Surface Areas and Volumes | Exercise 12.2

Question 2

  1. 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 cm3\text{ cm} and its length is 12 cm12\text{ cm}. If each cone has a height of 2 cm2\text{ cm}, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
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Solution
Understand the Question
  • The model consists of a central cylinder with two identical cones attached at both ends.
  • The volume of air contained in the model is the total volume of these parts: Total Volume=Volume of Cylinder+2×Volume of a Cone\text{Total Volume} = \text{Volume of Cylinder} + 2 \times \text{Volume of a Cone}
  • The radius of both the cylinder and the cones is r=diameter2=32 cmr = \dfrac{\text{diameter}}{2} = \dfrac{3}{2}\text{ cm}.
  • The height of the cylinder (h2h_2) is obtained by subtracting the heights of the two cones from the total length of the model.

Step 1 · Find the Dimensions of the Model

Diagram 1

Given: Diameter=3 cm  ⟹  Radius (r)=32 cm\text{Diameter} = 3\text{ cm} \implies \text{Radius } (r) = \dfrac{3}{2}\text{ cm}

Total length of model=12 cm\text{Total length of model} = 12\text{ cm}

Height of each cone (h1)=2 cm\text{Height of each cone } (h_1) = 2\text{ cm}

Height of the cylindrical part (h2h_2):

h2=12−(2×h1)=12−(2×2)=12−4=8 cm\begin{aligned} h_2 &= 12 - (2 \times h_1) \\[0.6em] &= 12 - (2 \times 2) \\[0.6em] &= 12 - 4 \\[0.6em] &= 8\text{ cm} \end{aligned}

Step 2 · Calculate the Total Volume of Air

Using the formulas for the volume of a cylinder (πr2h2\pi r^2 h_2) and cone (13πr2h1\frac{1}{3}\pi r^2 h_1): Volume=Volume of cylinder+2×Volume of cone\text{Volume} = \text{Volume of cylinder} + 2 \times \text{Volume of cone}

Volume=πr2h2+2×13πr2h1\text{Volume} = \pi r^2 h_2 + 2 \times \dfrac{1}{3}\pi r^2 h_1

Substitute the values:

Volume=π(32)2(8)+2×13π(32)2(2)=π(94)(8)+23π(94)(2)=π(9×2)+23π(92)=18π+3π=21π\begin{aligned} \text{Volume} &= \pi \left(\dfrac{3}{2}\right)^2 (8) + 2 \times \dfrac{1}{3} \pi \left(\dfrac{3}{2}\right)^2 (2) \\[0.6em] &= \pi \left(\dfrac{9}{4}\right) (8) + \dfrac{2}{3} \pi \left(\dfrac{9}{4}\right) (2) \\[0.6em] &= \pi (9 \times 2) + \dfrac{2}{3} \pi \left(\dfrac{9}{2}\right) \\[0.6em] &= 18\pi + 3\pi \\[0.6em] &= 21\pi \end{aligned}

Substitute π=227\pi = \dfrac{22}{7}:

Volume=21×227=3×22=66 cm3\begin{aligned} \text{Volume} &= 21 \times \dfrac{22}{7} \\[0.6em] &= 3 \times 22 \\[0.6em] &= 66\text{ cm}^3 \end{aligned}
Answer

66 cm366\text{ cm}^3

Common Mistakes
  • Subtracting Only One Cone's Height: Subtracting 2 cm2\text{ cm} instead of 2×2=4 cm2 \times 2 = 4\text{ cm} from the total length, which leads to an incorrect cylinder height of 10 cm10\text{ cm} instead of 8 cm8\text{ cm}.
  • Diameter vs Radius: Using r=3 cmr = 3\text{ cm} instead of r=1.5 cm=32 cmr = 1.5\text{ cm} = \dfrac{3}{2}\text{ cm} in the volume formulas.
  • Single Cone Volume: Forgetting that there are two identical cones attached to the ends, adding the volume of only one cone.

More questions in Exercise 12.2

Q1

Unless stated otherwise, take π=227\pi = \dfrac{22}{7}.

  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 is equal to its radius. Find the volume of the solid in terms of π\pi.
Q2
  1. 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 cm3\text{ cm} and its length is 12 cm12\text{ cm}. If each cone has a height of 2 cm2\text{ cm}, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Q3

A gulab jamun, contains sugar syrup up to about 30%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 length 5 cm5\text{ cm} and diameter 2.8 cm2.8\text{ cm} (see Fig. 12.15).

Q4

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. 12.16).

Q5

A vessel is in the form of an inverted cone. Its height is 8 cm8\text{ cm} and the radius of its top, which is open, is 5 cm5\text{ cm}. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm0.5\text{ cm} are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Q6

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 cm31\text{ cm}^3 of iron has approximately 8 g8\text{ g} mass. (Use π=3.14\pi = 3.14)

Q7

A solid consisting of a right circular cone of height 120 cm120\text{ cm} and radius 60 cm60\text{ cm} standing on a hemisphere of radius 60 cm60\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 cm60\text{ cm} and its height is 180 cm180\text{ cm}.

Q8

A spherical glass vessel has a cylindrical neck 8 cm long, 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 cm3345\text{ cm}^3. Check whether she is correct, taking the above as the inside measurements, and π=3.14\pi = 3.14.

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