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

Given:

Radius (r) = 50 cm

Formula used:

Curved Surface Area of Hemisphere = 2πr²

Calculation:

Curved Surface Area = 2 × 3.14 × 50²

⇒ Curved Surface Area = 2 × 3.14 × 2500

⇒ Curved Surface Area = 15700 cm²

∴ The correct answer is option (4).

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).

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.

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.

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).

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.

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 × 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 \(675\frac{3}{7}\).

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.

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.

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.

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:

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 × r2 :  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.

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.

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:

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]

⇒ √[122 + {(39 - 21)/2}2]

⇒ √(122 + 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\(\frac{{2}}{7}\).

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.

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.