Sphere MCQ Quiz - Objective Question with Answer for Sphere - Download Free PDF

Given:

The diameter of a sphere is increased by 30%.

Formula used:

Surface area of a sphere = 4πr²

Where r is the radius of the sphere.

Calculations:

Let the original diameter of the sphere be D, and the original radius be r = D/2.

So, the original surface area = 4πr².

Now, if the diameter is increased by 30%, the new diameter is:

New diameter = D × (1 + 30/100) = 1.30D

The new radius is:

New radius = 1.30 × r = 1.30r.

The new surface area is:

New surface area = 4π(1.30r)² = 4π × 1.69r² = 1.69 × (4πr²).

The increase in the surface area is:

Increase = 1.69 × (original surface area) - original surface area

⇒ Increase = (1.69 - 1) × (original surface area)

⇒ Increase = 0.69 × (original surface area).

∴ The increase in the surface area is 69%.

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

From the previous calculation, the radius of the circular opening (r) = 10$\sqrt{3}$ cm

From the previous calculation, the perpendicular distance from the center of the sphere to the plane of the circular opening (d) = 10 cm

Calculations:

The perpendicular distance from the center of the sphere to the plane of the circular opening (d) as the adjacent side to the angle θ (this is the vertical line segment from the center to the center of the opening).

The inner radius of the circular opening (r) as the opposite side to the angle θ.

We can use trigonometric ratios:

cos(θ) = Adjacent / Hypotenuse = d / R

cos(θ) = d / R = 10 cm / 20 cm = 1/2

The angle whose cosine is 1/2 is 60°.

⇒ θ = 60° = π/3

∴ The angle made by the line joining the center of the sphere and any point on the rim of the circular opening with a vertical line passing through the center is 60°.

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

Calculations:

Let 'r' be the inner radius of the circular opening of the pot.

Draw a vertical line from the center of the sphere to the center of the circular opening. This line segment will be perpendicular to the plane of the circular opening.

Let the center of the sphere be O. Let the center of the circular opening be C'.

The radius of the sphere (R) goes from O to any point on the surface of the sphere, including the edge of the circular opening.

Consider a right-angled triangle formed by:

The radius of the sphere (R) as the hypotenuse, from the center of the sphere to any point on the edge of the circular opening.

d = |R - h| = |20 - 30| = |-10| = 10 cm.

Now, substitute the values into the Pythagorean theorem:

R² = r² + d²

20² = r² + 10²

400 = r² + 100

r² = 400 - 100

r² = 300

r = $\sqrt{300}$

r = $\sqrt{100\times 3}$

r = 10$\sqrt{3}$ cm

∴ The inner radius of the circular opening of the pot is 10$\sqrt{3}$ cm.

Given:

Volume of the sphere (V) = 36π cm³

Formula used:

V = (4/3)πr³

Where, r = radius of the sphere

Calculation:

36π = (4/3)πr³

⇒ r³ = (36π × 3) / 4π

⇒ r³ = 108 / 4

⇒ r³ = 27

⇒ r = ∛27

⇒ r = 3

∴ The correct answer is option (1).

Given:

Radius (r) = 32 cm

π = 3.14

Formula Used:

Curved Surface Area (CSA) of a sphere = 4 × π × r²

Calculation:

CSA = 4 × 3.14 × 32²

⇒ CSA = 4 × 3.14 × (32 × 32)

⇒ CSA = 4 × 3.14 × 1024

⇒ CSA = 12861.44 cm²

∴ The correct answer is option (2).

GIVEN:

The surface area of a sphere = 1386 $c{m}^2$ 

FORMULA USED:

The surface area of a sphere = 4πr²where r is the radius of the sphere.

CALCULATION:

The surface area of a sphere =4πr2 = 1386 

⇒  4 × (22/7) × r² = 1386      ....(value of  $\pi$ is $\frac{22}{7}$)

⇒ r2 = 110.25 

⇒ r2 = $\frac{11025}{100}$  

⇒ r = $\sqrt{\frac{11025}{100}}$ = $\frac{105}{10}$ = 10.5 cm.

∴ The radius of the sphere is 10.5 cm.

Given:

The volume of a solid sphere = 36π m3 

The surface area of a smaller sphere = 4π m2

Formula used:

(1.) Volume of solid sphere = $\frac{4}{3}$πr³

(2.) Surface area of solid sphere = 4πr2

Calculation:

According to the question,

Surface area of solid sphere = 4πr2

So,

⇒ 4πr2 = 4π

⇒ r2 = 1

⇒ r = 1 m

Volume of a small sphere = $\frac{4}{3}$πr3 = $\frac{4}{3}$π m³

Let N be the number of smaller spherical balls that can be drawn out from the larger solid sphere.

⇒ N = $\frac{36\pi }{\frac{4}{3}\pi }$

⇒ N = 27

Therefore, '27' is the required answer.

Additional Information(1.) Total surface area of solid sphere = 4πr2

(2.) Lateral surface area of solid sphere = 4πr2

Mistake PointsWe can not divide the Volume of the sphere and Surface area of the sphere

Concept used:

• Sum of the volumes of small droplets = volume of the big droplet

• Volume of sphere = 4/3 × π × r³

Calculation:

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

Let the radius of big bubble be R mm

⇒ 1728 × 4/3 × π × (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 of a sphere = 64πcm2  

FORMULA USED:

The surface area of a sphere = 4πr²    

The volume of a sphere = $\frac{4\pi r^3}{3}$ 

CALCULATION:

The surface area of a sphere = 64π

⇒ 4πr2 = 64π

⇒ r2 = 16 

⇒ r = 4cm

Now, volume = 4/3 $\pi$$r^3$   = 4/3 ×π × 4 × 4 × 4 = $\frac{256\pi }{3}$ $c{m}^3$. 

∴ The volume of the sphere is $\frac{256\pi }{3}$ $c{m}^3$.

Given:

The radius of the sphere, r = 15√3 cm

Concept:

The total surface area of a cube = 6 ×  (edge length)².

Length of the main diagonal of  cube = (edge length)√3

Solution:

The diameter of the sphere = Length of the main diagonal of the cube.

2 × $15\surd 3$ = a√3 

a = 30 cm

Total surface area of the cube = 6 ×  (edge length)²

Total surface area of the cube = 6 × (30)² = 5400 cm².

Hence, the total surface area of the largest possible cube cut out from the sphere will be 5400 cm².

Given:

Diameter of spherical ball (D)= 3 cm

Diameter of 1st small ball (D₁)= 1.5 cm

Diameter of 2nd small ball (D₂)= 2 cm

Concept used:

Total volume of small spherical balls = Volume of large spherical balls

Formula used:

Volume of spherical ball = (4/3) × π × R³

Calculation:

Let, the diameter of 3rd small spherical ball = D₃

Volume of (1st small spherical ball + 2nd spherical ball + 3rd spherical ball) = volume of large spherical ball

⇒ 4/3 π × (D1/2)³ +  4/3 π × (D2/2)3 + 4/3 π × (D3/2)3 = 4/3 π (D/2)³

⇒ 4/3 π × [(1.5/2)3 + (2/2)3 + (D3/2)3 ]= 4/3 π (3/2)3

⇒ [(3.375/8) + 1 + (D3/2)3 ] = 3.375

⇒ (D3/2)3 = 2.375 - (3.375/8)

⇒ (D3/2)3 = (19 - 3.375)/8

⇒ D3 = ³√15.625 = 2.5

∴ The correct answer is 2.5.

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

⇒ 2πr(h + r)/4πr² = 1/2

⇒ 2 × π × 4(h + 4)/(4 × π × 8²) = 1/2

⇒ 8(h + 4)/256 = 1/2

⇒ h + 4/32 = 1/2

⇒ h + 4 = 16

⇒ h = (16 – 4)

⇒ h = 12 cm

∴ The height of the cylinder is 12 cm

Given: 125 small spheres

Concept used: The surface area of a sphere is given by the formula 4πr^2, where r is the radius of the sphere.

Solution:

Diameter of the large sphere = 15 cm

Radius of the large sphere

⇒ 15 cm / 2 = 7.5 cm

Radius of each small sphere = Radius of large sphere / ∛125

⇒ 7.5 cm / 5 = 1.5 cm

The surface area of each small sphere

⇒ 4π(1.5 cm)² = 4π(2.25 cm²) = 9π cm²

Therefore, the surface area of each small sphere is 9π cm².

Given:

R = 10 cm

Formula used:

Volume = 4/3 x 22/7 x R x R x R

Solution:

Volume of bigger sphere = 4/3 x 22/7 x 10³ 

We have 8 smaller spheres of equal radius

Volume of smaller sphere = 4/3 x 22/7 x r3 

Volume of bigger sphere = 8 × Volume of smaller sphere

4/3 x 22/7 x 103 = 8 × 4/3 x 22/7 x r³ 

⇒ r³ = 1000/8

⇒ r = 5 cm

TSA(Sphere)= 4 x 22/7 x 5² 

Given:

Radius of big ladoo, R = 810 cm

Radius of small ladoo, r = 90 cm

Formula used:

The volume of a sphere = (4/3)πR³

The surface area of a sphere =  4πR²

Calculation:

Volume of big ladoo = Volume of all small ladoos together

⇒ (4/3)πR³ = n × (4/3)πr³

⇒ n = (R/r)³ (Where n is the number of small ladoos)

Total surface area of all small ladoos = n × 4πr² = (R/r)³ × 4πr²

The ratio of total surface area of all small ladoos to the surface area of big ladoo 

[(R/r)³ × 4πr²] : 4πR² = R/r = 810 : 90 = 9 : 1 

∴ The correct answer is 9 : 1