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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$^2$.
Solution :
Given:
- Height of the cylinder ($h$) = $2.4$ cm
- Diameter of the cylinder ($d$) = $1.4$ cm
- Radius of the cylinder ($r$) = $\frac{d}{2} = \frac{1.4}{2} = 0.7$ cm
- A conical cavity of the same height ($h = 2.4$ cm) and same radius ($r = 0.7$ cm) is hollowed out.
To Find:
The total surface area of the remaining solid, rounded to the nearest cm$^2$.
Step 1: Identify the components of the Total Surface Area (TSA)
When a conical cavity is hollowed out from a solid cylinder, the total surface area of the remaining solid consists of:
- The curved surface area of the cylinder ($2\pi rh$)
- The area of the circular base of the cylinder ($\pi r^2$)
- The curved surface area of the conical cavity ($\pi rl$)
Formula: $TSA = 2\pi rh + \pi r^2 + \pi rl$
Step 2: Calculate the slant height ($l$) of the cone
The slant height $l$ is given by the Pythagorean theorem: $l = \sqrt{r^2 + h^2}$
$l = \sqrt{(0.7)^2 + (2.4)^2}$
$l = \sqrt{0.49 + 5.76}$
$l = \sqrt{6.25} = 2.5$ cm
Step 3: Calculate the individual areas
Using $\pi \approx \frac{22}{7}$:
Curved Surface Area of Cylinder = $2 \times \frac{22}{7} \times 0.7 \times 2.4 = 2 \times 22 \times 0.1 \times 2.4 = 10.56$ cm$^2$
Area of circular base = $\pi r^2 = \frac{22}{7} \times 0.7 \times 0.7 = 22 \times 0.1 \times 0.7 = 1.54$ cm$^2$
Curved Surface Area of Cone = $\pi rl = \frac{22}{7} \times 0.7 \times 2.5 = 22 \times 0.1 \times 2.5 = 5.5$ cm$^2$
Step 4: Sum the areas
$TSA = 10.56 + 1.54 + 5.5$
$TSA = 17.6$ cm$^2$
Step 5: Rounding to the nearest cm$^2$
Since $17.6$ is closer to $18$ than $17$, we round to $18$ cm$^2$.
Final Answer: 18 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.
- 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.
- 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.)
- 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
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