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Q2:
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 to be nearly the same.)
Solution :
Given:
- The model is shaped like a cylinder with two cones attached at its ends.
- Diameter of the model ($d$) = $3$ cm.
- Total length of the model ($H_{total}$) = $12$ cm.
- Height of each cone ($h_{cone}$) = $2$ cm.
To Find:
The volume of air contained in the model.
Step 1: Determine the dimensions of the cylinder and cones.
Since the diameter of the model is $3$ cm, the radius ($r$) of the cylinder and the cones is the same:
$r = \frac{d}{2} = \frac{3}{2} = 1.5$ cm.
The height of the cylinder ($h_{cyl}$) is the total length of the model minus the heights of the two cones:
$h_{cyl} = H_{total} - (2 \times h_{cone})$
$h_{cyl} = 12 - (2 \times 2) = 12 - 4 = 8$ cm.
Step 2: Formulate the total volume of the model.
The total volume of air ($V_{total}$) is the sum of the volume of the cylinder ($V_{cyl}$) and the volumes of the two cones ($2 \times V_{cone}$):
$V_{total} = V_{cyl} + 2 \times V_{cone}$
Using the formulas for volume: $V_{cyl} = \pi r^2 h_{cyl}$ and $V_{cone} = \frac{1}{3} \pi r^2 h_{cone}$
$V_{total} = \pi r^2 h_{cyl} + 2 \times \left( \frac{1}{3} \pi r^2 h_{cone} \right)$
Step 3: Calculate the volume.
Substitute the values $r = 1.5$, $h_{cyl} = 8$, and $h_{cone} = 2$:
$V_{total} = \pi (1.5)^2 (8) + \frac{2}{3} \pi (1.5)^2 (2)$
$V_{total} = \pi (2.25)(8) + \frac{2}{3} \pi (2.25)(2)$
$V_{total} = 18\pi + \frac{2}{3} \pi (4.5)$
$V_{total} = 18\pi + 2\pi (1.5)$
$V_{total} = 18\pi + 3\pi = 21\pi$
Step 4: Final numerical evaluation.
Using $\pi \approx \frac{22}{7}$:
$V_{total} = 21 \times \frac{22}{7} = 3 \times 22 = 66$ cm$^3$.
Final Answer: The volume of air contained in the model is 66 cm$^3$.
More Questions from Class 10 Mathematics Surface Areas and Volumes EXERCISE 12.2
- Q1: 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 $\pi$.
- Q3: 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 length $5$ cm and diameter $2.8$ 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 $8$ 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.
- 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 $1$ cm$^3$ of iron has approximately $8$g mass. (Use $\pi = 3.14$)
- Q7: 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.
- 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 $345$ cm$^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.
CBSE Solutions for Class 10 Mathematics Surface Areas and Volumes
Chapters in CBSE - Class 10 Mathematics
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