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Q4:
A cubical block of side $7$ cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution :
Given:
A cubical block with side length $a = 7$ cm. A hemisphere is surmounted on top of this cube.
To Find:
1. The greatest diameter ($d$) the hemisphere can have.
2. The total surface area of the resulting solid.
Visual Representation:
Step 1: Determining the greatest diameter of the hemisphere.
The hemisphere is placed on the top face of the cube. For the hemisphere to be contained within the boundaries of the top face of the cube, its diameter cannot exceed the side length of the cube.
Since the side of the cube is $7$ cm, the maximum diameter $d$ is equal to the side of the cube.
$d = 7$ cm
Therefore, the radius $r$ of the hemisphere is:
$r = \frac{d}{2} = \frac{7}{2} = 3.5$ cm
Step 2: Formulating the Total Surface Area (TSA) of the solid.
The total surface area of the solid is composed of:
1. The total surface area of the cube ($6a^2$).
2. The curved surface area of the hemisphere ($2\pi r^2$).
3. Subtracting the area of the base of the hemisphere, as it is covered by the hemisphere and is not part of the external surface area ($\pi r^2$).
Formula: $TSA = (\text{Total Surface Area of Cube}) - (\text{Area of the circular base of hemisphere}) + (\text{Curved Surface Area of Hemisphere})$
$TSA = 6a^2 - \pi r^2 + 2\pi r^2$
$TSA = 6a^2 + \pi r^2$
Step 3: Calculating the values.
Substitute $a = 7$ cm and $r = 3.5$ cm (or $\frac{7}{2}$ cm) into the formula:
$TSA = 6(7)^2 + \pi (\frac{7}{2})^2$
$TSA = 6(49) + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$TSA = 294 + \frac{11 \times 7}{2}$
$TSA = 294 + \frac{77}{2}$
$TSA = 294 + 38.5$
$TSA = 332.5$ cm$^2$
Final Answer: The greatest diameter of the hemisphere is 7 cm, and the total surface area of the solid is 332.5 cm$^2$.
More Questions from Class 10 Mathematics Surface Areas and Volumes EXERCISE 12.1
- Q1: 2 cubes each of volume $64$ cm$^3$ are joined end to end. Find the surface area of the resulting cuboid.
- Q2: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14$ cm and the total height of the vessel is $13$ cm. Find the inner surface area of the vessel.
- Q3: A toy is in the form of a cone of radius $3.5$ cm mounted on a hemisphere of same radius. The total height of the toy is $15.5$ cm. Find the total surface area of the toy.
- Q5: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
- Q6: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is $14$ mm and the diameter of the capsule is $5$ mm. Find its surface area.
- Q7: A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1$ m and $4$ m respectively, and the slant height of the top is $2.8$ m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of $₹ 500$ per m$^2$. (Note that the base of the tent will not be covered with canvas.)
- Q8: From a solid cylinder whose height is $2.4$ cm and diameter $1.4$ cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm$^2$.
- Q9: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is $10$ cm, and its base is of radius $3.5$ cm, find the total surface area of the article.
CBSE Solutions for Class 10 Mathematics Surface Areas and Volumes
Chapters in CBSE - Class 10 Mathematics
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