Conic 3D MCQ Quiz - Objective Question with Answer for Conic 3D - Download Free PDF

Last updated on May 17, 2025

Latest Conic 3D MCQ Objective Questions

Conic 3D Question 1:

Comprehension:

Direction: Consider the following for the items that follow:

Let 2x2 + 2y2 + 2z2 + 3x + 3y + 3z - 6 = 0 be a sphere.

The centre of the sphere lies on the plane 

  1. 2x + 2y + 2z - 3 = 0
  2. 4x + 4y + 4z - 3 = 0
  3. 4x + 8y + 8z - 15 = 0
  4. 4x + 8y + 8z + 15 = 0

Answer (Detailed Solution Below)

Option 4 : 4x + 8y + 8z + 15 = 0

Conic 3D Question 1 Detailed Solution

Explanation:

The Centre of the sphere is C \((\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})\)

4x + 8y + 8z + 15 = 0,

Plane satisfy the centre of sphere – 3 –6 – 6 + 15 = 0 

⇒ 0 = 0

∴ Option (d) is correct

Conic 3D Question 2:

Comprehension:

Direction: Consider the following for the items that follow:

Let 2x2 + 2y2 + 2z2 + 3x + 3y + 3z - 6 = 0 be a sphere.

What is the diameter of the sphere? 

  1. \(\frac{5\sqrt3}{4}\)
  2. \(\frac{5\sqrt3}{2}\)
  3. \(\frac{3\sqrt5}{4}\)
  4. \(\frac{3\sqrt5}{2}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{5\sqrt3}{2}\)

Conic 3D Question 2 Detailed Solution

Explanation:

Given:

Sphere: 2x2 + 2y2 + 2z2 + 3x + 3y + 3z – 6 = 0

⇒ \(x^2 + y^2+z^2 + \frac{3}{2}x +\frac{3}{2}y + \frac{3}{2}z\) -3 = 0

Now Radius = \(\sqrt{(\frac{3}{4})^2+ (\frac{3}{4})^2+ (\frac{3}{4})^2 + 3}\)

\(\sqrt{\frac{27}{16} +3} = 5 \frac{\sqrt3}{4}\)

Diamter = \(5 \frac{\sqrt3}{2}\)

∴ Option (b) is correct

Conic 3D Question 3:

Reciprocal cone of cone x2 + 2y2 + 3z2 = 0 is

  1. 6x2 + 3y2 + 2z2 = 0
  2. 3x2 + 2y2 + z2 = 0
  3. \(\rm \frac{x^2}{2}+\frac{y^2}{2}+z^2=0\)
  4. 6x2 + 2y2 + 3z2 = 0

Answer (Detailed Solution Below)

Option 1 : 6x2 + 3y2 + 2z2 = 0

Conic 3D Question 3 Detailed Solution

Explanation:

We know that the reciprocal cone of cone ax2 + by2 + cz2 = 0 is \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\)

In the cone x2 + 2y2 + 3z2 = 0

a = 1, b = 2, c = 3.

Hence reciprocal cone is

\(\frac{x^2}{1}+\frac{y^2}{2}+\frac{z^2}3=0\)

i.e., 6x2 + 3y2 + 2z2 = 0

Option (1) is true.

Conic 3D Question 4:

Equation λ1S1 + λ2S2 = 0, does not represent an equation of sphere, that passes through the intersection of two spheres S1 = 0, S2 = 0, if-

  1. λ1 = -1
  2. λ2 = -1
  3. λ1 = -λ2
  4. λ1 = λ2

Answer (Detailed Solution Below)

Option 3 : λ1 = -λ2

Conic 3D Question 4 Detailed Solution

Explanation:

λ1S1 + λ2S2 = 0

⇒ λ1(x2 + y2 + z2 + ...) + λ2(x2 + y2 + z2 + ...) = 0

If λ1 = -λthen the above equation does not represent an sphere.

Hence option (3) is true. 

Conic 3D Question 5:

A cone is ____

  1. Plane figure
  2. Curve
  3. Solid Figure
  4. Line

Answer (Detailed Solution Below)

Option 3 : Solid Figure

Conic 3D Question 5 Detailed Solution

Concept:

A cone is a three-dimensional geometric shape that tapers smoothly from a flat

base (usually circular) to a point called the apex or vertex. It has one flat

surface and one curved surface.

Explanation:

Option 1: A plane figure is a flat, two-dimensional shape like a circle, triangle,

or square. Since a cone is three-dimensional, it cannot be classified as a plane

figure.

Option 2: A curve is a smoothly flowing, continuous line without sharp angles.

Although a cone has a curved surface, it is not solely a curve.

Option 3: Solid Figure: A solid figure is a three-dimensional object. A cone fits

this definition as it has volume and occupies space.

Option 4:  A line is a straight one-dimensional figure having no thickness and

extending infinitely in both directions. A cone is not a line as it is three-dimensional.

The correct option is option 3.

Top Conic 3D MCQ Objective Questions

A lead hemisphere of radius 7 cm is molded into a cone whose base diameter is 14 cm. Find the height of the cone.

  1. 12 cm
  2. 14 cm
  3. 18 cm
  4. 16 cm

Answer (Detailed Solution Below)

Option 2 : 14 cm

Conic 3D Question 6 Detailed Solution

Download Solution PDF

Given:

Radius of hemisphere = 7 cm

Diameter of cone = 14 cm

Formula Used:

Volume of hemisphere = 2/3 × πR3

Volume of cone = 1/3 × πr2h

Calculation:

⇒ 2/3 × πR1/3 × πr2h

⇒ 2/3 × π71/3 × π72h

⇒ 2 × 7= 1 × 72h

⇒ h = 14

The answer is 14 cm.

Comprehension:

Direction: Consider the following for the items that follow:

Let 2x2 + 2y2 + 2z2 + 3x + 3y + 3z - 6 = 0 be a sphere.

What is the diameter of the sphere? 

  1. \(\frac{5\sqrt3}{4}\)
  2. \(\frac{5\sqrt3}{2}\)
  3. \(\frac{3\sqrt5}{4}\)
  4. \(\frac{3\sqrt5}{2}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{5\sqrt3}{2}\)

Conic 3D Question 7 Detailed Solution

Download Solution PDF

Explanation:

Given:

Sphere: 2x2 + 2y2 + 2z2 + 3x + 3y + 3z – 6 = 0

⇒ \(x^2 + y^2+z^2 + \frac{3}{2}x +\frac{3}{2}y + \frac{3}{2}z\) -3 = 0

Now Radius = \(\sqrt{(\frac{3}{4})^2+ (\frac{3}{4})^2+ (\frac{3}{4})^2 + 3}\)

\(\sqrt{\frac{27}{16} +3} = 5 \frac{\sqrt3}{4}\)

Diamter = \(5 \frac{\sqrt3}{2}\)

∴ Option (b) is correct

What is the diameter of the sphere?

x2 + y2 + z2 − 16x + 12y − 2√dz + d = 0

  1. 40
  2. 20
  3. 10
  4. 5

Answer (Detailed Solution Below)

Option 2 : 20

Conic 3D Question 8 Detailed Solution

Download Solution PDF

 Given:

 The equation of the sphere is given.

 Concept Used:

 Comparing the given equation of sphere to the standard form of equation of spheres.

 Solution:

 We have, standard equation of sphere as: (x - a)2 + (y - b)2 + (z - c)2 = r2

 ⇒ x2 + y2 + z2 - 2ax - 2by - 2cz + a2 + b2 + c2 - r2 = 0

 ⇒ comparing the coefficient of x, y, z and constant.

 we have, - 2a = - 16 {coefficient of x}

 ⇒ a = 8 

 we have, - 2b = 12 {coefficient of y}

 ⇒ b = - 6

 we have, - 2c = -2\( \sqrt{d}\)

 ⇒ c = \( \sqrt{d}\)

  we have, a2 + b2 + c2 - r2 = d 

 ⇒ (8)2 + (- 6)2 + (\( \sqrt{d}\))2 - r2 = d

 ⇒ 64 + 36 + d - r2 = d

 ⇒ r2 =100

 ⇒ r = 10

 we know that diameter of sphere is 2 times radius of sphere.

 D = 2r 

 ⇒ D = 20

 \(\therefore\)Option 2 is correct.

Comprehension:

Direction: Consider the following for the items that follow:

Let 2x2 + 2y2 + 2z2 + 3x + 3y + 3z - 6 = 0 be a sphere.

The centre of the sphere lies on the plane 

  1. 2x + 2y + 2z - 3 = 0
  2. 4x + 4y + 4z - 3 = 0
  3. 4x + 8y + 8z - 15 = 0
  4. 4x + 8y + 8z + 15 = 0

Answer (Detailed Solution Below)

Option 4 : 4x + 8y + 8z + 15 = 0

Conic 3D Question 9 Detailed Solution

Download Solution PDF

Explanation:

The Centre of the sphere is C \((\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})\)

4x + 8y + 8z + 15 = 0,

Plane satisfy the centre of sphere – 3 –6 – 6 + 15 = 0 

⇒ 0 = 0

∴ Option (d) is correct

Conic 3D Question 10:

A lead hemisphere of radius 7 cm is molded into a cone whose base diameter is 14 cm. Find the height of the cone.

  1. 12 cm
  2. 14 cm
  3. 18 cm
  4. 16 cm

Answer (Detailed Solution Below)

Option 2 : 14 cm

Conic 3D Question 10 Detailed Solution

Given:

Radius of hemisphere = 7 cm

Diameter of cone = 14 cm

Formula Used:

Volume of hemisphere = 2/3 × πR3

Volume of cone = 1/3 × πr2h

Calculation:

⇒ 2/3 × πR1/3 × πr2h

⇒ 2/3 × π71/3 × π72h

⇒ 2 × 7= 1 × 72h

⇒ h = 14

The answer is 14 cm.

Conic 3D Question 11:

A cone is ____

  1. Plane figure
  2. Curve
  3. Solid Figure
  4. Line

Answer (Detailed Solution Below)

Option 3 : Solid Figure

Conic 3D Question 11 Detailed Solution

Concept:

A cone is a three-dimensional geometric shape that tapers smoothly from a flat

base (usually circular) to a point called the apex or vertex. It has one flat

surface and one curved surface.

Explanation:

Option 1: A plane figure is a flat, two-dimensional shape like a circle, triangle,

or square. Since a cone is three-dimensional, it cannot be classified as a plane

figure.

Option 2: A curve is a smoothly flowing, continuous line without sharp angles.

Although a cone has a curved surface, it is not solely a curve.

Option 3: Solid Figure: A solid figure is a three-dimensional object. A cone fits

this definition as it has volume and occupies space.

Option 4:  A line is a straight one-dimensional figure having no thickness and

extending infinitely in both directions. A cone is not a line as it is three-dimensional.

The correct option is option 3.

Conic 3D Question 12:

Equation of the sphere that passes through the points (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c) is -

  1. \((x - \frac{a}{2})^2\) + \((y - \frac{b}{2})^2\) + \((z - \frac{c}{2})^2\) = \(\frac{1}{2}\) (a2 + b2 + c2)
  2. x2+ y2 + z2 = a2 + b2 + c2
  3. \((x - \frac{a}{2})^2\) + \((y - \frac{b}{2})^2\) + \((z - \frac{c}{2})^2\) = \(\frac{1}{4}\) (a2 + b2 + c2)
  4. (x – a)2 + (y – b)2 + (z – c)2\(\frac{1}{2}\)\(\sqrt{(a^2 + b^2 + c^2)}\)

Answer (Detailed Solution Below)

Option 3 : \((x - \frac{a}{2})^2\) + \((y - \frac{b}{2})^2\) + \((z - \frac{c}{2})^2\) = \(\frac{1}{4}\) (a2 + b2 + c2)

Conic 3D Question 12 Detailed Solution

Concept:

General equation of sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + c = 0

Explanation:

Now, equation of sphere is  x2 + y2 + z2 + 2ux + 2vy + 2wz + c = 0 ....... (i)

Passes through (0,0,0), (a,0,0), (0,b,0), (0,0,c) 

Now, when (i) equation passes through (0,0,0) ⇒ c = 0 

Now, when (i) equation passes through (a,0,0) ⇒ u = \(\frac{-a}{2}\) 

Similarly, For (0,b,0) we get v = \(\frac{-b}{2}\) and for (0,0,c) we get w = \(\frac{-c}{2}\) 

By substituting all value we get

 x2 + y2 + z2 - ax - by - cz = 0

Now, By completing the square we get

\((x - \frac{a}{2})^2\) + \((y - \frac{b}{2})^2\) + \((z - \frac{c}{2})^2\) =  \(\frac{1}{4}\) (a2 + b2 + c2)

Hence, Option (3) is true

Conic 3D Question 13:

Comprehension:

Direction: Consider the following for the items that follow:

Let 2x2 + 2y2 + 2z2 + 3x + 3y + 3z - 6 = 0 be a sphere.

What is the diameter of the sphere? 

  1. \(\frac{5\sqrt3}{4}\)
  2. \(\frac{5\sqrt3}{2}\)
  3. \(\frac{3\sqrt5}{4}\)
  4. \(\frac{3\sqrt5}{2}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{5\sqrt3}{2}\)

Conic 3D Question 13 Detailed Solution

Explanation:

Given:

Sphere: 2x2 + 2y2 + 2z2 + 3x + 3y + 3z – 6 = 0

⇒ \(x^2 + y^2+z^2 + \frac{3}{2}x +\frac{3}{2}y + \frac{3}{2}z\) -3 = 0

Now Radius = \(\sqrt{(\frac{3}{4})^2+ (\frac{3}{4})^2+ (\frac{3}{4})^2 + 3}\)

\(\sqrt{\frac{27}{16} +3} = 5 \frac{\sqrt3}{4}\)

Diamter = \(5 \frac{\sqrt3}{2}\)

∴ Option (b) is correct

Conic 3D Question 14:

What is the diameter of the sphere?

x2 + y2 + z2 − 16x + 12y − 2√dz + d = 0

  1. 40
  2. 20
  3. 10
  4. 5

Answer (Detailed Solution Below)

Option 2 : 20

Conic 3D Question 14 Detailed Solution

 Given:

 The equation of the sphere is given.

 Concept Used:

 Comparing the given equation of sphere to the standard form of equation of spheres.

 Solution:

 We have, standard equation of sphere as: (x - a)2 + (y - b)2 + (z - c)2 = r2

 ⇒ x2 + y2 + z2 - 2ax - 2by - 2cz + a2 + b2 + c2 - r2 = 0

 ⇒ comparing the coefficient of x, y, z and constant.

 we have, - 2a = - 16 {coefficient of x}

 ⇒ a = 8 

 we have, - 2b = 12 {coefficient of y}

 ⇒ b = - 6

 we have, - 2c = -2\( \sqrt{d}\)

 ⇒ c = \( \sqrt{d}\)

  we have, a2 + b2 + c2 - r2 = d 

 ⇒ (8)2 + (- 6)2 + (\( \sqrt{d}\))2 - r2 = d

 ⇒ 64 + 36 + d - r2 = d

 ⇒ r2 =100

 ⇒ r = 10

 we know that diameter of sphere is 2 times radius of sphere.

 D = 2r 

 ⇒ D = 20

 \(\therefore\)Option 2 is correct.

Conic 3D Question 15:

Comprehension:

Direction: Consider the following for the items that follow:

Let 2x2 + 2y2 + 2z2 + 3x + 3y + 3z - 6 = 0 be a sphere.

The centre of the sphere lies on the plane 

  1. 2x + 2y + 2z - 3 = 0
  2. 4x + 4y + 4z - 3 = 0
  3. 4x + 8y + 8z - 15 = 0
  4. 4x + 8y + 8z + 15 = 0

Answer (Detailed Solution Below)

Option 4 : 4x + 8y + 8z + 15 = 0

Conic 3D Question 15 Detailed Solution

Explanation:

The Centre of the sphere is C \((\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})\)

4x + 8y + 8z + 15 = 0,

Plane satisfy the centre of sphere – 3 –6 – 6 + 15 = 0 

⇒ 0 = 0

∴ Option (d) is correct

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