MCQ: Sphere - SSC CGL MCQ

Given:

Edge of the cube = 5.6 cm

Formula used:

The volume of the sphere = (4/3) × π × r³

Calculation:

Diameter of the sphere = Edge of the cube
⇒ 2 × radius of sphere = 5.6 cm
⇒ The radius of sphere = 5.6/2 = 2.8 cm
The volume of the sphere = (4/3) × (22/7) × (2.8)³
⇒ 88/21 × 21.952
⇒ 91.98
∴ The volume of the largest sphere that can be taken out of the cube is 91.98 cm³

Given:

Spheres of radius r₁ = 6 cm, r₂ = 8 cm, r₃ = 10 cm molten in single one. 

Formula used:

The volume of Sphere = 4/3π r³

Calculations:

Let the Radius of the larger bead be 'R' 
According to the question,
The volume of smaller beads = Volume of the larger bead
⇒ 4/3π [(6)³ + (8)³ + (10)³] = 4/3π R³
⇒ R³ = [(216) + (512) + (1000)]
⇒ R³ = 1728 = 12 cm.
∴ The radius (in cm) of the larger bead will be 12 cm.

Given:

Radius of sphere = 8 cm
Radius of cylinder = 4 cm
The total surface area of the cylinder is half the surface area of the sphere

Formula used:

Total surface area of cylinder = 2πr(h + r)
Surface area of sphere = 4πr²

Calculation:

According to the question
The total surface area of the cylinder is half the surface area of the sphere

∴ The height of the cylinder is 12 cm

Given:

The sides of the cuboid are 44 cm, 32 cm, and 36 cm
The radius of the sphere is 12 cm

Concept Used:

The volume of a cuboid of sides l, b and h = l x b x h
The volume of the sphere of radius r = (4/3)πr³

Calculation:

The volume of the metallic cuboid is (44 x 32 x 36) cm³
The volume of the sphere is (4/3) x π x 123
Let, the total number of such sphere is n
Accordingly,

∴ Such 7 spheres can be made by given metallic cuboid.

GIVEN:

The surface area of a sphere = 64π cm²  

FORMULA USED:

The surface area of a sphere = 4πr²    
The volume of a sphere = 4πr³/3

CALCULATION:

The surface area of a sphere = 64π
⇒ 4πr² = 64π
⇒ r2 = 16 
⇒ r = 4cm

Given:

A copper sphere of diameter 18 cm is drawn into a wire of diameter 6 mm.

Concept used:

Volume of sphere = 4/3πr³
Volume of cylinder = πr²h

Calculation:
Radius of sphere = 9 cm
Radius of wire = 0.3 cm  [As 1 cm = 10 mm]
According to the question,
Volume of sphere = Volume of cylinder

∴ The length of the wire is 108 m.

GIVEN:

The surface area of a sphere = 1386 cm²

FORMULA USED:

The surface area of a sphere = 4πr² where r is the radius of the sphere.

CALCULATION:

The surface area of a sphere =4πr² = 1386 

∴ The radius of the sphere is 10.5 cm.

Formula used:

Area of Sphere = 4πr²
Here,
r = radius

Calculation:

It is given that, the diameter of a sphere is doubled so its radius will also get doubled.
Let the initial and final radius of sphere be r and R.

Now change in surface area = 4A - A = 3A 

∴ Its surface area will increase three times.

Given:

A hemisphere of lead of radius 4 cm is cast into a right circular cone of height 72 cm.

Concept used:

3. The volume of the hemisphere must be equal to the volume of the cone.

Calculation:

The volume of the hemisphere 

Let the radius of the base of the right circular cone be R cm.

​According to the concept,

∴ The radius of the base of the cone is 1.33 cm.

Given:
The volume of a solid spherical ball (V₁) = 972π cm²

Formula used:
The volume of the sphere = 4/3πr³
The total surface area of the sphere = 4πr²
Where, r = radius

Calculations:
According to the question,
It was melted into 27 smaller spheres of radius r and Volumes (V₂)
V₁ = V₂

Now, the total surface area of 27 spheres:

∴ The total surface area of the 27 smaller spheres taken together will be 972π cm².

Given:

A metallic spherical ball S of radius 9 cm.
The metallic spherical ball S is melted and recast into N number of spherical balls of radius 3 cm.

Concept used:

The formula of the sphere:

Where,

R, is the radius of the sphere

Calculation:

The radius of bigger metallic sphere S, R = 9 cm
The radius of the smaller metallic sphere, r = 3 cm
Let N be the number of smaller spheres.
Let VB and VS be the volume of bigger and smaller spheres.
According to the question,

∴ The required number of smaller spheres is 27.

Given:

For hollow sphere,
Outer radius (ro) = 4 cm.
Inner radius (ri) = 1 cm.

Concept used:
Volume of hollow sphere

Calculation:

Volume of hollow sphere = (4/3)p x (4³ - 1³)
⇒ (4/3)p x (64 - 1)

⇒ (4/3)px 63

⇒ 84p 

∴ The volume of the hollow sphere is 84p cm³.

Concept used:

• Sum of the volumes of small droplets = volume of the big droplet
• Volume of sphere = 4/3 x π x r³

Calculation:

Total number of small droplets are  0.1% of 1.728 x 106 = 1728

Let the radius of big bubble be R mm
⇒ 1728 x 4/3 x π x (2/2)³ = 4/3 × π × R³
⇒ R³ = 1728
⇒ R = 12 mm or 1.2 cm
Then diameter is 2 × 1.2 = 2.4 cm
∴ The correct answer is 2.4 cm

Given:

The surface area increased by 4840/7 cm² by the increase of the radius 5 cm

Formula used:

A = 4 x π x R²     Where, V = Volume of the sphere, A = Surface area of the sphere and R = Radius of the sphere 
 

Calculation:

Let the radius of the sphere be R

According to the question

The volume of the sphere 

= 113.14 ≈ 113 cm³

Given:

A solid cone of radius 7 cm and height 7 cm was melted along with two solid spheres of radius 7 cm each to form a solid cylinder of radius 7 cm.

Formula used:

The volume of a cone = 1/3πr²h
Volume of a cylinder = πr2h
The volume of a sphere = 4/3πr³
The curved surface area of a cylinder = 2πrh
Here,
r = radius
h = height

Calculation:

The volume of the cone = 1/3π × 72 × 7
⇒ 1/3π x 73

The volume of each sphere = 4/3π × 73
Let the height of the formed cylinder be H
According to the question,

So, the height of the formed cylinder is 21 cm
Now,
The curved surface area of the cylinder = 2 x 22/7 x 7 x 21
⇒ 924
∴ The curved surface area (in cm2) of the cylinder is 924.