Exercise 12.2 Practice
Volume of Combination of Solids
Q1: Volume of Solid
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A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 2 cm and the height of the cone is equal to its diameter. Find the volume of the solid in terms of $\pi$.
Given: Radius $r = 2$ cm.
Height of cone $h = \text{diameter} = 2r = 4$ cm.
Height of cone $h = \text{diameter} = 2r = 4$ cm.
Volume of Solid = Volume of Cone + Volume of Hemisphere
$= \frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3$.
$= \frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3$.
$= \frac{1}{3}\pi(2)^2(4) + \frac{2}{3}\pi(2)^3$
$= \frac{16\pi}{3} + \frac{16\pi}{3} = \frac{32\pi}{3}$.
$= \frac{16\pi}{3} + \frac{16\pi}{3} = \frac{32\pi}{3}$.
Volume is $32\pi/3$ cm³.
Q2: Engineering Model
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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 cm and its total length is 14 cm. If each cone has a height of 3 cm, find the volume of air contained in the model.
Dimensions: $r = 2$ cm.
Height of each cone $h_{cone} = 3$ cm.
Height of cylinder $h_{cyl} = 14 - 3 - 3 = 8$ cm.
Height of each cone $h_{cone} = 3$ cm.
Height of cylinder $h_{cyl} = 14 - 3 - 3 = 8$ cm.
Total Volume = Vol Cylinder + 2(Vol Cone)
$= \pi r^2 h_{cyl} + 2(\frac{1}{3}\pi r^2 h_{cone})$
$= \pi r^2 (h_{cyl} + \frac{2}{3}h_{cone})$.
$= \pi r^2 h_{cyl} + 2(\frac{1}{3}\pi r^2 h_{cone})$
$= \pi r^2 (h_{cyl} + \frac{2}{3}h_{cone})$.
$= \frac{22}{7} \times 4 \times (8 + \frac{2}{3}(3)) = \frac{88}{7} \times (8 + 2)$
$= \frac{880}{7} \approx 125.71$.
$= \frac{880}{7} \approx 125.71$.
Volume is approx 125.71 cm³.
Q3: Gulab Jamun
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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 cm and diameter 3 cm.
One Gulab Jamun: $r = 1.5$ cm. Total L = 6 cm.
Length of cylinder $h = 6 - 1.5 - 1.5 = 3$ cm.
Length of cylinder $h = 6 - 1.5 - 1.5 = 3$ cm.
Volume = Vol Cylinder + 2(Vol Hemisphere)
$= \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2(h + \frac{4}{3}r)$
$= \frac{22}{7} \times 1.5^2 (3 + \frac{4}{3} \times 1.5) = \frac{22}{7} \times 2.25 \times 5 \approx 35.36$ cm³.
$= \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2(h + \frac{4}{3}r)$
$= \frac{22}{7} \times 1.5^2 (3 + \frac{4}{3} \times 1.5) = \frac{22}{7} \times 2.25 \times 5 \approx 35.36$ cm³.
Total Syrup: 30% of Volume of 50 pieces.
$= 0.30 \times 50 \times 35.36 = 530.4$.
$= 0.30 \times 50 \times 35.36 = 530.4$.
Approx 530 cm³ of syrup.
Q4: Pen Stand
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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 cm by 12 cm by 5 cm. The radius of each of the depressions is 0.6 cm and the depth is 1.5 cm. Find the volume of wood in the entire stand.
Volume of Cuboid: $L \times B \times H = 20 \times 12 \times 5 = 1200$ cm³.
Volume of 4 Cones: $4 \times \frac{1}{3}\pi r^2 h$
$= 4 \times \frac{1}{3} \times \frac{22}{7} \times (0.6)^2 \times 1.5$
$= 4 \times \frac{1}{3} \times 3.14 \times 0.36 \times 1.5 \approx 2.26$ cm³.
$= 4 \times \frac{1}{3} \times \frac{22}{7} \times (0.6)^2 \times 1.5$
$= 4 \times \frac{1}{3} \times 3.14 \times 0.36 \times 1.5 \approx 2.26$ cm³.
Wood Volume: $1200 - 2.26 = 1197.74$.
Volume is 1197.74 cm³.
Q5: Dropping Lead Shots
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A vessel is in the form of an inverted cone. Its height is 10 cm and the radius of its top, which is open, is 6 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.6 cm are dropped into the vessel, one-third of the water flows out. Find the number of lead shots dropped in the vessel.
Volume of Cone (Water): $\frac{1}{3}\pi R^2 H = \frac{1}{3}\pi (6)^2 (10) = 120\pi$.
Water Displaced: $\frac{1}{3}$ of Total Vol $= \frac{1}{3} \times 120\pi = 40\pi$.
Volume of 1 Shot: $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.6)^3 = \frac{4}{3}\pi (0.216) = 0.288\pi$.
Number of Shots (n): $n \times 0.288\pi = 40\pi \Rightarrow n = \frac{40}{0.288} \approx 138.8$. (Integer: 139)
Approx 139 lead shots.
Q6: Iron Pole Mass
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A solid iron pole consists of a cylinder of height 200 cm and base diameter 20 cm, which is surmounted by another cylinder of height 50 cm and radius 6 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use $\pi = 3.14$)
Volume 1 (Big Cyl): $\pi (10)^2 (200) = 20000\pi$.
Volume 2 (Small Cyl): $\pi (6)^2 (50) = 1800\pi$.
Total Volume: $21800\pi = 21800 \times 3.14 = 68452$ cm³.
Mass: $68452 \times 8 = 547616$ g $= 547.6$ kg.
Mass is 547.6 kg.
Q7: Solid in Cylinder
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A solid consisting of a right circular cone of height 100 cm and radius 50 cm standing on a hemisphere of radius 50 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 cm and its height is 160 cm.
Volume of Cylinder: $\pi (50)^2 (160) = 400000\pi$.
Volume of Solid (Cone + Hemi):
Cone: $\frac{1}{3}\pi (50)^2 (100) = \frac{250000\pi}{3}$.
Hemisphere: $\frac{2}{3}\pi (50)^3 = \frac{250000\pi}{3}$.
Total Solid = $\frac{500000\pi}{3}$.
Cone: $\frac{1}{3}\pi (50)^2 (100) = \frac{250000\pi}{3}$.
Hemisphere: $\frac{2}{3}\pi (50)^3 = \frac{250000\pi}{3}$.
Total Solid = $\frac{500000\pi}{3}$.
Water Left: $400000\pi - \frac{500000\pi}{3} = \frac{700000\pi}{3}$ cm³.
Volume left is approx 733,333 cm³.
Q8: Spherical Glass Vessel
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A spherical glass vessel has a cylindrical neck 10 cm long, 3 cm in diameter; the diameter of the spherical part is 9 cm. By measuring the amount of water it holds, a child finds its volume to be 450 cm³. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.
Volume Sphere: Radius $R = 4.5$. $V_1 = \frac{4}{3}\pi (4.5)^3 \approx 381.7$ cm³.
Volume Neck: Radius $r = 1.5$, Length $h = 10$.
$V_2 = \pi (1.5)^2 (10) = 3.14 \times 2.25 \times 10 = 70.65$ cm³.
$V_2 = \pi (1.5)^2 (10) = 3.14 \times 2.25 \times 10 = 70.65$ cm³.
Total Volume: $381.7 + 70.65 = 452.35$ cm³.
She is incorrect (452.35 $\neq$ 450).