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

Last updated on Jun 3, 2025

Testbook provides Hemisphere MCQ Quizwith logical and easy explanations to all the questions. Detailed solutions for all the Hemisphere Objective questions have been provided so that the candidates can get the strategies and shortcuts to approach a question and solve it in less time. The Hemisphere Question Answers will help the candidates understand the concept better and grasp faster making it easier for them to solve the problems.

Latest Hemisphere MCQ Objective Questions

Hemisphere Question 1:

Find the curved surface area of the hemisphere (in cm2) whose radius is 60 cm and

  1. 25608
  2. 22608 
  3. 23608 
  4. 24608 

Answer (Detailed Solution Below)

Option 2 : 22608 

Hemisphere Question 1 Detailed Solution

Given:

Radius (r) = 60 cm

Formula used:

Curved Surface Area of Hemisphere = 2πr2

Calculation:

Curved Surface Area = 2 × 3.14 × 602

⇒ Curved Surface Area = 2 × 3.14 × 3600

⇒ Curved Surface Area = 22608 cm2

∴ The correct answer is option (2).

Hemisphere Question 2:

Find the Total surface area of the hemisphere whose radius is 33 cm and π = 3.14(cm2).

  1. 10358.38
  2. 10258.38
  3. 10558.38
  4. 10458.38

Answer (Detailed Solution Below)

Option 2 : 10258.38

Hemisphere Question 2 Detailed Solution

Given:

Radius (r) = 33 cm

Formula used:

Total Surface Area of a Hemisphere = 3πr2

Calculation:

Total Surface Area = 3 × 3.14 × 332

⇒ Total Surface Area = 3 × 3.14 × 1089

⇒ Total Surface Area = 10258.38 cm2

∴ The correct answer is option 2.

Hemisphere Question 3:

Find the Total surface area of the hemisphere whose radius is 30cm and π = 3.14(In cm2).

  1. 8478
  2. 8678
  3. 8778
  4. 8578

Answer (Detailed Solution Below)

Option 1 : 8478

Hemisphere Question 3 Detailed Solution

Given:

Radius (r) = 30 cm

Formula used:

Total surface area of hemisphere = 3 × π × r2

Calculations:

Total surface area = 3 × π × r2

⇒ Total surface area = 3 × 3.14 × 302

⇒ Total surface area = 3 × 3.14 × 900

⇒ Total surface area = 8478 cm2

∴ The correct answer is option 1.

Hemisphere Question 4:

Find the Total surface area of the hemisphere (In cm2) whose radius is 36cm and π = 3.14.

  1. 13208.32
  2. 15208.32
  3. 12208.32
  4. 14208.32

Answer (Detailed Solution Below)

Option 3 : 12208.32

Hemisphere Question 4 Detailed Solution

Given:

Radius (r) = 36 cm

π = 3.14

Formula used:

Total Surface Area of Hemisphere = 3πr2

Calculation:

Total Surface Area = 3 × 3.14 × 362

⇒ Total Surface Area = 3 × 3.14 × 1296

⇒ Total Surface Area = 12208.32 cm2

∴ The correct answer is option (3).

Hemisphere Question 5:

Find the Total surface area of the hemisphere (In cm2) whose radius is 39 cm and π = 3.14

  1. 14627.82
  2. 14427.82
  3. 14327.82
  4. 14527.82

Answer (Detailed Solution Below)

Option 3 : 14327.82

Hemisphere Question 5 Detailed Solution

Given:

Radius (r) = 39 cm

Formula used:

Total Surface Area of Hemisphere = 3πr2

Calculation:

Total Surface Area = 3πr2

⇒ Total Surface Area = 3 × 3.14 × (39)2

⇒ Total Surface Area = 3 × 3.14 × 1521

⇒ Total Surface Area = 14327.82 cm2

∴ The correct answer is option (3).

Top Hemisphere MCQ Objective Questions

A spherical metal of radius 10 cm is molten and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by:

  1. 1000 times
  2. 100 times
  3. 9 times
  4. No change

Answer (Detailed Solution Below)

Option 3 : 9 times

Hemisphere Question 6 Detailed Solution

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Formula Used:

Volume of sphere = \(\frac{4}{3}\)πr3

Surface area of sphere = 4πr2

Calculation:

If the radius of a smaller sphere be 'r cm' then

Acoording to the question:

\(\frac{4}{3}\)π(10)3 = 1000\(\frac{4}{3}\)π(r)3

r = 1 cm

Surface area of the larger sphere = 4π(10)2 = 400π

Total surface area of 1000 smaller spheres = 1000 × 4π(1)2 = 4000π

Net increase in the surface area = 4000π − 400π = 3600π

Hence, surface area of the metal is increased by 9 times.

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).

(Use \(\pi = \frac{22}{7}\))

  1. \(900\frac{4}{7}\)
  2. \(675\frac{3}{7}\)
  3. \(112\frac{4}{7}\)
  4. \(713\frac{1}{7}\)

Answer (Detailed Solution Below)

Option 2 : \(675\frac{3}{7}\)

Hemisphere Question 7 Detailed Solution

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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,

 

F1 Savita SSC 17-11-22 D1

 

 

 

 

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 × 102 - 6 × π22 + 6 × 2π22

⇒ Required surface area = 600 + 6π22

⇒ Required surface area = 600 + 24 × 22/7

⇒ Required surface area = 600 + 528/7

⇒ Required surface area = 600 + \(75\frac{3}{7}\) = \(675\frac{3}{7}\)

∴ The surface area of the remaining solid (in cm2) is \(675\frac{3}{7}\).

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

  1. 1.63 cm
  2. 1.35 cm
  3. 1.33 cm
  4. 1.45 cm

Answer (Detailed Solution Below)

Option 3 : 1.33 cm

Hemisphere Question 8 Detailed Solution

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Given:

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

Concept used:

1. Volume of hemisphere = \(\frac {2\pi \times Radius^3 }{3}\)

2. Volume of a cone = \(\frac {\pi \times Radius^2 \times Height}{3}\)

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

Calculation:

The volume of the hemisphere = \(\frac {2\pi \times 4^3 }{3}\) = \(\frac {128\pi}{3}\) cm3

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

​According to the concept,

\(\frac {128\pi}{3}\) = \(\frac {\pi \times R^2 \times 72}{3}\)

⇒ R2 = 16/9

⇒ R = 4/3

⇒ R ≈ 1.33

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

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 \(\frac{2}{3}\)part of the tank, if the internal radius of the tank is 10.5 m?

  1. \(\frac{185}{3}\)
  2. \(\frac{175}{3}\)
  3. \(\frac{185}{6}\)
  4. \(\frac{175}{2}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{175}{3}\)

Hemisphere Question 9 Detailed Solution

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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 × r3

1 m3 = 1000 L

1000 cm3 = 1 L

Calculation: 

The volume of the hemispherical tank is 

⇒ 2/3 × 22/7 × 10.5 × 10.5 × 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 × 2425500) ÷ 7.7 sec

⇒ 210,000 sec

Time in hours 

⇒ 210,000/3600

⇒ 175/3 hours 

∴ The required time is 175/3 hours.

A hemispherical bowl made of brass has an inner diameter of 14 cm. Find the cost of painting it from the inside at the rate of Rs. 15 per cm2. (Use π = \(\frac{22}{7}\))

  1. Rs. 4,120
  2. Rs. 3,170
  3. Rs. 4,620
  4. Rs. 2,670

Answer (Detailed Solution Below)

Option 3 : Rs. 4,620

Hemisphere Question 10 Detailed Solution

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Given:

The inner diameter of the bowl = 14 cm

Rate of painting = Rs. 15 per cm²

Formula used:

The surface area of the hemisphere = 2πr2

Cost of painting = Surface area × Rate of painting

Calculation:

Radius (r) = Diameter/2 = 14 cm/2 = 7 cm

Surface area = 2 × (22/7) × 72 = 308 cm²

Cost of painting = 308 × 15 = Rs. 4620

∴ The cost of painting the bowl from the inside is Rs. 4620.

The bases of a hemisphere and a cone are equal. If their height is also the same, then what is the ratio between their curved surface areas?

  1. √3 ∶ 1
  2. √2 ∶ √3
  3. 1 ∶ √3
  4. √2 ∶ 1

Answer (Detailed Solution Below)

Option 4 : √2 ∶ 1

Hemisphere Question 11 Detailed Solution

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Given:

The radius of the cone is equal to the radius of the hemisphere

The height of the cone is equal to the height of the hemisphere

Formula used:

The curved surface area of the hemisphere = 2πr2

The curved surface area of the Cone = πrl

l = √ (h2 + r2)

Where 'l' is a slant height

Calculations:

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Let the radius of the cone is equal to the radius of the hemisphere be r.

So the height of the cone, h = r.

According to the question,

The ratio of the areas of curved surfaces of hemispheres and cone

⇒ 2π × r2 :  π × r × l

⇒  2π × r π × r × √ (r2 + r2)

⇒  2π × r2  :  π × r × √ (2r2)

⇒  2 : \(\sqrt{2}\)

⇒  \(\sqrt{2}\) : 1

∴ The ratio of the areas of curved surfaces of hemisphere and cone is \(\sqrt{2}\) : 1.

A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 154 cm, then find the cost of painting it if the cost of painting is Rs. 4 per 100 cm2 (use π = \(\frac{22}{7}\)).

  1. Rs. 150.92
  2. Rs. 150.66
  3. Rs. 105.29
  4. Rs. 105.66

Answer (Detailed Solution Below)

Option 1 : Rs. 150.92

Hemisphere Question 12 Detailed Solution

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Given:

Circumference of the base of dome = 154 cm

Rate of painting = Rs.4/100cm2

Concept used:

Circumference of circle = 2 × π × r

Curved surface area of hemisphere = 2 × π × r2  

Calculation:

Circumference of the base of dome = 154 cm

⇒ 2 × π × r = 154

⇒ 2 × (22/7) × r = 154

⇒ r = (154 × 7)/44

⇒ r = 49/2

Curved surface area of hemispherical dome = 2 × π × r2

⇒ 2 × π × r × r

⇒ 154 × (49/2)

⇒ 77 × 49 = 3773 cm2

The rate of painting of 3773 cm2 = (4 × 3773)/100

⇒ Rs.150.92

∴ The correct answer is Rs.150.92.

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 = \(\frac{{22}}{7}\))

  1. 1635\(\frac{{6}}{7}\)
  2. 2800\(\frac{{2}}{7}\)
  3. 2107\(\frac{{2}}{7}\)
  4. 787\(\frac{{2}}{7}\)

Answer (Detailed Solution Below)

Option 3 : 2107\(\frac{{2}}{7}\)

Hemisphere Question 13 Detailed Solution

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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

R = Lower radius

r = upper radius and the radius of the hemisphere

Calculation:

F5 Madhuri SSC 09.09.2022 D1

In the given figure,

Height of the frustum = 12 m and height of the hemisphere = 10.5 m

Upper radius = 21/2

Lower radius = 39/2

Now,

Slant height (l) of the frustum = √[122 + {(39/2) - (21/2)}2]

⇒ √[12+ {(39 - 21)/2}2]

(12+ 92)

⇒ √(144 + 81)

⇒ √225

⇒ 15

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

⇒ π[15 × (60/2) + 2 × 441/4]

⇒ π[15 × 30 + 441/2]

⇒ 22/7 × [450 + 441/2]

⇒ 22/7 × [(900 + 441)/2]

⇒ 11/7 × 1341

⇒ 14751/7

⇒ 2107\(\frac{{2}}{7}\)

So, the required area of the cloth = 2107\(\frac{{2}}{7}\) m2

∴ The area of the cloth (in m2) used to make the tent is 2107\(\frac{{2}}{7}\).

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

  1. 145232 cm3
  2. 140232 cm3
  3. 150032 cm3
  4. 155232 cm3

Answer (Detailed Solution Below)

Option 4 : 155232 cm3

Hemisphere Question 14 Detailed Solution

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Given:

TSA of Hemisphere = 16632 cm2

Formula used:

Volume of solid hemisphere = \(\frac{2}{3}\pi r^3\)

Calculation:-

It is given that the total surface area of the hemisphere is 16632 cm2

Total surface area 3 π r2

⇒ 3 × π × r2 = 16632

⇒ r2 = 5544/π 

⇒ r2 = 1764 

⇒ r = 42 cm

Volume = \(\frac{2}{3}\pi r^3\)

V = \(\frac{2}{3}\times \frac{22}{7} \times r^3\)

⇒ \(\frac{2}{3}\times \frac{22}{7} \times 42^3\)

On solving,

V = 155232 cm3

∴ The volume of the solid hemisphere is 155232 cm3.

The inner and outer radii of a hemispherical wooden bowl are 6 cm and 8 cm, respectively. Its entire surface has to be polished and the cost of polishing π cm2 is Rs. 50. How much will it cost to polish the bowl?

  1. Rs. 11,400
  2. Rs. 10,000
  3. Rs. 12,000
  4. Rs. 11,600

Answer (Detailed Solution Below)

Option 1 : Rs. 11,400

Hemisphere Question 15 Detailed Solution

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Given:

Inner radius (r1) = 6 cm

Outer radius (r2) = 8 cm

Cost to polish π cm2 = ₹50

Formula used:

Surface area of hemispherical bowl = Outer curved surface area + Inner curved surface area + Area of circular rim

Outer curved surface area = 2πr22

Inner curved surface area = 2πr12

Area of circular rim = π(r22 - r12)

Total surface area = 2πr22 + 2πr12 + π(r22 - r12)

Calculation:

Total surface area = 2π(8)2 + 2π(6)2 + π[(8)2 - (6)2]

⇒ Total surface area = 2π(64) + 2π(36) + π(64 - 36)

⇒ Total surface area = 128π + 72π + 28π

⇒ Total surface area = 228π cm2

Cost per π cm2 = ₹50

Cost = 228 × ₹50

⇒ Cost = ₹11,400

∴ The cost to polish the bowl is ₹11,400.

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