Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 12 Exercise 12.2 on calculating the Volumes of Combinations of Solids. Learn to find the volume of composite shapes like a cone on a hemisphere (Q.1) and a cylinder with two cones (Q.2). Solutions cover real-world applications such as finding the syrup volume in gulab jamuns (Q.3), calculating wood volume in a pen stand with conical depressions (Q.4), and solving problems involving volume displacement (Q.5, Q.7). Also includes finding the mass of a composite iron pole (Q.6) and verifying volume measurements of a cylindrical neck and spherical vessel (Q.8). Essential for mastering volume formulas for 3D shapes.
This exercise focuses on calculating the volume of solids formed by combining two or more basic geometrical shapes. Unless stated otherwise, use $\pi = 22/7$.
1. Volume of the Solid in Terms of $\pi$
1. A solid is in the shape of a cone standing on a hemisphere, with both their radii being equal to 1 cm and the height of the cone being equal to its radius. Find the volume of the solid in terms of π.
- Answer:
- Identify Dimensions:Radius ($r$): $1 \text{ cm}$.Height of cone ($h$): $r = 1 \text{ cm}$.
- Calculate Volume ($V$):$$V = V_{\text{cone}} + V_{\text{hemisphere}}$$$$V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$$Substitute $r=1$ and $h=1$:$$V = \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3 = \frac{1}{3}\pi + \frac{2}{3}\pi = \pi \text{ cm}^3$$The volume of the solid is $\mathbf{\pi \text{ cm}^3}$.
2. Volume of Air in the Model
2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm, and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model are nearly the same.)
- Answer:
- Identify Dimensions:Radius ($r$): $3/2 = 1.5 \text{ cm}$.Height of each cone ($h_{\text{cone}}$): $2 \text{ cm}$.Height of cylinder ($h_{\text{cylinder}}$): $12 – (2 + 2) = 8 \text{ cm}$.
- Calculate Volume ($V$):$$V = V_{\text{cylinder}} + 2 \times V_{\text{cone}}$$$$V = \pi r^2 h_{\text{cylinder}} + 2 \times \frac{1}{3}\pi r^2 h_{\text{cone}} = \pi r^2 \left(h_{\text{cylinder}} + \frac{2}{3}h_{\text{cone}}\right)$$$$V = \frac{22}{7} \times (1.5)^2 \times \left(8 + \frac{2}{3} \times 2\right) = \frac{22}{7} \times 2.25 \times \frac{28}{3} = 66 \text{ cm}^3$$The volume of air contained in the model is $\mathbf{66 \text{ cm}^3}$.
3. Volume of Syrup in 45 Gulab Jamuns
3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length of 5 cm and a diameter of 2.8 cm (see figure).
- Answer:
- Identify Dimensions of one gulab jamun:Radius ($r$): $2.8/2 = 1.4 \text{ cm}$.Height of cylindrical part ($h$): $5 – (2 \times 1.4) = 2.2 \text{ cm}$.
- Calculate Volume of one gulab jamun ($V_{\text{one}}$):$$V_{\text{one}} = V_{\text{cylinder}} + 2 \times V_{\text{hemisphere}} = \pi r^2 h + \frac{4}{3}\pi r^3$$$$V_{\text{one}} = \frac{22}{7} \times (1.4)^2 \times \left(2.2 + \frac{4}{3} \times 1.4\right) \approx 25.05 \text{ cm}^3$$
- Find Volume of Syrup:$$V_{\text{total}} = 45 \times V_{\text{one}} = 45 \times 25.05 \approx 1127.25 \text{ cm}^3$$$$V_{\text{syrup}} = 30\% \text{ of } V_{\text{total}} = 0.30 \times 1127.25 \approx 338.175 \text{ cm}^3$$Approximately, the volume of syrup is $\mathbf{338 \text{ cm}^3}$.
4. Volume of Wood in the Pen Stand
4. 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 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm, and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig.).
- Answer:
- Identify Dimensions:Cuboid ($L, B, H$): $15 \text{ cm}, 10 \text{ cm}, 3.5 \text{ cm}$.Conical depressions (4): Radius ($r$): $0.5 \text{ cm}$, Depth ($h$): $1.4 \text{ cm}$.
- Calculate Volume of Wood ($V_{\text{wood}}$):$$V_{\text{wood}} = V_{\text{cuboid}} – 4 \times V_{\text{cone}}$$$$V_{\text{cuboid}} = 15 \times 10 \times 3.5 = 525 \text{ cm}^3$$$$4 \times V_{\text{cone}} = 4 \times \frac{1}{3}\pi r^2 h = \frac{4}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 \approx 1.47 \text{ cm}^3$$$$V_{\text{wood}} = 525 – 1.47 = 523.53 \text{ cm}^3$$The volume of wood in the entire stand is approximately $\mathbf{523.53 \text{ cm}^3}$.
5. Number of Lead Shots Dropped
5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
- Answer:
- Calculate Volume of Water that Flowed Out ($V_{\text{out}}$):$$V_{\text{cone}} = \frac{1}{3}\pi R^2 H = \frac{1}{3}\pi (5)^2 (8) = \frac{200}{3}\pi \text{ cm}^3$$$$V_{\text{out}} = \frac{1}{4} V_{\text{cone}} = \frac{50}{3}\pi \text{ cm}^3$$
- Calculate Volume of one Lead Shot ($V_{\text{shot}}$):$$V_{\text{shot}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.5)^3 = \frac{0.5}{3}\pi \text{ cm}^3$$
- Find the Number of Lead Shots ($n$):The volume of the $n$ shots equals the volume of water displaced ($V_{\text{out}}$).$$n = \frac{V_{\text{out}}}{V_{\text{shot}}} = \frac{50\pi/3}{0.5\pi/3} = \frac{50}{0.5} = 100$$The number of lead shots dropped is $\mathbf{100}$.
6. Mass of the Solid Iron Pole ($\pi = 3.14$)
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass.
- Answer:
- Identify Dimensions:Cylinder 1 (Large): $R_1=12 \text{ cm}, H_1=220 \text{ cm}$.Cylinder 2 (Small): $R_2=8 \text{ cm}, H_2=60 \text{ cm}$.
- Calculate Total Volume ($V_{\text{total}}$):$$V_{\text{total}} = V_1 + V_2 = \pi (R_1^2 H_1 + R_2^2 H_2)$$$$V_{\text{total}} = 3.14 \times ((12)^2 \times 220 + (8)^2 \times 60) = 3.14 \times 35520 = 111532.8 \text{ cm}^3$$
- Calculate Mass:$$\text{Mass} = V_{\text{total}} \times \text{Density} = 111532.8 \times 8 = 892262.4 \text{ g}$$$$\text{Mass (kg)} = \frac{892262.4}{1000} = 892.2624 \text{ kg}$$The mass of the pole is $\mathbf{892.26 \text{ kg}}$.
7. Volume of Water Left in the Cylinder
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 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 cm and its height is 180 cm.
- Answer:
- Volume of Cylinder ($V_{\text{cyl}}$):$R_{\text{cyl}} = 60 \text{ cm}, H_{\text{cyl}} = 180 \text{ cm}$.$$V_{\text{cyl}} = \pi (60)^2 (180) = 648000\pi \text{ cm}^3$$
- Volume of Solid ($V_{\text{solid}}$):$r=60 \text{ cm}$. Cone $h_{\text{cone}} = 120 \text{ cm}$. Hemisphere $r=60 \text{ cm}$.$$V_{\text{solid}} = V_{\text{cone}} + V_{\text{hemisphere}} = \frac{1}{3}\pi (60)^2 (120) + \frac{2}{3}\pi (60)^3 = 288000\pi \text{ cm}^3$$
- Volume of Water Left ($V_{\text{left}}$):$$V_{\text{left}} = V_{\text{cyl}} – V_{\text{solid}} = 648000\pi – 288000\pi = 360000\pi \text{ cm}^3$$Using $\pi = 22/7$:$$V_{\text{left}} = 360000 \times \frac{22}{7} \approx 1131428.57 \text{ cm}^3$$The volume of water left is approximately $\mathbf{1131428.57 \text{ cm}^3}$.
8. Check Volume Measurement ($\pi = 3.14$)
8. A spherical glass vessel has a cylindrical neck 8 cm long and 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and π = 3.14.
- Answer:
- Volume of Cylindrical Neck ($V_{\text{cyl}}$):$L=8 \text{ cm}, r_{\text{cyl}} = 1 \text{ cm}$.$$V_{\text{cyl}} = 3.14 \times (1)^2 \times 8 = 25.12 \text{ cm}^3$$
- Volume of Spherical Part ($V_{\text{sph}}$):$D=8.5 \text{ cm} \implies r_{\text{sph}} = 4.25 \text{ cm}$.$$V_{\text{sph}} = \frac{4}{3}\pi r_{\text{sph}}^3 = \frac{4}{3} \times 3.14 \times (4.25)^3 \approx 321.39 \text{ cm}^3$$
- Total Calculated Volume ($V_{\text{total}}$):$$V_{\text{total}} = V_{\text{cyl}} + V_{\text{sph}} = 25.12 + 321.39 = 346.51 \text{ cm}^3$$
- Conclusion:The calculated volume ($\mathbf{346.51 \text{ cm}^3}$) is not equal to the child’s measured volume ($345 \text{ cm}^3$). Therefore, the child’s measurement is incorrect.