NCERT Solutions for Class 9 Maths Exercise 13.8: Volume of Sphere and Hemisphere

Core Formulas and Mathematical Concepts

To solve problems in Exercise 13.8, you must understand two critical solid geometric figures: the Sphere and the Hemisphere.

A Sphere is a perfectly round, three-dimensional solid figure where every point on its boundary surface is equidistant from its center. The distance from the center to any point on the boundary is the radius ($r$). The formula to find the volume of a sphere of radius $r$ is:
$\text{Volume of a Sphere} = \frac{4}{3}\pi r^3$

A Hemisphere is exactly half of a sphere. When a sphere is cut right through its center, two equal hemispheres are formed. Since it is exactly half, its volume is also half of a sphere's volume. The formula is:
$\text{Volume of a Hemisphere} = \frac{2}{3}\pi r^3$

Important Unit conversions: Volume is expressed in cubic units like $\text{cm}^3$ or $\text{m}^3$. Frequently, questions require you to convert this capacity into litres. Keep these standard relationships handy:

• $1 \text{ m}^3 = 1000 \text{ litres}$
• $1000 \text{ cm}^3 = 1 \text{ litre}$

Methodical Steps for Solving Exercise 13.8 Problems

1. Step 1: Identify the Shape and Given Values — Read the problem carefully to verify if it features a full sphere or a hemispherical bowl. Extract given values like radius ($r$), or diameter ($d$). If the diameter is given, divide it by 2 to obtain the radius ($r = \frac{d}{2}$).
2. Step 2: Check and Unify Units — Ensure all dimensions are in the same system of units (either all in cm or all in m). Convert if necessary before plugging values into formulas.
3. Step 3: Choose the Right Formula — Apply $V = \frac{4}{3}\pi r^3$ for a sphere, or $V = \frac{2}{3}\pi r^3$ for a hemisphere. Use $\pi = \frac{22}{7}$ unless specified otherwise in the question.
4. Step 4: Execute Calculations and State Final Units — Simplify equations carefully. If working with decimals, retain precision to at least 2 decimal places. Append the correct cubic units (e.g., $\text{cm}^3$ or litres) to your final answer.

Common Pitfalls & Board Exam Tips

Here are a few quick tips straight from our experienced CBSE educators to keep you from losing silly marks on exam day:

1. The Radius Trap: Always verify whether the question gives you the diameter or the radius. Using the diameter instead of the radius in the volume formula is the most common student error.
2. The Exponent Mistake: For volume, we cube the radius ($r^3$). Do not confuse it with the surface area formula which uses $r^2$. Write down $r \times r \times r$ explicitly during calculations to prevent errors.
3. Calculation Optimization: Do not multiply out $22 \times r^3$ immediately. Write it as fractions first and see if you can cancel out factors with $7$ or $3$ in the denominator.

Practice Questions with Solutions

• Q: Find the volume of a sphere whose radius is 7 cm. A: Step 1: Identify the shape and given value. The shape is a sphere, and radius $r = 7\text{ cm}$. Step 2: State the formula for the volume of a sphere. $\text{Volume } (V) = \frac{4}{3}\pi r^3$ Step 3: Substitute the values into the formula. $V = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$ $V = \frac{4}{3} \times 22 \times 1 \times 49$ $V = \frac{4312}{3}\text{ cm}^3$ Step 4: Convert the fraction into a decimal representation. $V \approx 1437.33\text{ cm}^3$ Final answer: The volume of the sphere is $\frac{4312}{3}\text{ cm}^3$ (or approximately $1437.33\text{ cm}^3$).
• Q: Find the amount of water displaced by a solid spherical ball of diameter 28 cm. A: Step 1: When a solid ball is fully immersed in water, it displaces a volume of water equal to its own volume (Archimedes' Principle). Step 2: Given the diameter $d = 28\text{ cm}$. Therefore, radius $r = \frac{28}{2} = 14\text{ cm}$. Step 3: Apply the volume of a sphere formula. $V = \frac{4}{3}\pi r^3$ $V = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14$ $V = \frac{4}{3} \times 22 \times 2 \times 196$ $V = \frac{4}{3} \times 8624$ $V = \frac{34496}{3}\text{ cm}^3$ Step 4: Represent the final calculated volume as a mixed fraction or decimal. $V = 11498\frac{2}{3}\text{ cm}^3$ Final answer: The amount of water displaced is $11498\frac{2}{3}\text{ cm}^3$.
• Q: A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it can contain? A: Step 1: Identify the shape and the radius. The shape is a hemisphere, and the radius $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$ (converting to fraction simplifies calculation). Step 2: State the formula for the volume of a hemisphere. $V = \frac{2}{3}\pi r^3$ Step 3: Substitute $r = \frac{7}{2}\text{ cm}$ and $\pi = \frac{22}{7}$ into the equation. $V = \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$ $V = \frac{2}{3} \times 11 \times \frac{7}{2} \times \frac{7}{2}$ (by cancelling 2 and 7) $V = \frac{11 \times 49}{3 \times 4}$ $V = \frac{539}{12}\text{ cm}^3$ Step 4: Solve the final division. $V \approx 44.92\text{ cm}^3$ Final answer: The hemispherical bowl can contain $44.92\text{ cm}^3$ of water.
• Q: The diameter of a metallic ball is 4.2 cm. If the density of the metal is $8.9\text{ g/cm}^3$, find the mass of the ball. A: Step 1: Determine the radius from the diameter. Given diameter $d = 4.2\text{ cm}$, so radius $r = \frac{4.2}{2} = 2.1\text{ cm}$. Step 2: Calculate the volume of the metallic ball (sphere). $V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$ $V = 4 \times 22 \times 0.1 \times 2.1 \times 0.7$ $V = 88 \times 0.441 = 38.808\text{ cm}^3$ Step 3: Compute the mass using the density formula: $\text{Mass} = \text{Volume} \times \text{Density}$. $\text{Mass} = 38.808\text{ cm}^3 \times 8.9\text{ g/cm}^3$ $\text{Mass} \approx 345.39\text{ grams}$ Final answer: The mass of the ball is approximately $345.39\text{ grams}$.

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