These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4

Question 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:

Diameters of two circular ends are 4 cm and 2 cm

∴ Radius are 2 cm and 1 cm.

Volume of the furstum of the cone

Question 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 curved surface area of the frustum.

Solution:

Let r_{1} and r_{1} be the radii of the circular bases of the frustum, l be the slant height and h be its height.

We have, l = 4 cm, 2πr_{1} = 18 and 2πr_{2} = 6

⇒ l = 4 cm, r_{1} = — and r_{2} = —

Curved surface area = l

= π\(\left(\frac{9}{\pi}+\frac{3}{\pi}\right)\) x 4 cm²

= π\(\left(\frac{9+3}{\pi}\right)\) x 4 cm²

= (12 x 4) cm²

= 48 cm².

Question 3.

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). 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.

Solution:

Radius of open side (r_{1}) = 10 cm

Radius of upper base (r_{2}) = 4 cm

Slant height (l) = 15 cm

Area of material used for making = Curved surface area of frustum of cone + area of closed side

Area of material used for cap = 710\(\frac { 2 }{ 3 }\) cm².

Question 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 ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm². (Take π = 3.14)

Solution:

Radius of the lower end (r_{1}) = 8 cm

Radius of the upper end (r_{2}) = 20 cm

Height of the frustum (h) = 16 cm

Cost of milk at the rate of ₹ 20 per litre = ₹(20 x 10.45) = ₹ 209

Now, Total surface area of the frustum = π(r_{1}) + r_{2}) l + πr_{2}² [∵ Top is open]

= [3.14 (20 + 8) x 20 + 3.14 x 8²] cm²

= 3.14 x (560 + 64) cm²

= 3.14 x 624 cm²

= 1959.36 cm²

Cost of metal used = ₹\(\left(\frac{1959.36 \times 8}{100}\right)\) = ₹ 156.75

Question 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 be drawn into a wire of diameter \(\frac { 1 }{ 16 }\) cm, find the length of the wire.

Solution:

Height of right circular cone = 20 cm

In ∆ABP,

Diameter of wire = \(\frac { 1 }{ 16 }\) cm

Radius of wire = \(\frac { 1 }{ 32 }\) cm

Volume of wire = πr²h (h = 1)

= π(\(\frac { 1 }{ 32 }\))² x l … (2)

From (1) and (2)

π(\(\frac { 1 }{ 32 }\))² x l = \(\frac { 7000π }{ 9 }\)

l = \(\frac{7000 \times 32 \times 32}{9}\)

= 79644.44 cm = 7964.4 m

∴ Length of the wire = 7964.4 m