HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Haryana State Board HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Take π = \(\frac{22}{7}\) unless stated otherwise

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:
We have,
Radius of glass of upper end (r₁) = \(\frac{4}{2}\) = 2 cm
Radius of glass of lower end (r₂) = \(\frac{2}{2}\) = 1 cm
Height of the glass (h) = 14cm
Capacity (volume) of the glass = \(\frac{\pi h}{3}\) (r₁² + r₂² + r₁r₂)
= \(\frac{22 \times 14}{7 \times 3}\) (2² + 1² + 2 × 1)
= \(\frac{44}{3}\) (4 + 1 + 2)
= \(\frac{44 \times 7}{3}\)
= 10267 cm³
Hence, capacity of the glass = 10267 cm³.

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:
We have,
Perimeter (circumference) of the upper end = 18 cm
2πr₁ = 18
r₁ = \(\frac{18}{2 \pi}=\frac{9}{\pi}\) cm
Perimeter (circumference) of the lower end = 6 cm
2πr₂ = 6
r₂ = \(\frac{6}{2 \pi}=\frac{3}{\pi}\)
Slant height of frustum of cone (l) = 4 cm
Curved surface area of frustum of cone = πl (r₁ + r₂)
= π × 4\(\left(\frac{9}{\pi}+\frac{3}{\pi}\right)\)
= π × 4 × \(\frac{12}{\pi}\)
= 4 × 12 = 48 cm².
Hence, curved surface area of frustum of cone = 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:
We have,
Radius of fez at the upper base (closed base) = r₂ = 4 cm
Radius of fez at lower base (open base) = r₁ = 10 cm

Slant height of the fez (l) = 15 cm
∴ Area of material used = C.S.A. + Area of closed base
= π l (r₁ + r₂) + π r₂²
= \(\frac{22}{7}\) × 15 (10 + 4) + \(\frac{22}{7}\) × 4
= \(\frac{22}{7}\) × 15(14) + \(\frac{22}{7}\) × 16
= \(\frac{22}{7}\) × 15 × 14 + 50\(\frac{2}{7}\)
= 660 + 50\(\frac{2}{7}\)
= 710\(\frac{2}{7}\) cm²
Hencte, area of material used = 710\(\frac{2}{7}\) 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).
OR
A bucket is in the form of a frustum of cone of height 16 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the bucket, at the rate of ₹ 40 per liotre. (use π = 3.14)
Solution:
We have,
Radius of container of upper end (r₁) = 20 cm
Radius of container of lower end (r₂) = 8 cm
Height of container (h) = 16 cm

Volume of container = \(\frac{\pi h}{3}\) (r₁² + r₂² + r₁r₂)
= \(\frac{3.14 \times 16}{3}\) (20² + 8² + 20 × 8)
= \(\frac{50 \cdot 24}{3}\) × 624
= 50.24 × 208
= 10449.92 cm³.
Milk in the container = \(\frac{10449.92}{1000}\) litres
= 10.45 litres (approx)
Cost of the milk = 10.45 × 20 = ₹ 209.
Area of the metal sheet used = πl (r₁ + r₂) + πr₂²
= 3.14 × 20(20 + 8) + 3.14 × 8²
= 62.8 × 28 + 200.96
= 1758.4 + 200.96 = 1959.36 cm².
Cost of metal sheet used = ₹ \(\frac{1959 \cdot 36 \times 8}{100}\)
= ₹ 156.75
Hence, cost of milk = ₹ 209
and cost of metal sheet used = ₹ 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:
We have,
Height of the cone = 20 cm
Since cone is cut into parts at the middle.
So, height of the frustum on , (ACDB) = h = \(\frac{20}{2}\) = 10 cm
In right ∆VOB tan 30° = \(\frac{\mathrm{BO}}{\mathrm{VO}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{r_1}{20}\)
⇒ r₁ = \(\frac{20}{\sqrt{3}}\) cm
and in right triangle VO’D
tan 30° = \(\frac{\mathrm{O}^{\prime} \mathrm{D}}{\mathrm{VO}^{\prime}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{r_2}{10}\)
⇒ r₂ = \(\frac{10}{\sqrt{3}}\) cm

Radius of wire = \(\frac{1}{2 \times 16} \Rightarrow \frac{1}{32}\) cm
Let length of wire be x cm.
Volume of wire drawn = π × \(\left(\frac{1}{32}\right)^2\) × x
According to question,
Volume of wire drawn = Volume of metal used in the frustum
⇒ \(\pi \times\left(\frac{1}{32}\right)^2 \times x=\frac{7000 \pi}{9}\)
⇒ \(\pi \times \frac{1}{1024} \times x=\frac{7000 \pi}{9}\)
⇒ x = \(\frac{7000 \pi \times 1024}{\pi \times 9}\)
⇒ x = 796444.44 cm
⇒ x = \(\frac{796444 \cdot 44}{100}\) m
⇒ x = 7964.4 m.
Hence, length of wire = 7964.4 m.