Cone: Right Circular Cone MCQs Quiz to help you practise for the Quantitative Aptitude section of your exam. This will help you clear many interview rounds, competitive and entrance exams. This includes frequently asked Right Circular Cone Objective Questions. Your final target should be to solve all the questions with maximum accuracy. Regular practise of the Right Circular Cone Question Answer will help you get a good score. q

Option 4 : 770 cm^{3}

**Given****:**

Height of cone = 15 cm

Radius of cone = 7 cm

**Formula:**

Volume of cone = πr^{2}h/3

**Calculation:**

Volume of cone

⇒ [1/3] × π × r^{2} × h

⇒ [1/3] × [22/7] × 7 × 7 × 15

⇒ 22 × 7 × 5

⇒ 770 cm^{3}

Option 2 : 1936 cm^{2}

**Given****:**

Slant height of cone l = 30 cm

Radius of cone r = 14 cm

**Formula:**

Total surface area of cone = πr (l + r)

**Calculation:**

Total surface area of cone

⇒ πr(r + l)

⇒ [22/7] × 14 × (14 + 30)

⇒ 22 × 2 × 44

⇒ 1936 cm^{2}

Option 1 : 16π cm^{3}

**Given:**

Radius = 4 cm

CSA = 20π cm^{2}

**Formula used:**

CSA of cone = π × r × l where, l → slant height

Volume of cone = 1/3 × πr^{2}h

**Calculation:**

CSA = 20π cm^{2}

⇒ π × r × l = 20π

⇒ 4 × l = 20

⇒ l = 5 cm

So, h = 3 cm [Using pythagorian triplet]

Volume = 1/3 × πr2h

⇒ Volume = 1/3 × π × 4 × 4 × 3

⇒ Volume = 16π cm^{3}

**∴ Volume of cone is 16π cm ^{3}**.

Option 1 : 14784 sq cm

**Given:**

Radius of Cone = 42 cm

Height of Cone = 56 cm

**Formula Used:**

Slant Height of cone ( l ) = √(r^{2} + h^{2})

Total surface area of a cone = πrl + πr^{2}

⇒ πr(l + r)

where, r = Radius of the cone,

h = Height of the cone, l = Slant Height of the cone

**Calculation:**

Observe the diagram below carefully.

Here,

Slant Height of the cone, l = √(r^{2} + h^{2})

⇒ √{(42)2 + (56)2}

⇒ √(1764 + 3136)

⇒ √(4900)

⇒ 70 cm

Now,

Total surface area of a cone = πr(l + r)

⇒ (22/7) × 42 × (70 + 42)

⇒ 22 × 6 × 112

⇒ 14784 cm^{2}

**∴ The total surface area of the cone is 14784 cm2.**

Option 1 : 1200π cm^{3}

**Given:**

The sides of the triangle are 15 cm, 20 cm.

**Formula used:**

The volume of cone = (1/3)πr^{2}h

Where,

r = radius of cone

h = height of the cone

Pythagoras theorem

H^{2} = P^{2} + B^{2}

Where,

H = hypotenuse of the triangle

P = Perpendicular of the triangle

B = Base of the triangle

**Calculation:**

Pythagoras theorem

⇒ H^{2} = P^{2} + B^{2}

⇒ H^{2} = 15^{2} + 20^{2}

⇒ H = 25 cm

If the triangle is rotated along the 25 cm hypotenuse, it will form a double cone.

The radius of cone-formed = (Perpendicular × Base)/Height

⇒ (15 × 20)/25

⇒ 12 cm

Height of the cone formed = 25 cm

The volume of cone-formed

⇒ (1/3)πr^{2}h

⇒ (1/3)π × (12)^{2} × 25

⇒ 1200π cm^{3}

**∴ The volume of the shape formed is 1200π cm ^{3}.**

Option 3 : 1232 cm^{3}

**Given:**

The curved surface area of a cone = 550 cm^{2}

The area of its base =154 cm^{2 }

**Formula used:**

Curved surface area of cone = πrl

Area of base = πr^{2}

Volume of cone = (1/3) × πr^{2}h (where r is the radius of the circle)

**Calculation:**

πrl = 550

πr^{2} = 154

r = 7 cm

⇒ l = 550/πr = 550/((22/7) × 7) = 25 cm

Now, \(h = \sqrt {\left( {{l^2} - {r^2}} \right)} \)

\(h = \sqrt {\left( {{{25}^2} - {7^2}} \right)} = \sqrt {625 - 49} = 24\)

⇒ Volume of cone = (1/3) × πr^{2}h

Option 1 : 1500π

**Given****:**

Height of cone = 20 cm

Slant height of cone = 25 cm

**Formula:**

Volume of cone = [1/3]πr^{2}h

l^{2} = r^{2} + h^{2}

**Calculation:**

According to the question

25^{2} = r^{2} + 20^{2}

⇒ 625 = r^{2} + 400

⇒ r^{2} = 625 – 400

⇒ r^{2} = 225

⇒ r = 15

∴ Volume of cone = [1/3] × π × 15 × 15 × 20 = 1500πOption 2 : 770 cm^{2}

**Given: **

Slant height = 28 cm

Diameter = Slant height/2

**Formula Used:**

Radius = Diameter/2

Total surface area of cone = πr(r + l)

**Calculation:**

Diameter = 28/2 = 14 cm

Radius = 14/2 = 7 cm

Total surface area of cone = πr(r + l)

⇒ Total surface area of cone = π × 7(7 + 28)

⇒ Total surface area of cone = 22 × 35 = 770 cm^{2}

∴ Total surface area of cone is 770 cm2.

The correct option is 2 i.e. 770 cm2

Option 1 : 2310 cm^{2}

**GIVEN:**

Radius = 21 cm and Height = 28 cm

**FORMULA USED:**

C.S.A of Cone = πRL Where L = √R^{2} + H^{2}

**CALCULATION:**

C.S.A of Cone = πRL

**⇒ L **= (√21 ^{2} + 28^{2}) = 35 cm

⇒ Curved surface area of cone = (22/7) × 21 × 35 = 2310 cm^{2}

∴ **Curved surface area of cone = 2310 cm ^{2}**

Slant height of cone is 15 cm and height of cone is 12 cm. Total surface area of cone is

Option 3 : 216π cm^{2}

**Given****:**

Slant height of cone l = 15 cm

Height of cone h = 12 cm

**Formula:**

Total surface area of cone = πr(r + l)

l^{2} = h^{2} + r^{2}

**Calculation:**

l^{2} = h^{2} + r^{2}

⇒ 15^{2} = r^{2} + 12^{2}

⇒ 225 = r^{2} + 144

⇒ r^{2} = 225 - 144

⇒ r = √81

⇒ r = 9

∴ Total surface area of cone = π × 9 × (9 + 15) = π × 9 × 24 = 216π cmOption 4 : Rs. 22880

**Concept:**

In this type of question we have to remember that painting is done on lateral surface area of the cone.

**Formula Used: **

Lateral surface area of cone = π × (radius) × slant height

**Given: **

Height of cone = 24 cm

Radius of cone = 10 cm

Rate of painting of cone = 28 / cm^{2}

**Calculation: **

By Pythagoras theorem

Hypotenous^{2} = perpendicular^{2} + base^{2}

Hypotenuse^{2} = 24^{2} + 10^{2}

Hypotenuse^{2} = 676

Hypotenuse = √676 = 26 cm = slant height

According to question,

Lateral surface area = π × (10) × 26

⇒ π × 260

Price of painting = rate of painting × lateral surface area

⇒ Price of painting = 28 × {(22 / 7) × 260}

∴ Price of painting = Rs. 22880Option 3 : 14 cm

**Given:**

Volume of cone = 2464 cm^{3}

Height of cone = 12 cm

**Formula used:**

Volume of a cone = πR^{2}H / 3

Where R = base radius of cone and H = height of cone

**Calculation:**

Let the base radius of cone be r

So 2464 = πr^{2} × 12 / 3

⇒ πr^{2} = 616

⇒ (22 / 7) × r^{2} = 616

⇒ r^{2} = 28 × 7

⇒ r = 14 cm

∴ Base radius of cone = 14 cm

A conical vessel has a slant height of 25 cm, and its Curved surface area 175π cm^{2}. Find the volume of the conical vessel.

Option 1 : 1232

**Given: **

Slant height = 25 cm

Curved surface area = 175π cm^{2}

**Formula used:**

C.S.A = πrl

Volume of cone = 1/3 πr^{2}h

Where r = radius, h = height, l = slant height, C.S.A = Curved Surface Area

**Calculation:**

175π = πr × 25

⇒ r = 175π/25π = 7 cm

Using triplet (7, 24, 25)

⇒ h = 24 cm

Volume of cone = (1/3) × (22/7) × (7)^{2} × 24

⇒ Volume = 1232 cm^{3}

Option 2 : 3

**Formula used:**

Volume of cone = 1/3 πr^{2}h

Area of circular base = πr^{2}

**Calculation:**

Height = H

V = 1/3 πr^{2}H

A = πr^{2}

\({AH\over V} =\ { \pi r^2 \times H \over 1/3 \pi r^2 H} \)

⇒ AH/V = 3

**∴ The value of \(\frac{{AH}}{V}\) is 3.**

A number of conical tools of radius 5 cm and height 8 cm are melted to form a cylindrical tool of radius and height both equal to 10 cm. Find the number of cones melted to form a new cylinder.

Option 2 : 15

**Given:**

Radius of cone = 5 cm

Height of cone = 8 cm

Radius of cylinder = Height of cylinder = 10 cm

**Concepts used:**

Volume of cone = πr^{2}h/3

Volume of cylinder = πr^{2}h

**Calculation:**

Volume of cone = πr^{2}h/3

⇒ Volume of cone = π(5)^{2} × 8/3 = 200π/3 cm^{3}

⇒ Volume of cylinder = π × (10)^{2} × 10 cm^{3} = 1,000π cm^{3}

Let number of cones melted be y.

⇒ y × volume of one cone = Volume of cylinder

⇒ y × 200π/3 = 1,000π

⇒ y = 1,000 × 3/200 = 15

**∴ 15 cones are melted to form new cylinder.**

Option 2 : 1078 cm3

Given:

The radius of the cone = 7 cm

Height of the cone = 21 cm

Formula used:

Volume of a cone = (1/3) × π × r2 × h

Where, r = radius of the cone

h = Height of the cone

Calculations:

Volume of the cone = (1/3) × π × r2 × h

⇒ (1/3) × π × 72 × 21

⇒ (1/3) × (22/7) × 49 × 21

⇒ 1078 cm3

∴ The volume of the cone is 1078 cm3.

Option 2 : 18 : 13

**Given****:**

Height of cone = 12

Radius of cone = 5 cm

**Formula:**

l^{2} = r^{2} + h^{2 }(l = slant height, r = radius, h = height)

Total surface area of cone = πr(r + l)

Curved surface area of cone = πrl

**Calculation:**

l^{2} = r^{2} + h^{2}

⇒ l^{2} = 5^{2} + 12^{2}

⇒ l^{2} = 25 + 169

⇒ l = √169

⇒ l = 13

∴ Required Ratio = πr(r + l) : πrl

⇒ (r + l) : l

⇒ (5 + 13) : 13

⇒ 18 : 13

Option 1 : 63 m^{2}

**Given:**

Height = 3 m

Radius = 4 m

**Concept used:**

CSA of cone = π × r × l

Where slant height, l = √(r^{2} + h^{2})

**Calculation:**

l = √(3^{2} + 4^{2}) = 5

CSA of the cone = (22 × 4 × 5)/7 = 62.85

**∴ CSA of the cone ≈ 63 m ^{2}**.

Option 3 : 12936

**Given: **

CSA of the cone = 2310 cm^{2}

ratio of height and slant height = 4 : 5.

**Concept:**

CSA of cone = πrl

Volume of cone = 1/3πr^{2}h

\({l^2} = \;{r^2} + \;{h^2}\)

**Calculation: **

Let slant height 5k and height 4k

then, Radius = \(\sqrt {25{k^2} - 16{k^2}} \)

⇒ 3k

Now, CSA = 2310

⇒ πrl = 2310

⇒ 22/7 × 3k × 5k = 2310

⇒ k^{2} = (2310 × 7)/(15 × 22)

⇒ k^{2} = 7^{2}

⇒ k = 7

So, Radius = 3 × 7 = 21 cm, Height = 4 × 7 = 28 cm

Volume = 1/3 × 22/7 × 21^{2} × 28

⇒ 12936 cm^{3}

**∴ 12936 cm ^{3}**

Option 4 : 4 cm

**Given****:**

Radius of cone = 4 cm

Height of cone = 3 cm

Radius of cylinder = 2 cm

**Formula:**

Volume of cone = [1/3] × πr^{2}h (r = radius, h = height)

Volume of cylinder = πr^{2}h

**Calculation:**

Let the level of water in the cylinder increased by x cm, then

According to the question

[1/3] × π × 4 × 4 × 3 = π × 2 × 2 × x

⇒ x = 16/4

⇒ x = 4

∴ Level of water increased by 4 cm