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

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

## 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

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 ∠BAC = 90°

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

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

BC^{2} = 25

BC = 5 cm

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

ΔADB – ΔCAB (by AA similarity)

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}

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:

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).

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. ΔLMQ and rt. ΔLNS

∠1 = ∠1 (common)

∠M = ∠N (each 900)

∴ ΔLMQ – ΔLNS (by AA similarity)

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

= πr_{1}l_{1} – πr_{2}(l_{1} – l)

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. ΔLMQ and rt. ΔLNS (Common)

∠M = ∠N (Each 900)

ΔLMN – ΔLNS (By AA similarity)

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

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