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**NCERT TEXTBOOK QUESTIONS SOLVED****Page No. 258****EXERCISE 13.5**

**Q 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 cm ^{3 }.**

âˆ´ Radius of the cylinder (r) = 10/2 cm = 5 cm

â‡’ Length of wire in completely one round

2Ï€r = 2 Ã— 3.14 Ã— 5 cm = 31.4 cm

âˆµ Diameter of wire = 3 mm = 3/10 cm

âˆ´ The thickness of cylinder covered in one round = 3/10 cm

â‡’ Number of rounds (turns) of the wire to cover 12 cm

âˆ´ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds

= 40 Ã— 31.4 cm = 1256 cm

Now, radius of the wire

âˆ´ Volume of wire = Ï€r

âˆµ Density of wire = 8.88 gm/cm

âˆ´Weight of the wire = [Volume of the wire] Ã— Density

= 788 g (approx.)

AB = 3 cm, AC = 4 cm

âˆ´ Hypotenuse

Obviously, we have obtained two cones on the same base AAâ€² such that the radius = DA or DA'

Now, volume of the double cone

Surface area of the double cone

âˆ´ Volume of the cistern = 150 Ã— 120 Ã— 110 cm

Volume of water contained in the cistern = 129600 cm

âˆ´ Free space (volume) which is not filled with water = 1980000 âˆ’ 129600 cm

Now, Volume of one brick = 22.5 Ã— 7.5 Ã— 6.5 cm

âˆ´ Volume of water absorbed by one brick

Let â€˜nâ€™ bricks can be put in the cistern.

âˆ´ Volume of water absorbed by â€˜nâ€™

âˆ´ [Volume occupied by â€˜nâ€™ bricks] = [(free space in the cistern) + (volume of water absorbed by n-bricks)]

Thus, 1792 bricks can be put in the cistern.

Volume of rainfall = (surface area) Ã— (height of rainfall)

Since, 0.7236 km

âˆ´ The additional water in the three rivers is not equivalent to the rainfall.

Diameter = 8 cm

â‡’ Radius (r) = 4 cm

Height = 10 cm

â‡’ Curved Surface area

For the frustum:

and

Height (H) = 22 âˆ’ 10 = 12 cm

âˆ´ Slant height (l)

âˆ´Surface area,

Area of tin required = [Area of the frustum] + [Area of cylindrical portion]

Curved surface area of the frustum PQRS

= Ï€r

Now, Î”OC

Now, from (1), curved surface area of the frustum

Now, the total surface area of the frustum = (curved surface area) + (base surface area) + (top surfaces area)

Since Î”OC_{1}Q ~ Î”OC_{2}S

From (1) and (2), we have

{volume of the frustum RPQS}

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