Solve the followings Questions.

Unless stated otherwise, use π =22/7

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:

Here, 

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 curved surface area of the frustum.

Answer:

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 = 3/π

Now, CSA of frustum = π(r₁+r₂) × l

= π(9/π+3/π) × 4

= 12×4 = 48 cm²

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.

Answer:

, 

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 to be used for making the fez = 210π + 16π = (226 x 22)/7 = 710 2/7

∴ 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 total cost of milk which can completely fill the container at the rate of Rs. 20 per liter. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm². ( take π = 3.14)

Answer:

Given,

r₁ = 20 cm,

r₂ = 8 cm and

h = 16 cm

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

= 1/3 ×3.14 ×16((20)²+(8)²+(20)(8))

= 1/3 ×3.14 ×16(400 + 64 + 160) = 10449.92 cm³ = 10.45 lit

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

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

= 20 × 10.45

= Rs. 209

Now, slant height will be

l = 20 cm

So, CSA of the container = π(r₁+r₂)×l

= 1758.4 cm²

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

= 1758.4+πr² = 1758.4+π(8)²

= 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 60⁰ 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 1/16 cm, find the length of the wire.

Answer:

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2