Let the rise in water level in the tank is x.

∴ According to the question,

Total displacement of 220 Articles = Increase in level of Water in tank

⇒ 220 × 4.5 × 100= 52 × 47 × x; (Since 1m = 100cm)

⇒

⇒ x = 40.5 cm

Given:

r_a + r_b = 14 cm      …(i)

⇒ 4π(r_a² – r_b²) = 112π

⇒ (r_a – r_b)(r_a + r_b) = 28

⇒ (r_a – r_b) =  = 2      …(ii)

On adding Equation (i) and (ii):

2r_a = 14 + 2 = 16

⇒ r_a =  = 8 cm

Now, r_b = 14 – 8 = 6 cm

Now, ratio of volumes of A and B =

⇒ 8³ : 6³

⇒ 512 : 216

⇒ 64 : 27

Volume of a cube = = = 5832

Side of a cube =  = 18 cm

Total Surface Area of the cuboid = 2 × 18 × 36 + 2 × 36 × 72 + 2 × 18 × 72

= 1296 + 5184 + 2592 = 9072 cm²

Since, a cube has 4 lateral surfaces.

Then, Lateral Surface Area of all 8 cubes with side length as 18 cm = 8 × 4 × 18² = 10368 cm²

Required ratio = 9072 : 10368 = 7 : 8

Area of three adjacent faces of a cuboid are 18 cm², 20 cm² and 40 cm².

Then let the length be denoted by l , breadth by b and height be h

Then l × b = 18

b × h = 20

h × l = 40

then multiplying the above equations :

(lbh)² = 18 × 20 × 40

= 144 × 100

lbh = 12 × 10

volume = lbh = 120 cm³

Required, Area of Base = 25 × 4 = 100 m²

Required Volume = 25 × 80 = 2000 m³

Volume of the conical tent = () × Area of base × height = 2000

⇒ Height =  = 60 m

Given that,

Volume of the prism= 288 cm³

Height of the prism = 24 cm.

We know that,

The volume of a Prism = Base area of prismHeight

Where,

Therefore,

288 = Base area of prism 24

Base area of prism =

Base area of prism = 12 cm.

Therefore, option D is the right answer.

Lateral height of pyramid with base as a square =

= =  = 5√5 cm

Lateral surface area of pyramid =

=  = 100√5 cm²

Ratio of r₁ : r₂ = 1 : 4 (ratio of the diameters of A and B is 1 : 4)

r₂ = 4r₁

Given:

Volume of cylinder A = Volume of cylinder B

πr₁²h₁ = πr₂²h₂

r₁²h₁ = (4r₁)²h₂

h₁ = 16h₂

h₁ / h₂ = 16 / 1

Therefore, h₁ : h₂ = 16 : 1

Here, External diameter= 18 cm

So external radius(R) = 18/2 = 9 cm

∴ Internal radius(r) = (9 – 2) = 7 cm

Volume of a hollo pipe = πh(R² – r²)

= 3.14 × 10 × (9² – 7²)

= 31.4 × (81 – 49)

= 31.4 × 32 = 1004.8 cm³

∴ Weight of the hollow pipe (in Kg) = 1004.8 × 8.5/1000

= 8.54 kg

Volume of solid cylinder = πr²h; (r=radius and h=height)

Volume of hemi-sphere = πr³

Since hemi-spheres are cut from both ends,

Total volume cut out = 2 × πr³ = πr³

Now, required volume of the resultant object will be

= πr²h − πr³

= πr²(h – 4r/3)

= ; (h = 20 cm and diameter = 14 cm)

=

= 22×7×32/3

=