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

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

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 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
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 1

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 r1 and r1 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, 2πr1 = 18 and 2πr2 = 6
⇒ l = 4 cm, r1 = — and r2 = —
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.

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

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.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 2
Solution:
Radius of open side (r1) = 10 cm
Radius of upper base (r2) = 4 cm
Slant height (l) = 15 cm
Area of material used for making = Curved surface area of frustum of cone + area of closed side
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 3
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 (r1) = 8 cm
Radius of the upper end (r2) = 20 cm
Height of the frustum (h) = 16 cm
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 4
Cost of milk at the rate of ₹ 20 per litre = ₹(20 x 10.45) = ₹ 209
Now, Total surface area of the frustum = π(r1) + r2) l + πr2² [∵ 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

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

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,
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 5
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

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