NCERT Solutions for Exercise 13.1 Class 10 Maths Chapter 13 - Surface Area and Volumes

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.

Answer:

Since the vessel consists of a hemisphere and a cylinder, its area is given by :

Area of vessel = Inner area of the cylinder(curved) + Inner area of the hemisphere

The inner surface area of the hemisphere is :

$=2\pi r^2$

or $=2\times \left(\frac{22}{7}\right)\times 7^2$

or $=308c{m}^2$

And the surface area of the cylinder is :

$=2\pi rh$

or $=2\times \frac{22}{7}\times 7\times 6$

or $=264c{m}^2$

Thus, the inner surface area of the vessel is $=308+264=572c{m}^2.$

Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer:

The required surface area of the toy is given by :

Area of toy = Surface area of hemisphere + Surface area of the cone

Firstly, consider the hemisphere:

The surface area of a hemisphere is $=2\pi r^2$

or $=2\times \frac{22}{7}\times {\left(3.5\right)}^2$

or $=77c{m}^2$

Now, for the cone we have:

The surface area of a cone $=\pi rl$

Thus, we need to calculate the slant of the cone.

We know that:

$l^2=h^2+r^2$

or $={12}^2+{3.5}^2$

or $=\frac{625}{4}$

or $l=\frac{25}{2}=12.5cm$

Thus surface area of a cone $=\pi rl$

or $=\frac{22}{7}\times 3.5\times 12.5$

or $=137.5c{m}^2$

Hence, the total surface area of toy = $=77+137.5=214.5c{m}^2$

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.

Answer:

It is given that the hemisphere is mounted on the cuboid, thus, the hemisphere can take on complete as its diameter (which is maximum).

Thus, the greatest diameter of the hemisphere is 7 cm.

Now, the total surface area of the solid = Surface area of the cube + Surface area of the hemisphere - Area of the base of the hemisphere (as this is counted on one side of the cube)

The surface area of the cube is :

$=6a^3$

$=6\times 7^3=294c{m}^2$

Now the area of a hemisphere is

$=2\pi r^2$

$=2\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^2=77c{m}^2$

And the area of the base of a hemisphere is

$=\pi r^2=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^2=38.5c{m}^2$

Hence, the surface area of the solid is $=294+77-38.5=332.5c{m}^2$.

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.

Answer:

It is given that the diameter of the hemisphere is equal to the edge length of the cube.

The total surface area of the solid is given by :

The surface area of a solid = Surface area of the cube + Surface area of the hemisphere - Area of the base of the hemisphere

The surface area of the cube $=6l^2$

And the surface area of the hemisphere:

$=2\pi r^2=2\pi {\left(\frac{l}{2}\right)}^2$

Area of the base of the hemisphere:

$=\pi r^2=\pi {\left(\frac{l}{2}\right)}^2$

Thus, the area of the solid is:

$=6l^2+\pi {\left(\frac{l}{2}\right)}^{2}uni{t}^2$

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm, and the diameter of the capsule is 5 mm. Find its surface area.

Answer:

It is clear from the figure that the capsule has a hemisphere and cylinder structure.

The surface area of the capsule = 2 (Area of the hemisphere) + Area of the cylindrical part

Area of hemisphere $=2\pi r^2$

or $=2\pi \times {\left(\frac{5}{2}\right)}^2$

or $=\frac{25}{2}\pi m{m}^2$

And the area of the cylinder $=2\pi rh$

or $=2\pi \times \frac{5}{2}\times 9$

or $=45\pi m{m}^2$

Thus, the area of the solid is :

$=2\left(\frac{25}{2}\right)\pi +45\pi$

$=25\pi +45\pi$

$=70\pi$

$=220m{m}^2$

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 Rs 500 per m ². (Note that the base of the tent will not be covered with canvas.)

Answer:

The canvas will cover the cylindrical part as well as the conical part.

So, the area of canvas = Area of cylindrical part (curved) + Area of the conical part

Now, the area of the cylindrical part is $=2\pi rh$

or $=2\pi \times 2\times 2.1$

or $=8.4\pi m^2$

And the area of the cone is $=\pi rl$

or $=\pi \times 2\times 2.8$

or $=5.6\pi m^2$

Thus, the area of the canvas $=8.4\pi +5.6\pi$

or $=14\pi =44m^2$

Further, it is given that the rate of canvas per m ² is = Rs. 500.

Thus, the required money is $=500\times 44=Rs.22,000$

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.

Answer:

Firstly, we need to calculate the slant height of the cone :

$l^2=r^2+h^2$

or $=(0.7{)}^2+(2.4{)}^2$

or $l^2=6.25$

or $l=2.5cm$

Now, the total surface area of the solid can be calculated as :

The surface area of a solid = Surface area of cylinder + Surface area of cone + Area of base of the cylinder

The surface area of the cylinder is $=2\pi rh$

or $=2\pi \times 0.7\times 2.4$

or $=10.56c{m}^2$

Now, the surface area of a cone $=\pi rl$

or $=\pi \times 0.7\times 2.5$

or $=5.50c{m}^2$

And the area of the base of the cylinder is $=\pi r^2$

or $=\pi \times 0.7\times 0.7$

or $=1.54c{m}^2$

Thus required area of the solid = 10. 56 + 5.50 + 1.54 = 17.60 cm² .

Thus total surface area of the remaining solid to the nearest cm² is 18 cm².

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.

Answer:

The required surface area is given by :

The surface area of the article = Surface area of the cylindrical part + 2 (Surface area of the hemisphere)

Now, the area of the cylinder $=2\pi rh$

or $=2\pi \times 3.5\times 10$

or $=70\pi c{m}^2$

And the surface area of the hemisphere: $=2\pi r^2$

or $=2\pi \times 3.5\times 3.5$

or $=24.5\pi c{m}^2$

Thus, the required area $=70\pi +2(24.5\pi )=374c{m}^2$