NCERT Solutions for Class 10 Maths Exercise 13.5 Chapter 13 Surface Areas and Volumes- FREE PDF Download

NCERT Class 10 Maths Ch 13 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths
Chapter 13 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Surface Areas and Volumes solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes

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.

Ans. Number of rounds to cover 12 cm, i.e. 120 mm =

= 40

Here, Diameter = 10 cm, Radius cm

Length of the wire used in taking one round

= cm

Length of the wire used in taking 40 rounds

= cm

Radius of the copper wire = mm

= cm

Volume of wire =

= ^{$9 π^{2} c m^{3}$ ————–***}

Mass of the wire =

= 787.98 gm

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 as found appropriate)

Ans. Hypotenuse =

= 5 cm

In figure, ADB CAB [AA similarity]

AD = cm

Also,

DB = cm

CD = BC – DB = cm

Volume of the double cone

= =

Surface area of the double cone

= =

3. A cistern, internally measuring has 129600 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?

Ans. Volume of cistern =

Volume of water =

Volume of cistern to be filled

= 1980000 – 129600 =

Volume of a brick =

Let bricks be needed.

Then, water absorbed by bricks =

= 1792 (approx.)

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 km², 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.

Ans. Volume of rainfall =

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

= 0.728 km²

Volume of three rivers = = 0.7236 km²

Hence, the amount of rainfall is approximately equal to the amount of water in three rivers.

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).

Ans. Slant height of the frustum of the cone

= = 13 cm

Area of the tin sheet required

= CSA of cylinder + CSA of the frustum

= = =

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

Ans. According to the question, the frustum is the difference of the two cones OAB and OCD (in figure).

For frustum, height = slant height = and radii of the bases = and

OP = OA = OB =

Height of the cone OCD =

OQD OPB [ By, AA similarity]

……….(i)

height of the cone OCD =

= = ……….(ii)

V of the frustum

= V of cone OAB – V of cone OCD

[From eq. (i) & (ii)]

If are the surface areas of two circular bases, then

A₁ = and A₂ =

V of the frustum

7. 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.

Ans.

For frustum, height = slant height = and radii of the bases = and

OP = OA = OB =

Again, from DEB,