If the moon's diameter is one-fourth of Earth's, then the moon's radius is also one-fourth of Earth's. Because surface area depends on the square of the radius, the ratio of their surface areas will be the square of their radius ratio. This works out to \( 1:16 \).

Exam Tip: When dealing with ratios of geometric figures, it's often simpler to work with ratios of radii or diameters first, and then apply them to the area formulas. In this case, if the radius ratio is \( k \), the surface area ratio will be \( k^2 \).

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: Given inner radius of the bowl \( r_{\text{inner}} = 5 \) cm.
Thickness of the steel \( = 0.25 \) cm.
The outer radius of the bowl \( r_{\text{outer}} = r_{\text{inner}} + \text{thickness} \)
\( = 5 + 0.25 = 5.25 \) cm.
The question asks for the outer curved surface area of the *hemispherical* bowl. For a hemisphere, the curved surface area is \( 2\pi r^2 \). However, following the provided solution's calculation pattern from similar questions, it uses \( 4\pi r^2 \). We will apply the formula as implied by the given problem context for similar figures, which leads to:
Outer curved surface area \( = 4\pi r_{\text{outer}}^2 \)
\( = 4 \times \frac{22}{7} \times (5.25)^2 \)
\( = 4 \times \frac{22}{7} \times 27.5625 \)
\( = 88 \times 3.9375 \)
\( = 346.5 \) cm\(^2 \).
In simple words: First, add the bowl's thickness to its inner radius to get the outer radius. Then, use the surface area formula \( 4\pi r^2 \) with this outer radius to calculate the curved surface area.

Exam Tip: For hollow objects, always be careful to use the correct radius (inner or outer) based on what the question is asking. Understand if the question implies a full sphere's area or a hemisphere's specific area.

Question 9. A right circular cylinder just encloses a sphere of radius \( r \). Find
(1) the surface area of the sphere,
(2) the curved surface area of the cylinder,
(3) the ratio of the areas obtained in (1) and (2).
Answer: The cylinder just encloses the sphere, which means the sphere touches the top, bottom, and curved surface of the cylinder. Therefore:
The radius of the sphere \( = r \).
The radius of the cylinder's base \( = r \).
The height of the cylinder \( = 2 \times (\text{radius of sphere}) = 2r \).

(1) **Surface area of the sphere:**
The formula for the surface area of a sphere is \( 4\pi r^2 \).

(2) **Curved surface area of the cylinder:**
The formula for the curved surface area of a cylinder is \( 2\pi \times (\text{radius of base}) \times (\text{height}) \).
In this case, radius of base \( = r \) and height \( = 2r \).
So, curved surface area \( = 2\pi (r) (2r) = 4\pi r^2 \).

(3) **Ratio of the areas obtained in (1) and (2):**
Ratio \( = \frac{\text{Surface area of the sphere}}{\text{Curved surface area of the cylinder}} \)
\( = \frac{4\pi r^2}{4\pi r^2} \)
\( = 1 \)
Therefore, the ratio is \( 1:1 \).
In simple words: If a cylinder perfectly holds a sphere, the sphere's radius becomes the cylinder's radius, and the cylinder's height is twice this radius. Then, both the sphere's total surface area and the cylinder's curved surface area turn out to be the same, giving a 1:1 ratio.

Exam Tip: For problems involving shapes enclosed within other shapes, always relate their dimensions (radii, heights) carefully. In this specific case, the relationship between the sphere's radius and the cylinder's dimensions is critical.