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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.
Solution :
Given:
- A solid object consisting of a right circular cone mounted on a hemisphere.
- Height of the cone ($h_{cone}$) = $120$ cm.
- Radius of the cone ($r$) = $60$ cm.
- Radius of the hemisphere ($r$) = $60$ cm.
- A right circular cylinder containing water.
- Radius of the cylinder ($R$) = $60$ cm.
- Height of the cylinder ($H$) = $180$ cm.
To Find:
The volume of water left in the cylinder after the solid is submerged in it.
Step 1: Calculate the volume of the cylinder.
The formula for the volume of a cylinder is $V_{cyl} = \pi R^2 H$.
$V_{cyl} = \pi \times (60)^2 \times 180$
$V_{cyl} = \pi \times 3600 \times 180 = 648,000\pi \text{ cm}^3$.
Step 2: Calculate the volume of the solid.
The solid consists of a cone and a hemisphere. The total volume $V_{solid} = V_{cone} + V_{hemisphere}$.
Formula for volume of a cone: $V_{cone} = \frac{1}{3}\pi r^2 h_{cone}$.
$V_{cone} = \frac{1}{3} \times \pi \times (60)^2 \times 120 = \pi \times 3600 \times 40 = 144,000\pi \text{ cm}^3$.
Formula for volume of a hemisphere: $V_{hemisphere} = \frac{2}{3}\pi r^3$.
$V_{hemisphere} = \frac{2}{3} \times \pi \times (60)^3 = \frac{2}{3} \times \pi \times 216,000 = 2 \times \pi \times 72,000 = 144,000\pi \text{ cm}^3$.
Total volume of the solid $V_{solid} = 144,000\pi + 144,000\pi = 288,000\pi \text{ cm}^3$.
Step 3: Calculate the volume of water left in the cylinder.
When the solid is placed in the cylinder, it displaces a volume of water equal to its own volume. The volume of water left is the difference between the volume of the cylinder and the volume of the solid.
$V_{left} = V_{cyl} - V_{solid}$
$V_{left} = 648,000\pi - 288,000\pi = 360,000\pi \text{ cm}^3$.
Step 4: Convert to numerical value using $\pi \approx \frac{22}{7}$.
$V_{left} = 360,000 \times \frac{22}{7} \approx 360,000 \times 3.14159 \approx 1,131,428.57 \text{ cm}^3$.
Converting to cubic meters ($1 \text{ m}^3 = 1,000,000 \text{ cm}^3$):
$V_{left} \approx 1.131 \text{ m}^3$.
Final Answer: The volume of water left in the cylinder is $360,000\pi \text{ cm}^3$ or approximately $1.131 \text{ m}^3$.
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 $\pi$.
- 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.)
- 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 $\pi = 3.14$)
- 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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