## EXERCISE 13.5

#### Page No 258:

#### 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 cm^{3}.

#### Answer:

It can be observed that 1 round of wire will cover 3 mm height of cylinder.

Length of wire required in 1 round = Circumference of base of cylinder

= 2π*r* = 2π × 5 = 10π

Length of wire in 40 rounds = 40 × 10π

= 1257.14 cm = 12.57 m

Radius of wire

Volume of wire = Area of cross-section of wire × Length of wire

= π(0.15)^{2} × 1257.14

= 88.898 cm^{3}

Mass = Volume × Density

= 88.898 × 8.88

= 789.41 gm

#### 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 π as found appropriate.)

#### Answer:

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

Hypotenuse

= 5 cm

Area of ΔABC

Volume of double cone = Volume of cone 1 + Volume of cone 2

= 30.14 cm^{3}

Surface area of double cone = Surface area of cone 1 + Surface area of cone 2

= π*rl*_{1} + π*rl*_{2}

= 52.75 cm^{2}

#### Question 3:

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:

Volume of cistern = 150 × 120 × 110

= 1980000 cm^{3}

Volume to be filled in cistern = 1980000 − 129600

= 1850400 cm^{3}

Let *n* numbers of porous bricks were placed in the cistern.

Volume of *n* bricks = *n* × 22.5 × 7.5 × 6.5

= 1096.875*n*

As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these bricks

*n* = 1792.41

Therefore, 1792 bricks were placed in the cistern.

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

#### Answer:

Area of the valley = 7280 km^{2}

If there was a rainfall of 10 cm in the valley then amount of rainfall in the valley = Area of the valley × 10 cm

Amount of rainfall in the valley = 7280 km^{2} × 10 cm

=7280×1000m2×10100m=7280×105m3=7.28×108m3

Length of each river, *l* = 1072 km = 1072 × 1000 m = 1072000 m

Breadth of each river, *b* = 75 m

Depth of each river, *h* = 3 m

Volume of each river = *l* × *b* × *h*

= 1072000 × 75 × 3 m^{3}

= 2.412 × 10^{8 }m^{3}

Volume of three such rivers = 3 × Volume of each river

= 3 × 2.412 × 10^{8} m^{3}^{ }

= 7.236 × 10^{8 }m^{3}

Thus, the total rainfall is approximately same as the volume of the three rivers.

#### 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 the given figure).

#### Answer:

Radius (*r*_{1}) of upper circular end of frustum part

Radius (*r*_{2}) of lower circular end of frustum part = Radius of circular end of cylindrical

part

Height (*h*_{1}) of frustum part = 22 − 10 = 12 cm

Height (*h*_{2}) of cylindrical part = 10 cm

Slant height (*l*) of frustum part

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

#### Question 6:

Derive the formula for the curved surface area and total surface area of the frustum of cone.

#### Answer:

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let *r*_{1} and *r*_{2} be the radii of the ends of the frustum of the cone and *h* be the height of the frustum of the cone.

In ΔABG and ΔADF, DF||BG

∴ ΔABG ∼ ΔADF

CSA of frustum DECB = CSA of cone ABC − CSA cone ADE

CSA of frustum =

#### Question 7:

Derive the formula for the volume of the frustum of a cone.

#### Answer:

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base.

Let *r*_{1} and *r*_{2} be the radii of the ends of the frustum of the cone and *h* be the height of the frustum of the cone.

In ΔABG and ΔADF, DF||BG

∴ ΔABG ∼ ΔADF

Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE