Surface Areas and Volumes Ex 13.2 Class 10 NCERT

Core Concepts: Volume of Combined Solids

To solve problems in surface areas and volumes ex 13 2 class 10 ncert, we must understand the fundamental difference between calculating surface area and volume. While surface area requires us to ignore the surfaces that touch and merge, volume is strictly additive.

If a solid is formed by joining two or more basic solids, its total volume is equal to the sum of the volumes of the constituent solids. For instance, if a solid toy consists of a cone mounted on a hemisphere, the total volume of the toy is:
$\text{Total Volume} = \text{Volume of Cone} + \text{Volume of Hemisphere}$

Similarly, if a cylinder has hemispherical depressions at both ends, the volume of the remaining solid is obtained by subtracting the volume of the two hemispheres from the cylinder's volume:
$\text{Remaining Volume} = \text{Volume of Cylinder} - 2 \times \text{Volume of Hemisphere}$

By systematically identifying the individual shapes that form the combined solid, you can apply standard formulas to find the final result. Remember to factor out common variables like $\pi$ and radius $r$ before running calculations to keep your work clean and error-free.

Key Volume Formulas to Memorize

Volume of a Cylinder

$V = \pi r^2 h$, where $r$ is the radius of the base and $h$ is the height of the cylinder.

Volume of a Cone

$V = \frac{1}{3}\pi r^2 h$, where $r$ is the base radius and $h$ is the vertical height.

Volume of a Sphere

$V = \frac{4}{3}\pi r^3$, where $r$ is the radius of the sphere.

Volume of a Hemisphere

$V = \frac{2}{3}\pi r^3$, which is exactly half the volume of a complete sphere.

Step-by-Step Worked Solutions

• Example 1: Cone standing on a Hemisphere Question: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$. Step 1: Identify the given values. Radius of hemisphere ($r$) = $1\text{ cm}$. Radius of cone ($r$) = $1\text{ cm}$. Height of cone ($h$) = $1\text{ cm}$. Step 2: State the total volume formula. $\text{Total Volume} = \text{Volume of Cone} + \text{Volume of Hemisphere}$ $\text{Volume} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$ Step 3: Substitute the values. $\text{Volume} = \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3 = \frac{1}{3}\pi + \frac{2}{3}\pi$ Step 4: Simplify the expression. $\text{Volume} = \pi\text{ cm}^3$. Final Answer: The volume of the solid is $\pi\text{ cm}^3$.
• Example 2: Toy with Cylinder and Conical Ends Question: A solid toy is in the form of a cylinder with hemispherical ends. If the total length of the toy is $19\text{ cm}$ and the diameter of the cylinder is $7\text{ cm}$, find the volume of the toy. Step 1: Find the radius ($r$). Radius $r = \frac{7}{2} = 3.5\text{ cm}$. Step 2: Find the height of the cylindrical part ($h$). The hemispherical ends extend by $3.5\text{ cm}$ on each side. Thus, $h = 19 - (3.5 + 3.5) = 19 - 7 = 12\text{ cm}$. Step 3: Apply the total volume formula. $\text{Total Volume} = \text{Volume of Cylinder} + 2 \times \text{Volume of Hemisphere}$ $\text{Volume} = \pi r^2 h + 2 \times \left(\frac{2}{3}\pi r^3\right) = \pi r^2 \left(h + \frac{4}{3}r\right)$ Step 4: Substitute the values and compute. $\text{Volume} = \frac{22}{7} \times (3.5)^2 \times \left(12 + \frac{4}{3} \times 3.5\right) = \frac{22}{7} \times 12.25 \times \left(12 + 4.67\right) = 38.5 \times 16.67 \approx 641.8\text{ cm}^3$. Final Answer: The volume of the toy is approximately $641.8\text{ cm}^3$.

CBSE Board Exam Tips & Common Pitfalls

Here are some crucial tips directly from YoLearn's CBSE experts to help you score 100% in questions from this exercise:

• Don't substitute $\pi$ early: Keep $\pi$ as a symbol until the very final step of calculations. Often, $\pi$ values cancel out, or you can factor it out from all terms, saving you immense calculation time and reducing arithmetic errors.
• Match your units: Ensure all dimensions (radius, height, depth) are converted into the same unit (all in cm or all in m) before substituting them into the formulas.
• Height confusion: In combined shapes (like a cylinder with conical ends), distinguish clearly between the total height of the combined solid and the height of the individual cone or cylinder ($h_1$ vs $h_2$).
• Check the prompt for $\pi$ value: Use $\pi = \frac{22}{7}$ unless specified otherwise. If the question explicitly states to use $\pi = 3.14$, use exactly $3.14$.

Practice Questions with Solutions

• Q: A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15\text{ cm}$ by $10\text{ cm}$ by $3.5\text{ cm}$. The radius of each of the depressions is $0.5\text{ cm}$ and the depth is $1.4\text{ cm}$. Find the volume of wood in the entire stand. A: Step 1: Volume of cuboid = $l \times b \times h = 15 \times 10 \times 3.5 = 525\text{ cm}^3$. Step 2: Volume of one conical depression = $\frac{1}{3}\pi r^2 h_c = \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4 = 0.366\text{ cm}^3$. Step 3: Volume of four depressions = $4 \times 0.366 = 1.46\text{ cm}^3$. Step 4: Volume of wood left = Volume of cuboid - Volume of 4 depressions = $525 - 1.46 = 523.54\text{ cm}^3$. Final answer: The volume of wood in the entire stand is $523.54\text{ cm}^3$.
• Q: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends. The diameter of the model is $3\text{ cm}$ and its length is $12\text{ cm}$. If each cone has a height of $2\text{ cm}$, find the volume of air contained in the model. (Take $\pi = 22/7$). A: Step 1: Identify dimensions. Radius of cylinder and cones ($r$) = $1.5\text{ cm}$. Height of each cone ($h_1$) = $2\text{ cm}$. Height of cylinder ($h_2$) = Total length - $2 \times$ height of cone = $12 - 4 = 8\text{ cm}$. Step 2: Total volume of air = Volume of cylinder + $2 \times$ Volume of cone. Step 3: $\text{Volume} = \pi r^2 h_2 + 2 \times \left(\frac{1}{3}\pi r^2 h_1\right) = \pi r^2 \left(h_2 + \frac{2}{3}h_1\right)$. Step 4: Substitute values: $\text{Volume} = \frac{22}{7} \times 1.5^2 \times \left(8 + \frac{4}{3}\right) = \frac{22}{7} \times 2.25 \times \frac{28}{3} = 22 \times 2.25 \times 4 = 66\text{ cm}^3$. Final answer: The volume of air contained in the model is $66\text{ cm}^3$.
• Q: A solid consisting of a right circular cone of height $120\text{ cm}$ and radius $60\text{ cm}$ standing on a hemisphere of radius $60\text{ cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60\text{ cm}$ and its height is $180\text{ cm}$. A: Step 1: Volume of solid inserted = Volume of cone + Volume of hemisphere. $\text{Volume of Solid} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi (60)^2 (120) + \frac{2}{3}\pi (60)^3 = \pi (60)^2 [40 + 40] = 80 \times 3600 \pi = 288,000\pi\text{ cm}^3$. Step 2: Volume of cylinder = $\pi R^2 H = \pi (60)^2 (180) = 648,000\pi\text{ cm}^3$. Step 3: Volume of water left = Volume of cylinder - Volume of solid. $\text{Volume of water left} = 648,000\pi - 288,000\pi = 360,000\pi\text{ cm}^3$. Step 4: Convert using $\pi = 22/7$: $\text{Volume} = 360,000 \times \frac{22}{7} \approx 1,131,428\text{ cm}^3 = 1.131\text{ m}^3$. Final answer: The volume of water left in the cylinder is approximately $1.131\text{ m}^3$.
• Q: A vessel is in the form of an inverted cone. Its height is $8\text{ cm}$ and the radius of its top is $5\text{ cm}$. It is filled with water to the brim. When lead shots, each of which is a sphere of radius $0.5\text{ cm}$ are dropped, one-fourth of the water flows out. Find the number of lead shots dropped. A: Step 1: Volume of water in cone = $\frac{1}{3}\pi R^2 H = \frac{1}{3}\pi (5)^2 (8) = \frac{200\pi}{3}\text{ cm}^3$. Step 2: Volume of water that flowed out = $\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}\text{ cm}^3$. Step 3: Volume of one lead shot (sphere) = $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi (0.125) = \frac{\pi}{6}\text{ cm}^3$. Step 4: Number of lead shots ($n$) = Volume of water flowed out / Volume of one lead shot. $n = \frac{50\pi}{3} / \frac{\pi}{6} = \frac{50}{3} \times 6 = 100$. Final answer: The number of lead shots dropped in the vessel is $100$.

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