NCERT Solutions for Exercise 13.5 Class 10 Maths Chapter 13 - Surface Area and Volumes

Surface Areas and Volumes Class 10 Chapter 13 Excercise: 13.5

Q1 A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm ³ .

Answer:

A number of rounds are calculated by :

$=\ \frac{Height\ of\ cylinder }{Diameter\ of\ wire}$

$=\ \frac{12 }{0.3}\ =\ 40\ rounds$

Thus the length of wire in 40 rounds will be $=\ 40\times 2\pi \times 5\ =\ 400 \pi\ cm\ =\ 12.57\ m$

And the volume of wire is: Area of cross-section $\times$ Length of wire

$=\pi \times (0.15)^2\times 1257.14$

$=\ 88.89\ cm^3$

Hence the mass of wire is. _{$=\ 88.89\ \times 8.88\ =\ 789.41\ gm$}

Q2 A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\pi$ as found appropriate.)

Answer:

The volume of the double cone will be = Volume of cone 1 + Volume of cone 2.

$=\ \frac{1}{3}\ \pi r^2h_1\ +\ \frac{1}{3}\ \pi r^2h_2$

$=\ \frac{1}{3}\ \pi \times 2.4^2\times 5$ (Note that sum of heights of both the cone is 5 cm - hypotenuse).

$=\ 30.14\ cm^3$

Now the surface area of a double cone is :

$=\ \pi rl_1\ +\ \pi rl_2$

$=\ \pi \times 2.4 \left ( 4\ +\ 3 \right )$

$=\ 52.8\ cm^2$

Q3 A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm ³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Answer:

The total volume of the cistern is : $=\ 150\times 120\times 110\ =\ 1980000\ cm^3$

And the volume to be filled in it is $= 1980000 - 129600\ =\ 1850400\ cm^3$

Now let the number of bricks be n.

Then the volume of bricks : $= n\times 22.5\times 7.5\times 6.5 \ =\ 1096.87n\ cm^3$

Further, it is given that brick absorbs one-seventeenth of its own volume of water.

Thus water absorbed :

$=\ \frac{1}{17}\times 1096.87n\ cm^3$

Hence we write :

$1850400\ +\ \frac{1}{17}( 1096.87n)\ =\ 1096.87n$

$n\ =\ 1792.41$

Thus the total number of bricks is 1792.

Q4 In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km ² , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer:

Firstly we will calculate the volume of rainfall :

Volume of rainfall :

$=\ 7280\times \frac{10}{100\times 1000}$

$=\ 0.7280\ Km ^3$

And the volume of the three rivers is :

$=\ 3\left ( 1072\times \frac{75}{1000}\times \frac{3}{1000} \right )\ Km^3$

$=\ 0.7236\ Km ^3$

It can be seen that both volumes are approximately equal to each other.

Q5 An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.).

Answer:

From this, we can write the values of both the radius (upper and lower) and the height of the frustum.

Thus slant height of frustum is :

$=\ \sqrt{\left ( r_1\ -\ r_2 \right )^2\ +\ h^2}$

$=\ \sqrt{\left ( 9\ -\ 4 \right )^2\ +\ 12^2}$

$=\ 13\ cm$

Now, the area of the tin shed required :

= Area of frustum + Area of the cylinder

$=\ \pi \left ( r_1\ +\ r_2 \right )l\ +\ 2\pi r_2h$

$=\ \pi \left ( 9\ +\ 4 \right )13\ +\ 2\pi \times 4\times 10$

$=\ 782.57\ cm^2$

Q6 Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer:

In the case of the frustum, we can consider:- removing a smaller cone (upper part) from a larger cone.

So the CSA of frustum becomes:- CSA of bigger cone - CSA of the smaller cone

$\\=\ \pi r_1l_1\ -\ \pi r_2 (l_1\ -\ l)\\\\=\ \pi r_1\left ( \frac{lr_1}{r_1\ -\ r_2} \right )\ -\ \pi r_2\left ( \frac{r_1l}{r_1\ -\ r_2}\ -\ l \right )\\\\=\ \left ( \frac{\pi r_1^2 l}{r_1\ -\ r_2} \right )\ -\ \left ( \frac{\pi r_2^2 l}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( \frac{r_1^2\ -\ r_2^2}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( r_1\ +\ r_2 \right )$

And the total surface area of the frustum is = CSA of frustum + Area of upper circle and area of lower circle.

$=\ \pi l\left ( r_1\ +\ r_2 \right )\ +\ \pi r_1^2\ +\ \pi r_2^2$

Q7 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer:

Similar to how we find the surface area of the frustum.

The volume of the frustum is given by-

=Volume of the bigger cone - Volume of the smaller cone

$=\ \frac{1}{3} \pi r^2_1 h_1\ -\ \frac{1}{3} \pi r^2_2 (h_1\ -\ h)$

$=\ \frac{\pi}{3}\left ( r_1^2 h_1\ -\ r_2^2 (h_1\ -\ h) \right )$

$=\ \frac{\pi}{3}\left ( r_1^2 \left ( \frac{hr_1}{r_1\ -\ r_2} \right )\ -\ r_2^2 \left ( \frac{hr_1}{r_1\ -\ r_2}\ -\ h \right ) \right )$

$=\ \frac{\pi}{3}\left ( \frac{hr_1^3}{r_1\ -\ r_2}\ -\ \frac{hr_2^3}{r_1\ -\ r_2} \right )$

$=\ \frac{\pi}{3}h\left ( \frac{r_1^3\ -\ r_2^3}{r_1\ -\ r_2} \right )$

$=\ \frac{1}{3} \pi h\left ( r_1^2\ +\ r_2^2\ +\ r_1r_2 \right )$