Question

04) A disc is placed at the bottom of a water tank.
The thrust on it due to water is E. If another disc of
the double the area is placed at the bottom of the
same water tank, what will be the thrust on it?
A) 4F
B) 2F
C) 3F
D) F​

Answer 1

Answer:

I think option A is correct

Answer 2

Given:

The thrust on disc 1 with area A = F

The area of disc 2 = 2A

To find:

The thrust on disc 2 (F')

Solution:

Thrust experienced by an object inside a liquid is the buoyant force acting per unit area perpendicular to the surface of the object.

According to Archimedes Principle,

The buoyant force acting on an object equals the weight of the liquid displaced by the object.

Mathematically, Fb = V X d X g

where V is the volume of the liquid displaced and d is the density of the liquid.

Since the disc is completely immersed in the water,

⇒ The volume of the liquid displaced = The volume of the disc

Let t be the thickness of the disc, so its volume = Area X thickness

Fb on disc 1 = A x t X d X g

Fb on disc 2 = 2A x t X d X g

⇒ Fb 2 = 2 X Fb1     - (1)

Now thrust = Fb / Area

Thurst on disc 1 = E = Fb1 / A

Thurst on disc 2 (E') = Fb2 / 2X A

or $\frac{F}{F'} =\frac{Fb1 X 2A}{Fb2 X A}$

or F / F' = 2 X Fb1 / Fb2

From equation 1,

F / F' = 2 / 2

or F = F'

Hence, the thrust on the second plate is D) F​.