Gujarat Board GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.

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 cm3.

Solution:

In case of cylinder

Diameter = 10 cm

radius r = 5 cm

Diameter of wire which is rounded over cylinder = 3 mm

Length of cylinder = 12 cm

= 12 x 10 mm = 120 mm

Number of round to cover 120 mm

= \(\frac {120}{3}\) = 40

Length of wire in completing one round over the cylinder = 2Ï€r

= 2Ï€ x 5 = 10Ï€ cm

Length of wire in covering whole surface of the cylinder = Length of wire in completing 40 rounds

= 10 Ï€ x 40

= 400 Ï€ cm

= 400 x 3.14 = 1256 cm

Radius of copper wire = \(\frac {3}{2}\)mm

r_{1} = \(\frac {3}{20}\)cm

Volume of wire = Ï€r^{2}_{2}h

= Ï€ (\(\frac {3}{20}\)) x 400n = 9Ï€^{2} cm^{3}

Mass of wire = 9 Ï€^{2} x 8.88

= 9 x (3.14)^{2} x 8.88 g

= 787.98g â‰ˆ 788g

Question 2.

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 n as found appropriate.)

Solution:

Here

ABC is a right triangle in âˆ BAC = 90Â°

BC^{2} = AB^{2} + AC^{2}

= 3^{2} + 4^{2} = 9 + 16

BC^{2} = 25

BC = 5 cm

Here AD and Aâ€™D are radii of two cones which are formed by rotating the right triangle about the hypotenuse BC.

Î”ADB – Î”CAB (by AA similarity)

\(\frac {AD}{AC}\) = \(\frac {AB}{BC}\)

â‡’ \(\frac {AD}{4}\) = \(\frac {3}{5}\)

â‡’ AD = cm (r_{1} or r_{2} = AD let)

Also \(\frac {BD}{AB}\) = \(\frac {AB}{BC}\)

\(\frac {BD}{3}\) = \(\frac {3}{5}\)

â‡’ BD = \(\frac {3}{5}\) cm (h_{1} = BD say)

Hence CD = 5 – \(\frac {9}{5}\)

h_{2} = \(\frac {16}{5}\)cm (h_{2} CD say)

Volume of double cones

And surface area of the double cones

= Ï€r_{1}l_{1} + Ï€r_{2}l_{2} = Ï€ (r_{1}l_{1} + r_{2}l_{2})

= 3.14\(\left[\frac{12}{5} \times 3+\frac{12}{5} \times 4\right]\) [l_{1} = 3cm , l_{2} = 4cm ]

= 3.14 x \(\frac {12}{5}\) [3 + 4]

= \(\frac{3.14 \times 12 \times 7}{5}\) = 52.75 cm^{2}

Question 3.

A cistern, internally measuring 150 cm x 120 cm x 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 x 7.5 cm x 6.5 cm?

Solution:

Volume of cistern = l x b x h

= 150 x 120 x 110

= 1980000 cm^{3}

Volume of water in cistern = 129600 cm^{3}

Volume of cistern to be filled

= Volume of cistern – Volume of water

= 1980000 – 129600

= 1850400 cm^{3}

Let n bricks are needed to fill the cistern up to the brim.

Volume of one brick

= l x b x h

= 22.5 x 7.5 x 6.5

= 1096.875 cm^{3}

Water absorbed by one brick = \(\frac{1096.875}{17}\) cm^{3
}Water absorbed by n brick = \(\frac{n(1096.875)}{17}\) cm^{3
}According to the question,

n = 1792.41

Hence number of bricks cannot be put in the cistern = 1792

Question 4.

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 and each 1072 km long, 75 m wide and 3 m deep.

Solution:

Area of valley = 97280 km^{2}

Rainfall = 10 cm

= \(\frac {10}{100}\) metre = \(\frac{10}{100 \times 1000}\) km

Volume of rainfall = Area of river valley x Rainfall

= 7280 x \(\frac{10}{100 \times 1000}\)

= 0.728 km^{3}

Volume of three river = 3l x b x h

= 3 x 1072 x \(\frac {75}{1000}\) x \(\frac {3}{1000}\)

= 0.7236 km^{3}

Hence, total rainfall and normal water of three rivers are approximately equivalent.

Question 5.

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, 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 figure).

Solution:

Here mease of frustum

r_{1} = \(\frac {18}{2}\) = 9 cm

r_{1} = \(\frac {8}{2}\) = 4 cm

h = 22 – 10 = 12 cm

Slant height of frustum

l = \(\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}\)

=\(\sqrt{(9-4)^{2}+12^{2}}\) = \(\sqrt{5^{2}+12^{2}}\)

= \(\sqrt{25+144}\) = \(\sqrt{169}\) = 13cm

Area of tin sheet required = Curved surface area of the cylindrical portion + Curved surface area of the frustum portion

= 2Ï€r_{2}h_{2} + Ï€(r_{1} + r_{2})l

= 2 x Ï€ x 4 x 10 + Ï€(4 + 9) x 13

= 80Ï€ x 169Ï€ = 249Ï€

= 249 x \(\frac{22}{7}\) = \(\frac{5478}{7}\) = 782 \(\frac{4}{7}\) cm^{2}

Question 6.

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 are explained.

Solution:

Let r_{1} and r_{2} be the radii of circular base, where r_{1} > r_{2} and h be the height of the frustum of a cone.

Let h_{1} be the height of large cone LPQ

Height of small cone = h_{1} – h

Let l_{1} be the slant height of cone LPQ and l be the frustum of cone. The slant height of Cone LRS

= l_{1} – l

In rt. Î”LMQ and rt. Î”LNS

âˆ 1 = âˆ 1 (common)

âˆ M = âˆ N (each 900)

âˆ´ Î”LMQ – Î”LNS (by AA similarity)

(i) Lateral surface area or curved surface area of a frustum of cone = CSA of large cone – CSA of small cone

= Ï€r_{1}l_{1} – Ï€r_{2}(l_{1} – l)

= Ï€l(r_{1} + r_{2}) = Ï€(r_{1} + r_{2})l sq. units

(ii) Slant height = \(\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}\)

(iii) Total surface area of frustum of a cone + Ï€r_{1}r^{2} + Ï€r_{2}r^{2}

= CSA of frustum of cone + Ï€r^{2}_{1} + Ï€r^{2}_{2}

= Ï€(r_{1} + r_{2})l + Ï€r^{2}_{1} + Ï€r^{2}_{2} sq. units

Question 7.

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

Solution:

Let r_{1} and r_{2} be radii of two circular bases where r_{1} > r_{2}, h be the height of frustum of a cone.

Let h_{1} be the height of large cone LPD and height of small cone LRS = h_{1} – h

Let l_{1}, be the slant height of cone and t be the slant height of frustum of a cone.

Slant height of cone LRS = l_{1} – l

In rt. Î”LMQ and rt. Î”LNS (Common)

âˆ M = âˆ N (Each 900)

Î”LMN – Î”LNS (By AA similarity)

Volume of a frustum of a cone = Volume of large cone LPQ – Volume of small cone LRS