CLASS 10 MATH NCERT SOLUTIONS FOR CHAPTER – 13 SURFACE AREAS AND VOLUMES EX – 13.4

Surface Areas and Volumes

Page No: 257

1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution:

Radius (r1) of the upper base = 4/2 = 2 cm

Radius (r2) of lower the base = 2/2 = 1 cm

Height = 14 cm

Now, Capacity of glass = Volume of frustum of cone

So, Capacity of glass = (⅓)×π×h(r12+r22+r1r2)

= (⅓)×π×(14)(22+12+ (2)(1))

∴ The capacity of the glass = 102×(⅔) cm3

2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the surface area of the frustum.

Solution:

Given,

Slant height (l) = 4 cm

Circumference of upper circular end of the frustum = 18 cm

∴ 2πr1 = 18

Or, r1 = 9/π

Similarly, circumference of lower end of the frustum = 6 cm

∴ 2πr2 = 6

Or, r2 = 6/π

Now, CSA of frustum = π(r1+r2) × l

= π(9/π+6/π) × 4

= 12×4 = 48 cm2

3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Ncert solutions class 10 chapter 13-23

Solution:

Given,

For the lower circular end, radius (r1) = 10 cm

For the upper circular end, radius (r2) = 4 cm

Slant height (l) of frustum = 15 cm

Now,

The area of material to be used for making the fez = CSA of frustum + Area of upper circular end

CSA of frustum = π(r1+r2)×l

= 210π

And, Area of upper circular end = πr22

= 16π

∴ The area of material used = 710 × (2/7) cm2

4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.

Solution:

Given,

r1 = 20 cm,

r2 = 8 cm and

h = 16 cm

∴ Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

Ncert solutions class 10 chapter 13-24

It is given that the rate of milk = Rs. 20/litre

So, Cost of milk = 20×volume of the frustum

Ncert solutions class 10 chapter 13-25

= Rs. 209

Now, slant height will be

Ncert solutions class 10 chapter 13-26

So, CSA of the container = π(r1+r2)×l

Ncert solutions class 10 chapter 13-27

= 1758.4 cm2

Hence, the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)

= 1758.4+201 = 1959.4 cm2

∴ Total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157

5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution:

The diagram will be as follows

NCERT Solutions Class 10 chapter 13-27

Consider AEG

Ncert solutions class 10 chapter 13-29

Radius (r1) of upper end of frustum = (10√3)/3 cm

Radius (r2) of lower end of container = (20√3)/3 cm

Height (r3) of container = 10 cm

Now,

Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

Ncert solutions class 10 chapter 13-30

Solving this we get,

Volume of the frustum = 22000/9 cm3

The radius (r) of wire = (1/16)×(½) = 1/32 cm

Now,

Let the length of wire be “l”.

Volume of wire = Area of cross-section x Length

= (πr2)xl

= π(1/32)2x l

Now, Volume of frustum = Volume of wire

22000/9 = (22/7)x(1/32)2x l

Solving this we get,

l = 7964.44 m

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