A solid is in the shape of a cone standing on a hemisphere with radii equal to 2 cm and height of cone equal to its diameter. Find volume in terms of π.

Radius r = 2 cm, height h = 4 cm. Volume = (1/3)πr²h + (2/3)πr³ = (16π/3) + (16π/3) = 32π/3 cm³.

Find the volume of air contained in the engineering student model of cylinder with two cones.

Radius r = 2 cm, cylinder height h = 8 cm, each cone height = 3 cm. Volume = πr²(h + (2/3)h_cone) = (880/7) ≈ 125.71 cm³.

Class 10 Maths Chapter 12 Exercise 12.2 NCERT Solutions

Question 1: Cone on Hemisphere in Terms of $\pi$

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $2\text{ cm}$ and the height of the cone is equal to its diameter. Find the volume of the solid in terms of $\pi$.

Given Information

Radius of hemisphere and cone, $r = 2\text{ cm}$.
Height of the cone, $h = \text{diameter} = 2r = 2 \times 2 = 4\text{ cm}$.

Step-by-Step Volume Derivation

$$\text{Total Volume of Solid} = \text{Volume of Cone} + \text{Volume of Hemisphere}$$ $$= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 (h + 2r)$$ Substitute $r = 2\text{ cm}$ and $h = 4\text{ cm}$: $$= \frac{1}{3}\pi (2)^2 (4 + 2(2)) = \frac{1}{3}\pi (4)(4 + 4) = \frac{1}{3}\pi (4)(8) = \frac{32\pi}{3}\text{ cm}^3$$

$$\therefore \text{The volume of the solid is } \frac{32\pi}{3}\text{ cm}^3.$$

Question 2: Engineering Model with Two Cones

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 $4\text{ cm}$ and its total length is $14\text{ cm}$. If each cone has a height of $3\text{ cm}$, find the volume of air contained in the model. (Use $\pi = \frac{22}{7}$)

Step 1: Dimensions of Model Parts

Diameter $= 4\text{ cm} \implies$ Radius $r = 2\text{ cm}$.
Height of each cone, $h_{\text{cone}} = 3\text{ cm}$.
Total length $= 14\text{ cm}$.
Height of cylindrical part: $$h_{\text{cyl}} = 14 - (2 \times 3) = 14 - 6 = 8\text{ cm}$$

Step 2: Total Air Volume

$$\text{Volume} = \text{Volume of Cylinder} + 2 \times (\text{Volume of Cone})$$ $$= \pi r^2 h_{\text{cyl}} + 2 \times \left(\frac{1}{3}\pi r^2 h_{\text{cone}}\right) = \pi r^2 \left(h_{\text{cyl}} + \frac{2}{3}h_{\text{cone}}\right)$$ $$= \frac{22}{7} \times (2)^2 \times \left(8 + \frac{2}{3}(3)\right) = \frac{22}{7} \times 4 \times (8 + 2)$$ $$= \frac{88}{7} \times 10 = \frac{880}{7} \approx 125.71\text{ cm}^3$$

$$\therefore \text{The volume of air contained in the model is } \frac{880}{7}\text{ cm}^3 \approx 125.71\text{ cm}^3.$$

Question 3: Gulab Jamuns Sugar Syrup

A gulab jamun contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in 50 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length $6\text{ cm}$ and diameter $3\text{ cm}$. (Use $\pi = \frac{22}{7}$)

Step 1: Volume of One Gulab Jamun

Diameter $= 3\text{ cm} \implies$ Radius $r = 1.5\text{ cm}$.
Total length $= 6\text{ cm}$.
Length of cylindrical part: $$h = 6 - (1.5 + 1.5) = 6 - 3 = 3\text{ cm}$$ $$\text{Volume of 1 Gulab Jamun} = \text{Vol of Cylinder} + 2 \times (\text{Vol of Hemisphere})$$ $$= \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2 \left(h + \frac{4}{3}r\right)$$ $$= \frac{22}{7} \times (1.5)^2 \times \left(3 + \frac{4}{3}(1.5)\right) = \frac{22}{7} \times 2.25 \times (3 + 2)$$ $$= \frac{22}{7} \times 2.25 \times 5 = \frac{247.5}{7} \approx 35.357\text{ cm}^3$$

Step 2: Sugar Syrup in 50 Pieces

Syrup percentage $= 30\% = 0.30$.
$$\text{Volume of Syrup} = 50 \times 0.30 \times \text{Volume of 1 piece}$$ $$= 15 \times \frac{247.5}{7} = \frac{3712.5}{7} \approx 530.36\text{ cm}^3$$

$$\therefore \text{The volume of sugar syrup found in 50 gulab jamuns is approximately } 530\text{ cm}^3.$$

Question 4: Wooden Pen Stand

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 $20\text{ cm by } 12\text{ cm by } 5\text{ cm}$. The radius of each of the depressions is $0.6\text{ cm}$ and the depth is $1.5\text{ cm}$. Find the volume of wood in the entire stand.

Step 1: Volume of Cuboid

$$\text{Volume of Cuboid} = L \times B \times H = 20 \times 12 \times 5 = 1200\text{ cm}^3$$

Step 2: Volume of 4 Conical Depressions

Radius $r = 0.6\text{ cm}$, depth $h = 1.5\text{ cm}$.
$$\text{Volume of 4 Cones} = 4 \times \left(\frac{1}{3}\pi r^2 h\right)$$ $$= \frac{4}{3} \times \frac{22}{7} \times (0.6)^2 \times 1.5 = \frac{4}{3} \times \frac{22}{7} \times 0.36 \times 1.5$$ $$= \frac{4}{3} \times \frac{22}{7} \times 0.54 = 4 \times \frac{22}{7} \times 0.18 = \frac{15.84}{7} \approx 2.26\text{ cm}^3$$

Step 3: Volume of Wood

$$\text{Volume of Wood} = \text{Volume of Cuboid} - \text{Volume of 4 Depressions}$$ $$= 1200 - 2.26 = 1197.74\text{ cm}^3$$

$$\therefore \text{The volume of wood in the entire stand is } 1197.74\text{ cm}^3.$$

Question 5: Dropping Lead Shots into Inverted Cone

A vessel is in the form of an inverted cone. Its height is $10\text{ cm}$ and the radius of its top, which is open, is $6\text{ cm}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.6\text{ cm}$ are dropped into the vessel, one-third of the water flows out. Find the number of lead shots dropped in the vessel.

Step 1: Volume of Water in Vessel & Water Displaced

Radius $R = 6\text{ cm}$, Height $H = 10\text{ cm}$.
$$\text{Total Water Volume} = \frac{1}{3}\pi R^2 H = \frac{1}{3}\pi (6)^2 (10) = 120\pi\text{ cm}^3$$ Water displaced by lead shots: $$\text{Water Displaced} = \frac{1}{3} \times 120\pi = 40\pi\text{ cm}^3$$

Step 2: Volume of One Lead Shot

Radius of each spherical lead shot, $r = 0.6\text{ cm}$.
$$\text{Volume of 1 Shot} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.6)^3 = \frac{4}{3}\pi (0.216) = 0.288\pi\text{ cm}^3$$

Step 3: Number of Lead Shots

Let $n$ be the number of lead shots dropped: $$n \times \text{Volume of 1 Shot} = \text{Water Displaced}$$ $$n \times 0.288\pi = 40\pi \implies n = \frac{40}{0.288} = \frac{40000}{288} \approx 138.89 \approx 139$$

$$\therefore \text{The number of lead shots dropped into the vessel is approximately } 139.$$

Question 6: Mass of Solid Iron Pole

A solid iron pole consists of a cylinder of height $200\text{ cm}$ and base diameter $20\text{ cm}$, which is surmounted by another cylinder of height $50\text{ cm}$ and radius $6\text{ cm}$. Find the mass of the pole, given that $1\text{ cm}^3$ of iron has approximately $8\text{ g}$ mass. (Use $\pi = 3.14$)

Step 1: Volumes of Both Cylinders

Lower Cylinder: $h_1 = 200\text{ cm}$, radius $r_1 = \frac{20}{2} = 10\text{ cm}$.
$$V_1 = \pi r_1^2 h_1 = 3.14 \times (10)^2 \times 200 = 3.14 \times 100 \times 200 = 62800\text{ cm}^3$$ Upper Cylinder: $h_2 = 50\text{ cm}$, radius $r_2 = 6\text{ cm}$.
$$V_2 = \pi r_2^2 h_2 = 3.14 \times (6)^2 \times 50 = 3.14 \times 36 \times 50 = 3.14 \times 1800 = 5652\text{ cm}^3$$

Step 2: Total Volume & Mass Calculation

$$\text{Total Volume} = V_1 + V_2 = 62800 + 5652 = 68452\text{ cm}^3$$ $$\text{Total Mass} = 68452 \times 8\text{ g} = 547616\text{ g} = \frac{547616}{1000}\text{ kg} = 547.616\text{ kg}$$

$$\therefore \text{The mass of the iron pole is } 547.62\text{ kg (or } 547616\text{ g)}.$$

Question 7: Solid Submerged in Cylinder with Water

A solid consisting of a right circular cone of height $100\text{ cm}$ and radius $50\text{ cm}$ standing on a hemisphere of radius $50\text{ cm}$ 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 $50\text{ cm}$ and its height is $160\text{ cm}$. (Use $\pi = \frac{22}{7}$)

Step 1: Volume of Water Originally in Cylinder

Radius $r = 50\text{ cm}$, height $H = 160\text{ cm}$.
$$\text{Volume of Cylinder} = \pi r^2 H = \pi (50)^2 (160) = \pi (2500)(160) = 400000\pi\text{ cm}^3$$

Step 2: Volume of the Immersed Solid

The solid consists of a cone ($r = 50\text{ cm}, h = 100\text{ cm}$) and a hemisphere ($r = 50\text{ cm}$): $$\text{Vol of Cone} = \frac{1}{3}\pi (50)^2 (100) = \frac{250000\pi}{3}\text{ cm}^3$$ $$\text{Vol of Hemisphere} = \frac{2}{3}\pi (50)^3 = \frac{250000\pi}{3}\text{ cm}^3$$ $$\text{Total Vol of Solid} = \frac{250000\pi}{3} + \frac{250000\pi}{3} = \frac{500000\pi}{3}\text{ cm}^3$$

Step 3: Volume of Water Left

$$\text{Water Left} = 400000\pi - \frac{500000\pi}{3} = \frac{1200000\pi - 500000\pi}{3} = \frac{700000\pi}{3}$$ Substitute $\pi = \frac{22}{7}$: $$= \frac{700000}{3} \times \frac{22}{7} = \frac{100000 \times 22}{3} = \frac{2200000}{3} \approx 733333.33\text{ cm}^3 \approx 0.733\text{ m}^3$$

$$\therefore \text{The volume of water left in the cylinder is } \frac{2200000}{3}\text{ cm}^3 \approx 0.733\text{ m}^3.$$

Question 8: Spherical Glass Vessel

A spherical glass vessel has a cylindrical neck $10\text{ cm}$ long, $3\text{ cm}$ in diameter; the diameter of the spherical part is $9\text{ cm}$. By measuring the amount of water it holds, a child finds its volume to be $450\text{ cm}^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.

Step 1: Breakdown of Component Volumes

Cylindrical Neck: Diameter $= 3\text{ cm} \implies r = 1.5\text{ cm}$, height $h = 10\text{ cm}$.
$$V_1 = \pi r^2 h = 3.14 \times (1.5)^2 \times 10 = 3.14 \times 2.25 \times 10 = 70.65\text{ cm}^3$$ Spherical Body: Diameter $= 9\text{ cm} \implies R = 4.5\text{ cm}$.
$$V_2 = \frac{4}{3}\pi R^3 = \frac{4}{3} \times 3.14 \times (4.5)^3 = \frac{4}{3} \times 3.14 \times 91.125 = 381.51\text{ cm}^3$$

Step 2: Total Volume & Comparison

$$\text{Total Volume} = V_1 + V_2 = 70.65 + 381.51 = 452.16\text{ cm}^3$$ The child measured the volume to be $450\text{ cm}^3$. Since $452.16\text{ cm}^3 \neq 450\text{ cm}^3$, the child's measurement is incorrect.

$$\therefore \text{She is incorrect. The correct volume is } 452.16\text{ cm}^3.$$

Surface Areas and Volumes - Exercise 12.2

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