MCQ: Cone - 1 - SSC CGL MCQ

The height of a right circular cone is trisected by two planes parallel to its base at equal distances.

Formula used:

Cone:

The volume of frustum

Calculation:

According to the question, the required figure is:

Now, 

The volume of the cone AB''D'',

The volume of frustum cone B'D'BD,

The required ratio 

∴ 1 : 7 : 19 is the requried ratio.

Given:

Radius of the cone = 10 cm

Height of the cone = 24 cm

Formula:

Total surface of area of a cone = πR(L + R)

Where, R = Radius of the base, H = Height of the cone,

L = Slant height of the cone

Calculation:

Total surface area of the cone

⇒ π x 10 x (26 + 10)

⇒ 22/7 x 10 x 36

⇒ 7920/7 = 1131.428 ≈ 1131.43 cm²

∴ The total surface area of the cone is 1131.43 cm².

The slant height of cone (l) = 26 cm

The volume of cone = 800π cm³

The ratio of height (h) and radius (r) = 12 : 5

Concept used:

The volume of the cone = 1/3 x π x r² x h

l² = r² + h²

Calculation:

Let the height and radius be 12x and 5x


∴ Volume is 800π cm³

Given:

Radius = 14m

Height = 48m

Width of the canvas = 8m

Formula used:

Pythagoras Theorem,

Slant height² = Radius² + Height²

Curved Surface area of the cone = π r l [ where r is radius and l is slant height]

Calculation:

Curved Surface area of the cone = π r l

Canvas required if the width is 8m,

= 2200 / 8
= 275
Answer is 275.

Given:

Radius of cone = 5 cm
Height of cone = 12 cm

Formula:

Curved surface area of cone = πrl
Base area of cone = πr²
l² = r² + h²

Calculation:

In an Δ OAB, using Pythagoras' theorem,

⇒ l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l = √169

⇒ l = 13 cm

Required ratio = πrl : πr²

⇒ l : r

⇒ 13 : 5

Given:

The volume of frustum of a cone is 13244 m³.
The radius of the top circular surface is 9 m.
The radius of the other circular surface is 11 m.

Formula used:

Volume of frustum of a cone  

Where, 

The radius of the top circular surface is r.
The radius of the bottom circular surface is R.
The height of the frustum of a cone is h.

Calculation:

According to the question, the required figure is:

Let V be the volume of the frustum of a cone.

According to the question,

∴ The height of the frustum of a cone is 42 m.

Let the radius, height and slant height of cone be ‘r’, ‘h’ and ‘l’ units respectively
Given, curved surface area = 90% of Total surface area of cone
⇒ πrl = (9/10) × πr(l + r)
⇒ 10l = 9l + 9r
⇒ l = 9r
Also, l² = h² + r²
⇒ 81r² = h² + r²
⇒ h² = 80r²
⇒ r² = h2/80
Now,
Volume of cone = (1/3)πr²h
⇒ Volume of cone = π/3 x h²/80 c h
⇒ Volume of cone = (π/240) x h³
∴ The volume of cone is (π/240) times the cube of the height of the cone

Given:

Top radius of frustum (r1) = 20 cm
Bottom radius of frustum (r2) = 60/2 = 30 cm (as diameter is given)
Height of frustum (h1) = 40 cm

Calculation:

The ratio of height to radius

∴ The height of the cone that is to be filled should be 80 cm.

Given:

Height =  21 cm
R = 15 cm
r = 7 cm

Formula:

Calculation:

∴ The volume of a tumbler 8338 cm³ 

The slant height of a frustum of cone is 10 cm and the height of the frustum is 8 cm.

Calculation:

We know, the relationship between slant height, height and radius

Let R and r be the radius of two circular ends.

Slant height is l = 10 cm

Height is h = 8 cm

Therefore, The difference of radii of its two circular ends is 6 cm.