Surface Area And Volumes

Given, the radius of a sphere is 2r.

The volume of a sphere

Thus, the volume of the given sphere =

Thus, option d is correct.

The surface area of a cube is 96

Let the length of the cube is cm.

Thus,

[According to formula]

[Considering positive sign]

Thus the length of the cube is 4 cm.

Volume of a cube

= 64

Thus volume is 64 which is option c.

Height of cone, h = 8.4cm

Radius of base, r = 2.1cm

Thus volume of a cone =

Now when it is melted to form a sphere, say of radius r1 cm, the volumes of both are going to be equal.

Volume of sphere

Or

Or

Or

Thus the radius of the sphere is 2.1cm which is option b.

Let the radius of a cylinder be r unit and height be h unit. Now according to question,

Radius is doubled that is 2r and height is halved that is

Now curved surface area of a cylinder is

And according to the above condition, curved surface area

This is same as the curved surface area of the cylinder with radius r and height h.

Thus option c is correct.

Given, radius of a cone is and slant height is 2L.

Total surface area =

Which is option b.

Let the radius of first cylinder be 2x and second cylinder be 3x. the height of first cylinder be 5y and second cylinder be 3y.

Now volume of first cylinder be

And volume of first cylinder be

Thus ratio =

Thus the ratio is which option b is.

Let the side of cube be

Lateral surface area

[Given in the question]

⇒

⇒

Thus the length is 8 m.

Volume of a cube

= 512

Thus the volume is which is option a.

Dimension of plank = (4m50cm20cm)

= (4m0.5m0.2m)

Dimension of pit = (16m12m40m)

Thus no of planks that can be fitted into the pit

=

= 1920 m³

This is option b.

The dimension of a room give is (10m10m5m)

The length of the longest pole

= 15

The length of the longest pole is 15m.

This is option a.

Radius of hemisphere initially = 6cm

Radius of hemisphere finally = 12 cm

Surface area of hemisphere

Thus ratio of the surface areas

This is option a.

Let’s check:

We have a sphere and a cylinder.

Let height of the cylinder = diameter of the cylinder = x …(i)

Then according to the question, diameter of the sphere = x

This means, radius of the sphere = x/2 …(ii)

Also, radius of the cylinder = x/2 …(iii)

We know, volume of sphere is given by

Volume₁ (say) = 4/3 πr³

where r = radius of the sphere

⇒ Volume₁ =

[from (ii)] …(iv)

And Volume of cylinder is given by

Volume₂ (say) = πR²h

where R = radius of cylinder

⇒ Volume₂ = [from (i) & (iii)] …(v)

From equations (iv) & (v), we get

⇒ Volume of sphere = 2/3 (Volume of cylinder)

Thus, justified.

Hence, it is true.

Let original radius of right circular cone = R

And original height of right circular cone = H

Volume in this case is given by,

Volume = 1/3 πR²H …(i)

The radius is then halved, radius (say, r) = R/2

And height is doubled, height (say, h) = 2H

Now, volume is given by

Volume’ = 1/3 πr²h

⇒ …(ii)

By comparing equations (i) and (ii), we can clearly see that Volume ≠ Volume’

Thus, Volume will not remain unchanged.

Hence, it is false.

We have

This is a right circular cone, with

Height = h

Radius = r

And slant height = l

We know, ∠AOB = 90°

So by using Pythagoras theorem in ∆AOB, we can write

AB² = AO² + OB²

⇒ l² = h² + r²

This justifies that height, radius and slant height of cone can always be the sides of a right triangle.

Hence, it is true.

Let radius of the cylinder = r

Height of the cylinder = h

Then, curved surface area of the cylinder, CSA = 2πrh

According to the question, radius is doubled and curved surface area is not changed. We can write as,

New radius of the cylinder (say, R) = 2r

& new curved surface area of the cylinder, CSA’ = 2πrh [i.e., unchanged] …(i)

Alternate case: when R = 2r and CSA’ = 2πrh

But curved surface area of cylinder in this case, CSA’= 2πRh = 2π(2r)h = 4πrh …(ii)

Comparing equations (i) and (ii), we can say

equation (i) ≠ equation (ii)

as 2πrh ≠ 4πrh

Thus, if h = h/2 (height is halved) then,

CSA’ = 2π(2r)(h/2) = 2πrh

It matched CSA’ in equation (i).

Hence, it is true.

We have

The criteria for the largest cone to fit completely in a cube is that, the vertices of the cone will touch the face of the cube such that the diameter of the cone will be equal to the edge of the cube and height of the cone will be equal to the height of the cube.

Given: edge of cube, l = 2r

Then, diameter of the cone = 2r

⇒ radius of the cone = 2r/2 = r

Height of the cone, h = 2r [that is, height of the cube]

Volume of the cone is given by,

Volume of cone = 1/3 πr²h

= 1/3 πr²(2r) [∵, height of the cone = 2r]

= 2/3 πr³

= Volume of hemisphere of radius r

Hence, it is true.

According to the question,

Let diameter of cylinder = 2r, then diameter of cone = 2r

⇒ radius of cylinder = 2r/2 = r, and radius of cone = 2r/2 = r

Let height of cylinder = h, then height of cone = h

Now, volume of cylinder = πr²h …(i) [by using the above values]

And volume of cone = 1/3 πr²h

Or

Volume of cone = 1/3 (πr²h) = 1/3 (Volume of cylinder) [by equation (i)]

⇒ Volume of cone = 1/3 (Volume of cylinder)

⇒ Volume of cylinder = 3 × Volume of cone

Thus, the volume of cylinder is three times the volume of cone.

Hence, it true.

We have

According to the question, diameter of cone, hemisphere and cylinder are same.

So, radius of cone = radius of hemisphere = radius of cylinder = r

Also, height of cone, hemisphere and cylinder are same.

But in a hemisphere, radius and height always remain same.

So, height of cone = height of hemisphere = height of cylinder = r

Now,

Volume of cone = 1/3 π (radius)² (height)

= 1/3 π (r)² (r) = 1/3 πr³ …(i)

Volume of hemisphere = 2/3 π (radius)³

= 2/3 πr³ …(ii)

Volume of cylinder = π (radius)² (height)

= π r² r = πr³ …(iii)

Now, using equations (i), (ii) and (iii), we can write it in the ratio as

Volume of cone : Volume of hemisphere : Volume of cylinder = 1/3 πr³ : 2/3 πr³ : πr³

= 1/3 : 2/3 : 1

Taking L.C.M of the denominators (3, 3, 1), we get L.C.M as 3. Multiply 3 by each numerator,

Volume of cone : Volume of hemisphere : Volume of cylinder = 3/3 : 6/3 : 3

= 1 : 2 : 3

Hence, it is true.

Given: Length of the diagonal of a cube = 6√3 cm

We need to find length of the edge of the cube.

The relation between length of the edge of a cube and length of the diagonal of cube is given as,

Length of diagonal = √3 × length of the edge [of a cube]

⇒ 6√3 = √3 × length of the edge

⇒ length of the edge = (6√3)/√3 = 6 cm

Thus, length of the edge of the cube is 6 cm.

Hence, it is false.

If a sphere is inscribed in a cube, then length of the edge of the cube is equal to diameter of sphere.

So, let length of the edge of the cube = 2r

Then, diameter of the sphere = 2r

⇒ radius of the sphere = 2r/2 = r

Volume of the cube is given by,

Volume of the cube = (length of the edge)³

= (2r)³ = 8r³ …(i)

Volume of the sphere is given by,

Volume of the sphere = 4/3 π(radius)³

= 4/3 πr³ …(ii)

Using equations (i) & (ii),

Volume of the cube : Volume of the sphere = 8r³ : 4/3 πr³

= 8 : 4/3 π

= 2 : 1/3 π

Taking L.C.M (1,3) = 3. Multiply 3 by numerator of each term,

Volume of the cube : Volume of the sphere = 6 : 3/3 π

= 6 : π

Hence, it is true.

Let radius of cylinder = r

Let height of cylinder = h

Then, volume of cylinder is given by

V = πr²h …(i)

According to the question, now

Radius of the cylinder = 2r

Height of the cylinder = h/2

Then, volume of cylinder is given by

V’ = π(2r)²(h/2) = 4πr²(h/2) = 2πr²h = 2 (πr²h) = 2V

⇒ V’ = 2V

Thus, volume will be doubled if the radius of a cylinder is doubled and height is halved.

Hence, it is true.

Given: radius of each sphere, r = 2 cm

Then, volume of a sphere is given by

Volume of 1 sphere = 4/3 πr³

There are 16 spheres.

So, volume of 16 spheres = 16 × 4/3 πr³

= 16 × 4/3 × 3.14 × 2³

= 535.89 cm³

Volume of rectangular box = 16 × 8 × 8 = 1024 cm³ [∵, dimensions of rectangular box = 16 cm × 8 cm × 8 cm]

To find volume of the liquid that is filled in rectangular box, we need to find the space left in the rectangular box after the space occupied by the spheres.

So, Volume of the liquid = (Volume of the rectangular box) – (Volume of the 16 spheres)

⇒ Volume of the liquid = 1024 – 535.89 = 488.11 cm³

Thus, volume of this liquid is 488.11 cm³.

When the cubical tank is full of water:

Volume of water = Volume of cube = 15.625 m³

Then, Volume of cube = (length of edge of cube)³

⇒ (length of edge of cube)³ = 16.625

⇒ length of edge of cube =

= 2.5 m

Now we know the length of the edge of the cube, i.e., 2.5 m.

When the present depth of the water is 1.3 m:

Depth = 1.3 m

Length of tank = 2.5 m

Breadth of tank = 2.5 m

So, volume of water upto depth 1.3 m = length × breadth × depth

= 2.5 × 2.5 × 1.3

= 8.125 m³

Then, volume of water already used from the tank = (Volume of tank when it was full of water) – (Volume of water when depth is 1.3 m)

= 15.625 – 8.125

= 7.5 m³

Thus, volume of water already used from the tank is 7.5 m³.

When a solid spherical ball is immersed completely in water, it displaces volume of water equals to its own volume.

Given is, Diameter of spherical ball = 4.2 cm

⇒ radius of spherical ball = 4.2/2 = 2.1 cm

So, volume of a sphere is given by

Volume = 4/3 πr³

Thus, volume of water displaced is 38.81 cm³.

Dimensions of conical tent are:

Height = 3.5 m

Radius = 12 m

Canvas required to make a conical tent will only cover its curved surface, not the base.

So, we need to find the curved surface area of the conical tent to find out the area of canvas required to make that tent.

Curved surface area of cone = πr√(r² + h²)

= 22/7 × 12 √(12² + 3.5²)

= 22/7 × 12 × √156.25

= 22/7 × 12 × 12.5

= 471.43 m²

Since, area of canvas = curved surface area of conical tent

Thus, area of canvas required is 471.43 m².

Given is, two solid spheres of same metal. ⇒ Density of both spheres are same.

Mass (weight) of larger sphere, M = 5920 g

Mass (weight) of smaller sphere, m = 740 g

Diameter of smaller sphere = 5 cm

⇒ radius of smaller sphere, r = 5/2 = 2.5 cm

Volume of smaller sphere, v = 4/3 πr³

⇒

⇒ v = 1375/21 cm³

We know, density = mass/volume

⇒ density of smaller sphere = m/v

⇒ …(i)

And density of larger sphere = M/V

⇒ density of larger sphere = 5920/V g/cm …(ii)

By equations (i) & (ii), we get

Density of smaller sphere = Density of larger sphere

⇒

⇒

Volume of the larger sphere = 523.81 cm³

⇒ 4/3 πR³ = 523.81 [∵, volume of larger sphere = 4/3 πR³, where R = radius of the larger sphere]

⇒

⇒ R³ = 125

⇒ R = (125)^1/3

⇒ R = 5

Thus, radius of the larger sphere is 5 cm.

We have

Diameter of cylinder = 7 cm

⇒ Radius of cylinder, r = 7/2 = 3.5 cm

Height of the milk upto which it is filled in the glass, h = 12 cm

Then, volume of milk in cylindrical glass, V = πr²h

⇒ V = 22/7 × 3.5 × 3.5 × 12 = 462 cm³

This is the volume of milk in 1 cylindrical glass that serves just 1 student.

For 1600 students,

Volume of milk = 1600 × 462 cm³ = 739200 cm³

If 1 liter = 1000 cm³ that is, 1 cm³ = 0.001 liter

Then volume of milk for 1600 students = 739200 × 0.001 = 739.2 liter

Thus, 739.2 liters of milk is needed to serve 1600 students.

Given: length of cylindrical roller, i.e., height of the cylinder, h = 2.5 m

Radius of cylindrical roller, r = 1.75 m

The cylindrical roller will roll on the road only from its curved surface.

So, curved surface area of cylindrical roller, CSA = 2πrh

⇒ CSA = 2 × 22/7 × 1.75 × 2.5 = 27.5 m²

⇒ area of road covered in 1 revolution = 27.5 m²

And given that, total area of road covered = 5500 m²

So, number of revolutions made by road roller to cover 5500 m² = 5500/27.5 = 200

Thus, it made 200 revolutions.

If water required for 1 person per day = 75 l

Then water required for 5000 people per day = 75 × 5000 = 375000 l

= 375000 × 0.001 m³

= 375 m³

(as 1 liter = 0.001 m³)

Given, dimensions of overhead tank = 40 m × 25 m × 15 m

⇒ Volume of tank = 40 × 25 × 15 = 15000 m³

Days till the water if this tank will last = (volume of tank)/(water consumed by 5000 people per day)

= 15000/375

= 40

Thus, the water of this tank will last till 40 days.

Given that, radius of a spherical laddoo = 5 cm

Volume of that laddoo, v = 4/3 πr³

⇒

We have 523.81 cm³ of material of laddoo.

Volume of laddoo of radius 2.5 cm =

Then, laddoos that can be made out of 523.81 cm³ volume of laddoo material, of radius 2.5 cm can be found out as,

No. of laddoos = 523.81/65.48 = 8

Thus, 8 laddoos of radius 2.5 cm can be made out of the same amount of material.

As the right triangle is revolved about the side 8 cm, it forms a cone of height 8 cm and radius, 6 cm.

Then, volume of cone = 1/3 πr²h

Curved surface area of cone is given by

CSA = πrl = πr√(r² + h²)

⇒ CSA = 22/7 × 6 √(6² + 8²)

⇒ CSA = 22/7 × 6 × √100

⇒ CSA = 22/7 × 6 × 10

⇒ CSA = 188.57 cm²

Thus, volume of the cone is 301.71 cm³ and curved surface area of cone is 188.57 cm².

Given: Outer diameter d = 16cm

Thus,

Height = length = 100cm

Thickness of iron sheet = 2cm

To find: Volume of iron used

Volume of cylinder = πr²h

Where r= outer radius

And π = 3.14

Thus, Volume of cylinder = πr²h = 3.14 × (8)²× 100

= 20,096 cm³

Now, inner diameter = outer diameter – 2× thickness of iron sheet

inner diameter = 16-(2× 2) = 12cm

Thus, Volume of hollow space = πR² h

Where R= inner radius

And π = 3.14

Thus, Volume of hollow space = πR² h = 3.14 × (6)²× 100

= 11,304 cm³

Thus,

Volume of iron used = Volume of cylinder – Volume of hollow space

= (20,096 – 11,304 ) cm³

= 8800 cm³

Given: Diameter of semi circular sheet = 28cm

To find: Volume/ Capacity of open conical cup

Semi Circular sheet is bent to form an open conical cup

Thus, slant height of a conical cup (l)=radius of semicircular sheet

(r) = 14cm

We know that, Circumference of base of a cone = 2πR

Where, R = radius of a cone

Circumference of semi circle = πr

Thus, Circumference of base of a cone = Circumference of a

Semi circle

⇒ 2πR= πr

Thus, to find height of a cone we will use a formula

R² + h² = l² where R=radius of a cone

h=height of a cone

l=slant height of a cone

⇒ (7)² + h² = (14)²

⇒ 49+ h² = 196

⇒ h² = 196 – 49 = 147

⇒ h = √147 = 7√3 cm

Thus,

Where,

R = radius of a cone

h = height of a cone

Thus,

Given: Area of cloth = 165m²

Radius of conical tent = 5m

Curved surface area of cone = πrl

Where,

r = radius of a cone

l = slant height of a cone

Thus,

i)

Thus,

ii) Now, By using formula we can find height of a cone i.e.

r² + h² = l² where r=radius of a cone

h=height of a cone

l=slant height of a cone

⇒ (5)² + h² = (10.5)²

⇒ 25+ h² = 110.25

⇒ h² = 110.25 – 25 = 85.25

⇒ h = √85.25 = 9.23 m

Where

r = radius of base of a cone

h = height of a cone

Thus,

Given: Internal diameter of hemispherical tank = 14 m

Tank contains 50kl = 50m³ of water (since, 1kl = 1m³)

Where

r = radius of hemispherical tank

Thus,

Volume of water pumped into the tank = Volume of hemispherical

tank – 50m³

Volume of water pumped into the tank = 718.66 – 50 = 668.66 m³

Given: Ratio of volume of two sphere=64:27

We know that,

Let r₁ = radius of first sphere

r₂ = radius of second sphere

Thus,

We know that, Surface area of sphere = 4πr²

Where

r = radius of a sphere

Now,

Therefore, ratio of surface area of two sphere is 16:9

Given: length of a side of a cube “a” =4cm

Thus, Volume of a cube = a³

Where a = side of a cube

Thus, Volume of a cube = (4)³ = 64 cm³

Now, given that A cube contains a sphere touching side of a cube,

Now, By the figure drawn above, we can say that

Diameter of a sphere = side of a cube = 4cm

Where,

r = radius of a sphere

Thus,

Volume of a gap in between cube and sphere = Volume of a cube –

Volume of a sphere

Volume of a gap in between cube and sphere = (64 – 33.52)cm³

Volume of a gap in between cube and sphere = 30.48 cm³

We know that,

Where,

r = radius of a sphere

and, Volume of cylinder = πr²h

Where r= radius

And π = 3.14

Given: Volume of sphere = Volume of right circular cylinder

Thus,

Multiplying by 2 on both side

(since, diameter D=2r)

Let h be 100%

Thus,

Required difference = 150% - 100% = 50%

Thus, Diameter of a cylinder exceeds its height by 50%

Given: Total no. of circular plates = 30

Radius of each circular plate “r”= 14cm

Thickness of each circular plate = 3cm

We have placed circular plates one above the another to form the cylindrical solid

Thus, height of a cylindrical solid “h” = 30 × 3 = 90cm

i) Total surface area of cylinder = 2πrh + πr²

Where,

r = radius of cylinder

h= height of cylinder

ii) Volume of cylinder = πr²h

Where r= outer radius

And

Thus,