Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.4

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.

Answer:

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.

Answer:

Let R and r be the radii of two circular ends and slant height = l = 4 cm

2πR = 18, πR = \(\frac{18}{2}\) = 9cm

2πr = 6, πr = \(\frac{6}{2}\) = 3cm

Curved surface area of frustum

= π (R + r)l = (πR + πr)l

= (9 + 3) × 4 = 12 × 4

= 48 cm^{2}

Question 3.

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 15.24). 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.

Answer:

R = 10 cm, r = 4cm and l = 15.

Area of the material used for making the fez = Surface area of frustum + Area of top circular section

= π (R + r)l + π r^{2}

= π [(R + r)l + r^{2}]

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 cm^{2}. (Take π = 3.14)

Answer:

R = 20 cm, r = 8 cm, h = 16 cm.

Cost of milk to fill the container completely at the rate of ₹ 20 per litre.

= 20 × 10.44992

= ₹ 208.9984 = ₹ 209

Surface area

= π (R + r)l + πr^{2}

= 3.14 (20 + 8) 20 + 3.14 × (8)^{2}

= 3.14 × 28 × 20 + 3.14 × 64.

= 3.14 [560 + 64]

= 3.14 × 624

= 1959.36 cm^{2}

Total cost of metal sheet used to make the container at the rate of ₹ 8 Per 100 cm^{2} = ₹\(\frac{8}{100}\) × 1959.36 = ₹ 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

Answer:

Let VAB be the metallic right circular cone of height 20 cm.

Suppose this cone is cut by a plane parallel to its base at a point O_{1} such that VO_{1} = O_{1}O i.e. O_{1} is the midpoint of VO.

Let r_{1} and r_{2} be the radii of circular ends of the frustum ABB_{1}A_{1}

Let the length of wire of diameter = \(\frac{1}{16}\)cm be l cm.

Volume of metal used in wire

= π × \(\left(\frac{1}{32}\right)^{2}\) × 1

= \(\frac{\pi l}{1024}\) cm^{2}.

Since the frustum is recast into a wire of length l cm and diameter \(\) cm.

∴ Volume of metal used in wire = volume of the frustum