Question

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(diagram required)
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Answer 1

Answer:

sin θ = 1 / 2 √(ρ / ρ₀)

Explanation:

Given,

A uniform rod of density ρ is placed in a wide tank containing a liquid of density ρ₀(ρ₀>ρ).

The depth of liquid in the tank is half the length of the rod.

The rod is in equilibrium, with its lower end resting on the bottom of the tank.

To find: Value of sin theta.

Solution:

Let the length of rod be PQ = L as given in the figure attached.

Therby, SP = SQ = 1/2 PQ

at point S, Weight of rod = Alρg  

At mid-point of PR = l (say), Buoyancy Force Fb = Alρ₀g

To attain rotational equillibrium,

Alρ₀g * 1/2 cosθ  = ALρg * L/2 cosθ

l²/ L² = ρ / ρ₀

or, l / L = √(ρ / ρ₀)

From the diagram attached, we can find value of sinθ:

sin θ = h / l = L / 2l

sin θ = 1 / 2 √(ρ / ρ₀)

Answer 2

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•Refer the attachment!

@itztøxic!❤️