GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.8

Question 1. Find the volume of a sphere whose radius is:
(i) 7 cm
(ii) 0.63 m
Answer:
(i) Given, the radius \( r = 7 \) cm.
The volume of a sphere is calculated using the formula \( V = \frac{4}{3} \pi r^3 \).
So, \( V = \frac{4}{3} \times \frac{22}{7} \times (7)^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 343 \)
\( = \frac{4 \times 22 \times 49}{3} \)
\( = \frac{4312}{3} \)
\( = 1437 \frac{1}{3} \) cm³
(ii) Given, the radius \( r = 0.63 \) m.
The volume of a sphere is \( V = \frac{4}{3} \pi r^3 \).
So, \( V = \frac{4}{3} \times \frac{22}{7} \times (0.63)^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 0.250047 \)
\( = \frac{22004.136}{21} \)
\( = 1.05 \) m³ (approximately)
In simple words: To find the volume of a sphere, you use the formula with its radius. For 7 cm, the volume is \( 1437 \frac{1}{3} \) cm³, and for 0.63 m, it's about 1.05 m³.

Exam Tip: Remember to use the correct units (cm³ or m³) for volume and to carry out calculations accurately, especially with decimals.

Question 2. Find the amount of water displaced by a solid spherical ball of diameter:
1. 28 cm
2. 0.21 m
Answer:
1. When the diameter is 28 cm:
The diameter of the ball is 28 cm.
The radius \( (r) = \frac{28}{2} \) cm \( = 14 \) cm.
The volume of water displaced is equal to the ball's volume, calculated by \( \frac{4}{3} \pi r^3 \).
So, \( = \frac{4}{3} \times \frac{22}{7} \times (14)^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 2744 \)
\( = \frac{4 \times 22 \times 392}{3} \)
\( = \frac{34496}{3} \) cm³
\( = 11498 \frac{2}{3} \) cm³
2. When the diameter is 0.21 m:
The diameter of the ball is 0.21 m.
The radius \( (r) = \frac{0.21}{2} \) m \( = 0.105 \) m.
The volume of water displaced is \( \frac{4}{3} \pi r^3 \).
So, \( = \frac{4}{3} \times \frac{22}{7} \times (0.105)^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 0.001157625 \)
\( = 0.004851 \) m³
In simple words: The volume of water a spherical ball pushes out is the same as the ball's own volume. We find the radius from the diameter, then use the sphere volume formula to get the answer.

Exam Tip: Always remember that the amount of fluid displaced by an object is equal to the volume of the object itself. Pay close attention to units (cm vs. m) in your calculations.

Question 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:
Given, the diameter of the metal ball is 4.2 cm.
The radius \( (r) = \frac{4.2}{2} \) cm \( = 2.1 \) cm.
The volume of the ball \( V = \frac{4}{3} \pi r^3 \).
So, \( V = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 9.261 \)
\( = 38.808 \) cm³.
The density of the metal is given as 8.9 g per cm³.
The mass of the ball = Volume \( \times \) Density.
\( = 38.808 \times 8.9 \)
\( = 345.39 \) g (approximately).
In simple words: First, calculate the ball's volume using its diameter. Then, multiply this volume by the metal's density to find the ball's total mass.

Exam Tip: Ensure consistent units throughout your calculation. If density is in g/cm³, then volume should be in cm³ to get mass in grams.

Question 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 radius of the earth be \( r \).
Then, the diameter of the earth \( = 2r \).
The diameter of the moon is about one-fourth of the earth's diameter.
So, the diameter of the moon \( = \frac{1}{4} (2r) = \frac{r}{2} \).
The radius of the moon \( = \frac{1}{2} (\text{diameter of moon}) = \frac{1}{2} (\frac{r}{2}) = \frac{r}{4} \).
The volume of the earth \( (V_1) = \frac{4}{3} \pi r^3 \).
The volume of the moon \( (V_2) = \frac{4}{3} \pi (\frac{r}{4})^3 \)
\( = \frac{4}{3} \pi \frac{r^3}{64} \)
\( = \frac{1}{64} (\frac{4}{3} \pi r^3) \)
\( = \frac{1}{64} V_1 \).
Therefore, the volume of the moon is \( \frac{1}{64} \)th fraction of the volume of the earth.
In simple words: Since the moon's diameter is a quarter of the earth's, its radius is also a quarter. Because volume depends on the cube of the radius, the moon's volume ends up being \( \frac{1}{64} \) of the earth's volume.

Exam Tip: For problems involving ratios of volumes of similar shapes, express the dimensions (like radius or diameter) in terms of a common variable. Volume scales with the cube of the linear dimensions.

Question 5. How many liters of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer:
Given, the diameter of the hemispherical bowl is 10.5 cm.
The radius \( (r) = \frac{10.5}{2} = 5.25 \) cm.
The amount of milk the bowl can hold is equal to its volume.
The volume of a hemisphere \( = \frac{2}{3} \pi r^3 \).
So, \( V = \frac{2}{3} \times \frac{22}{7} \times (5.25)^3 \) cm³
\( = \frac{2}{3} \times \frac{22}{7} \times 144.703125 \)
\( = 303.1875 \) cm³ (approximately).
To convert to liters, we know that 1000 cm³ = 1 liter.
So, \( 303.1875 \) cm³ \( = \frac{303.1875}{1000} \) L \( = 0.303 \) L (approximately).
In simple words: We find the radius from the bowl's diameter and then use the formula for a hemisphere's volume. After calculating the volume in cm³, we convert it to liters by dividing by 1000.

Exam Tip: Remember the volume formula for a hemisphere is half the volume of a full sphere. Also, be careful with unit conversions, especially between cm³ and liters.

Question 6. A hemispherical tank is made up of an iron sheet. If the inner radius is 1 m, then find the volume of the iron used to make the tank. The thickness of the iron sheet is 1 cm.
Answer:
Given, the inner radius \( (r) = 1 \) m.
The thickness of the iron sheet \( = 1 \) cm.
To maintain consistent units, convert the thickness to meters: \( 1 \) cm \( = 0.01 \) m.
The outer radius \( (R) = \text{Inner radius} (r) + \text{Thickness of iron sheet} \).
\( = 1 \) m \( + 0.01 \) m \( = 1.01 \) m.
The volume of iron used to construct the tank is the difference between the outer and inner hemispherical volumes.
Volume of iron \( = \frac{2}{3} \pi (R^3 – r^3) \)
\( = \frac{2}{3} \times \frac{22}{7} \times [(1.01)^3 – 1^3] \)
\( = \frac{2}{3} \times \frac{22}{7} \times [1.030301 – 1] \)
\( = \frac{2}{3} \times \frac{22}{7} \times 0.030301 \)
\( = 0.06348 \) m³ (approximately).
In simple words: First, convert all measurements to the same unit. Then, calculate the outer radius by adding the inner radius and the sheet's thickness. The amount of iron used is found by subtracting the inner volume from the outer volume of the hemisphere.

Exam Tip: It's crucial to convert all dimensions to the same unit (meters in this case) before performing any calculations. The volume of material in a hollow sphere/hemisphere is the difference between the outer and inner volumes.

Question 7. Find the volume of a sphere whose surface area is 154 cm².
Answer:
Let the radius of the sphere be \( r \) cm.
The surface area of a sphere is given by the formula \( 4 \pi r^2 \).
We are given that the surface area \( = 154 \) cm².
So, \( 4 \pi r^2 = 154 \)
\( 4 \times \frac{22}{7} \times r^2 = 154 \)
\( r^2 = \frac{154 \times 7}{4 \times 22} \)
\( r^2 = \frac{11 \times 14 \times 7}{4 \times 22} \)
\( r^2 = \frac{154 \times 7}{88} \)
\( r^2 = \frac{1078}{88} \)
\( r^2 = \frac{49}{4} \)
\( r = \sqrt{\frac{49}{4}} \)
\( r = \frac{7}{2} \) cm.
Now, we need to find the volume of the sphere.
The volume of a sphere is \( \frac{4}{3} \pi r^3 \).
So, \( V = \frac{4}{3} \times \frac{22}{7} \times (\frac{7}{2})^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} \)
\( = \frac{1 \times 22 \times 49}{3 \times 2} \)
\( = \frac{1078}{6} \)
\( = \frac{539}{3} \) cm³
\( = 179 \frac{2}{3} \) cm³.
In simple words: First, use the given surface area to calculate the sphere's radius. Once you have the radius, plug it into the formula for the volume of a sphere to find your final answer.

Exam Tip: This problem requires a two-step approach: first find the radius using the surface area formula, then use that radius to find the volume. Make sure to correctly apply both formulas.

Question 8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of whitewashing is Rs. 2.00 per square meter, find the
1. inside surface area of the dome,
2. the volume of the air inside the dome.
Answer:
1. To find the inside surface area of the dome:
Total cost of whitewashing = Rs. 498.96.
Cost of whitewashing per square meter = Rs. 2.00.
Inside surface area \( = \frac{\text{Total cost}}{\text{Cost per square meter}} \)
\( = \frac{498.96}{2.00} \)
\( = 249.48 \) m².
2. To find the volume of the air inside the dome:
Let the radius of the hemisphere be \( r \) m.
The inside surface area of a hemisphere is given by \( 2 \pi r^2 \).
We know the inside surface area is \( 249.48 \) m².
So, \( 2 \pi r^2 = 249.48 \)
\( 2 \times \frac{22}{7} \times r^2 = 249.48 \)
\( r^2 = \frac{249.48 \times 7}{2 \times 22} \)
\( r^2 = \frac{1746.36}{44} \)
\( r^2 = 39.69 \)
\( r = \sqrt{39.69} \)
\( r = 6.3 \) m.
Now, we calculate the volume of the air inside the dome, which is the volume of the hemisphere.
Volume of a hemisphere \( = \frac{2}{3} \pi r^3 \).
So, \( V = \frac{2}{3} \times \frac{22}{7} \times (6.3)^3 \)
\( = \frac{2}{3} \times \frac{22}{7} \times 250.047 \)
\( = 523.9 \) m³ (approximately).
In simple words: First, divide the total whitewashing cost by the cost per square meter to get the dome's surface area. Then, use this surface area to find the dome's radius. Finally, calculate the volume of the hemispherical space using that radius.

Exam Tip: Be careful with the surface area formula for a hemisphere; it's \( 2 \pi r^2 \) for the curved part. The volume of a hemisphere is half the volume of a full sphere.

Question 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
1. radius r' of the new sphere,
2. the ratio of S and S'.
Answer:
1. To find the radius \( r' \) of the new sphere:
The volume of one solid iron sphere \( = \frac{4}{3} \pi r^3 \).
The total volume of 27 solid iron spheres \( = 27 \times (\frac{4}{3} \pi r^3) = 36 \pi r^3 \).
When these spheres are melted to form a new single sphere, its volume will be the same as the combined volume of the 27 smaller spheres.
Let the radius of this new sphere be \( r' \).
The volume of the new sphere \( = \frac{4}{3} \pi (r')^3 \).
According to the problem, the volume of the new sphere is equal to the total volume of the smaller spheres:
\( \frac{4}{3} \pi (r')^3 = 36 \pi r^3 \)
Divide both sides by \( \pi \):
\( \frac{4}{3} (r')^3 = 36 r^3 \)
Multiply both sides by \( \frac{3}{4} \):
\( (r')^3 = 36 r^3 \times \frac{3}{4} \)
\( (r')^3 = 9 r^3 \times 3 \)
\( (r')^3 = 27 r^3 \)
Take the cube root of both sides:
\( r' = (27 r^3)^{1/3} \)
\( r' = 3r \).
Therefore, the radius \( r' \) of the new sphere is \( 3r \).
2. To find the ratio of \( S \) and \( S' \):
The surface area of one original iron sphere \( S = 4 \pi r^2 \).
The surface area of the new, larger sphere \( S' = 4 \pi (r')^2 \).
Since \( r' = 3r \), substitute this into the formula for \( S' \):
\( S' = 4 \pi (3r)^2 \)
\( S' = 4 \pi (9r^2) \)
\( S' = 36 \pi r^2 \).
Now, we find the ratio of \( S \) to \( S' \):
\( \frac{S}{S'} = \frac{4 \pi r^2}{36 \pi r^2} \)
Cancel \( 4 \pi r^2 \) from the numerator and denominator:
\( \frac{S}{S'} = \frac{1}{9} \).
So, the ratio of \( S \) and \( S' \) is \( 1:9 \).
In simple words: When 27 small spheres melt into one big one, their total volume stays the same. This means the new sphere's radius is three times bigger than the small ones. Because surface area depends on the square of the radius, the new sphere's surface area will be nine times larger than one small sphere, making the ratio 1:9.

Exam Tip: Remember that when objects are melted and reshaped, their total volume remains constant. Surface area, however, changes. Pay attention to how radius affects volume (cubed) and surface area (squared).

Question 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?
Answer:
Given, the diameter of the medicine capsule is 3.5 mm.
The radius of the capsule \( (r) = \frac{3.5}{2} \) mm \( = 1.75 \) mm.
The amount of medicine needed to fill the capsule is its volume.
The volume of a sphere \( = \frac{4}{3} \pi r^3 \).
So, \( V = \frac{4}{3} \times \frac{22}{7} \times (1.75)^3 \) mm³
\( = \frac{4}{3} \times \frac{22}{7} \times 5.359375 \)
\( = 22.4583 \) mm³ (approximately).
Therefore, approximately 22.46 mm³ of medicine is required to fill this capsule.
In simple words: To find out how much medicine fills the spherical capsule, first find its radius from the diameter. Then, use the sphere volume formula to calculate the exact amount in cubic millimeters.

Exam Tip: Make sure to correctly halve the diameter to get the radius. Precision in calculation is important, especially when dealing with cubic measurements for small items like capsules.