MCQ: Hemispheres - SSC CGL MCQ

# MCQ: Hemispheres - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Hemispheres

MCQ: Hemispheres for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Hemispheres questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Hemispheres MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Hemispheres below.
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MCQ: Hemispheres - Question 1

### A 22.5 m high tent is in the shape of a frustum of a cone surmounted by a hemisphere. If the diameters of the upper and the lower circular ends of the frustum are 21 m and 39 m, respectively, then find the area of the cloth (in m2) used to make the tent (ignoring the wastage). (Use π = 22/7)

Detailed Solution for MCQ: Hemispheres - Question 1

Given:

Height of the tent = 22.5 m
the diameters of the upper and the lower circular ends of the frustum are 21 m and 39 m, respectively

Formula used:

Slant height of a frustum (l) = √[H2 + (R - r)2]
Curved surface area of a frustum = πl(R + r)
Curved surface area of a hemisphere = 2πr2
Here,
H = Height of the frustum

Calculation:

In the given figure,
Height of the frustum = 12 m and height of the hemisphere = 10.5 m
Now,
Slant height (l) of the frustum = √[122 + {(39/2) - (21/2)}2]

Surface area of the tent = Curved surface area of the frustum + curved surface area of the hemisphere
So,
Surface area of the tent = 15π(39/2 + 21/2) + 2π × (21/2)2

So, the required area of the cloth

∴ The area of the cloth (in m2) used to make the tent is

MCQ: Hemispheres - Question 2

### The outer diameter of a rubber ball's cross-section is 22 inches. The rubber has a thickness of 0.5 inches. To the nearest square inch, what is the area of the ball's interior surface? (Use π = 3.14).

Detailed Solution for MCQ: Hemispheres - Question 2

Given:

Outer diameter of rubber ball = 22 inches.

Thickenss = 0.5 inches

Formula used:

Surface area of sphere = 4πr2 [where, r is the radius of sphere]

Calculations:

Outer diameter = 22

Inner radius of rubber ball = 11 - 0.5 = 10.5 inches
Now, surface area of inner side of rubber ball = 4 x 3.14 x 10.52
= 4 × 3.14 × 110.25
= 1384.74 sq. inches
∴ The answer is 1384.74 sq. inches.

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MCQ: Hemispheres - Question 3

### Total surface area of hemisphere is 462 cm2. What is the diameter of hemisphere?

Detailed Solution for MCQ: Hemispheres - Question 3

Given:

Total surface area of hemisphere = 462 cm2

Formula:

Total surface area of hemisphere = 3πr2

Calculation:

According to the question

Total surface area of hemisphere = 462

∴ Diameter of hemisphere = 7 × 2 = 14 cm

MCQ: Hemispheres - Question 4

A hemispherical bowl made of iron has inner diameter 84 cm. Find the cost of tin plating it on the inside at the rate of Rs. 21 per 100 cm2  (Use π = 22/7) correct to two places of decimal.

Detailed Solution for MCQ: Hemispheres - Question 4

Given:

Inner Diameter = 84 cm
Cost of tin plating it on the inside = Rs. 21 per cm2.

Concept Used:

Curved Surface area of Hemisphere = 2π r2

Calculation:

Inner Diameter = 84 cm
Inner Radius = 84/2 cm = 42 cm
Curved Surface area of Hemisphere = 2π r2

The cost of tin-plating 100 cm2 of the bowl = Rs. 21
The cost of tin-plating 1 cm2 of the bowl = Rs. 21/100
The cost of tin-plating 11088 cm2 area of the bowl = (21/100) × 11088 = Rs. 2,328.48
Thus, the cost of tin-plating is Rs. 2,328.48.
∴ Option 1 is the correct answer.

MCQ: Hemispheres - Question 5

If a solid hemisphere of radius 3.5 cm is converted into four small spheres then find the radius of each small spheres.

Detailed Solution for MCQ: Hemispheres - Question 5

Given:

Radius of the hemisphere = 3.5 cm

Formula used:

Volume of a hemisphere = (2/3)πr3
Volume of a sphere = (4/3)πr3

Calculation:
Let the radius of spheres be r
According to the question:

∴ Radius of small spheres = 7/4 cm

MCQ: Hemispheres - Question 6

A cone, a hemisphere and a cylinder stand on equal bases and have the same height, then the ratio of their volumes is

Detailed Solution for MCQ: Hemispheres - Question 6

Given:

A cone, a hemisphere and a cylinder stand on equal bases and have same height.

Formula used:

Volume of cone = 1/3 πr2h

Volume of hemisphere = 2/3πr3

Volume of cylinder = πr2h

Calculation:

A cone, a hemisphere and a cylinder stand on equal bases and have same height.

They also have same height = h
Height of cone = h
Height of cylinder = h
Height of hemisphere = r = h
∴ Volume of cone = V1= 1/3π r2h
Volume of cylinder = V2 = πr2h
Volume of hemisphere = V3 = 2/3π r2 × r
⇒ 2/3 π r2 × h = 2/3π r2h
Ratio of volume of cone : hemisphere : cylinder

⇒  1/3π r2h :  2/3π r2h :  πr2h
⇒ 1/3 : 2/3 : 1
⇒ 1 : 2 : 3
∴ Option 1 is correct.

MCQ: Hemispheres - Question 7

A solid hemisphere has radius 21 cm. It is melted to form a cylinder such that the ratio of its curved surface area to total surface area is 2 ∶ 5. What is the radius (in cm) of its base (take π = 22/7)

Detailed Solution for MCQ: Hemispheres - Question 7

Given:

The radius of a solid hemisphere is 21 cm.
The ratio of the cylinder's curved surface area to its Total surface area is 2/5.

Formula used:

The curved surface area of the cylinder = 2πRh
The total surface area of cylinder = 2πR(R + h)
The volume of the cylinder = πR2h
The volume of the solid hemisphere = 2/3πr³
(where r is the radius of a solid hemisphere and R is the radius of a cylinder)

Calculations:
According to the question,
CSA/TSA = 2/5
⇒ [2πRh]/[2πR(R + h)] = 2/5
⇒ h/(R + h) = 2/5

⇒ 5h = 2R + 2h

⇒ h = (2/3)R .......(1)

The cylinder's volume and the volume of a solid hemisphere are equal.

⇒ πR2h = (2/3)πr3
⇒ R2 × (2/3)R = (2/3) × (21)3
⇒ R3 = (21)3
⇒ R = 21 cm
∴ The radius (in cm) of its base is 21 cm.

MCQ: Hemispheres - Question 8

A hemispherical tank full of water is emptied by a pipe at the rate of 7.7 litres per second. How much time (in hours) will it take to empty 2/3 part of the tank, if the internal radius of the tank is 10.5 m?

Detailed Solution for MCQ: Hemispheres - Question 8

Given:

The internal radius of the tank is 10.5 m
A pipe at the rate of 7.7 liters per second emptied the tank.

Concept:
Volume of hemisphere = 2π/3 x r3
1 m3 = 1000 L
1000 cm3 = 1 L

Calculation:
The volume of the hemispherical tank is
⇒ 2/3 × 22/7 x 10.5 x 10.5 x 10.5
⇒ 2425.5 m3

The capacity of the tank is
⇒ 2425.5 × 1000 L
⇒ 2425500 L
So,
Time taken by the pipe emptied 2/3 part of the tank is
⇒ (2/3 x 2425500) ÷ 7.7 sec
⇒ 210,000 sec
Time in hours
⇒ 210,000/3600
⇒ 175/3 hours
∴ The required time is 175/3 hours.

MCQ: Hemispheres - Question 9

A hemispherical depression of diameter 4 cm is cut out from each face of a cubical block of sides 10 cm. Find the surface area of the remaining solid (in cm2). (take π = 22/7)

Detailed Solution for MCQ: Hemispheres - Question 9

Given:

Diameter of the hemispherical depression = 4 cm
Each face of the cube = 10 cm

Concept used:

TSA of the cube = 6a2
CSA of the hemisphere = 2πr2
Area of a circle = πr2
a = side or edge of a cube
r = radius of circle or hemisphere

Calculation:

The figure below of the solid is created as per the given information along with the top view of the solid,

This hemispherical shape will be formed on each face of the cube

Radius of the hemisphere = 4/2 = 2 cm
According to the question,
Required surface area = TSA of the cube - 6 × areas of circle form on the top of each face of the cube + 6 × CSA of hemisphere formed on each face of a cube
⇒ Required surface area = 6 x 102 - 6 x π22 + 6 x 2π22
⇒ Required surface area = 600 + 6π22
⇒ Required surface area = 600 + 24 x 22/7
⇒ Required surface area = 600 + 528/7

∴ The surface area of the remaining solid (in cm2) is

MCQ: Hemispheres - Question 10

If the diameter of a hemisphere is 28 cm, then what is the volume of hemisphere?

Detailed Solution for MCQ: Hemispheres - Question 10

Given:

The diameter of a hemisphere is 28 cm.

Concept used:

1. Volume of a hemisphere = 2πR3/3 (R = Radius)
2. Diameter = Radius × 2

Calculation:

Radius of the hemisphere = 28/2 = 14 cm
Now, the volume of the hemisphere

∴ The volume of the hemisphere is 5749.33 cm3.

MCQ: Hemispheres - Question 11

The volume of a solid hemisphere is  What is its total surface area (in cm²)? (Take π = 22/7)

Detailed Solution for MCQ: Hemispheres - Question 11

Given:

The volume of a solid hemisphere is
Take π = 22/7

Formula used:

The volume of the solid hemisphere = 2/3πr3
The total surface area of the solid hemisphere = 3πr2

Where,
r, is the radius of the hemisphere

Calculation:

According to the question, the required figure is:

The volume of the solid hemisphere,

Now,

The total surface area of the solid hemisphere

∴ The total surface area of the solid hemisphere is 594/7 cm2

MCQ: Hemispheres - Question 12

The outer and inner radius of a hemispherical bowl are 45 cm and 41 cm, respectively. Find the total surface area of the bowl.

Detailed Solution for MCQ: Hemispheres - Question 12

Given:

Outer radius (R) of hemispherical bowl = 45 cm.
Inner radius (r) of hemispherical bowl = 41 cm.

Formula used:

Total surface area (T.S.A) of hemispherical shell

Calculation:

Using the above formula we get:

∴Total surface area of the bowl is 7756π cm2

MCQ: Hemispheres - Question 13

The total surface area of a solid hemisphere is 16632 cm2. Its volume is: (Take π = 22/7)

Detailed Solution for MCQ: Hemispheres - Question 13

Given:

TSA of Hemisphere = 16632 cm2

Formula used:

Volume of solid hemisphere

Calculation:-

It is given that the total surface area of the hemisphere is 16632 cm.

Total surface area = 3 π r2

∴ The volume of the solid hemisphere is 155232 cm3.

MCQ: Hemispheres - Question 14

Steel is used to make a hemispherical bowl that is 0.37 cm thick. The bowl's inner radius is 6 cm. Find the bowl's outside curved surface area  (Use π = 22/7)

Detailed Solution for MCQ: Hemispheres - Question 14

Given::

Thickness = 0.37 cm

Formula used:

Curved surface area of the hemisphere = 2πr2

Calculations:

According to the formula,
⇒ Outer radius = 6  + 0.37 = 6.37 cm

⇒ Hence, Curved surface area of the hemisphere is 255.0548 cm2

MCQ: Hemispheres - Question 15

The radius of large hemisphere is thrice times the radius of small hemisphere. Find the ratio of difference between the volume of large hemisphere and the volume of small hemisphere to the volume of smaller hemisphere?

Detailed Solution for MCQ: Hemispheres - Question 15

Given:

Concepts used:

Volume of hemisphere = (2/3) × π × (radius)3
Ratio = [(Volume of large hemisphere – volume of small hemisphere)/Volume of small hemisphere]

Calculation:

Let radius of small hemisphere be r units.
⇒ Radius of large hemisphere = 3r
Volume of small hemisphere = 2πr3/3 units
⇒ Volume of large hemisphere = 2π/3 × (3r)3 units = 18πr3 units
Ratio = [{18πr3 – (2πr3/3)}/2πr3/3] = 26 ∶ 1
∴ Ratio of difference between the volume of large hemisphere and the volume of small hemisphere to the volume of smaller hemisphere is 26 ∶ 1.

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