Question

A body of density rho is dropped from rest from a height h into a lake of density sigma sigma is greater than roop neglecting all dissipative force is the maximum depth to which the body sings before returning to float on surface is

Answer 1

Given:

Denisty of Body = p

Density of lake = d , d > p

Initial height of body = h

Initial velocity = 0m/s

To Find:

The maximum depth to which the body sings before returning to float on surface

Solution:

The velocity of the ball just before touching the surface of water can be found out.

• V² - 0² = 2gh
• V = √2gh -  (1)

Inside the water the acceleration of the body will be ,

• a = (weight - upthrust )/mass

This is a retardation. It will be upwards.

• a = (upthrust - weight )/ mass
• a =  (Vdg - Vpg)/Vp

Where V = volume of the body.

• a = -(p - d )g/p m/s²

Therefore at maximum depth, final velocity = 0

• V²- U² = 2as
• X = maximum depth
• - 2gh = 2 x a (p -d )g/p x  X
• X = ph/(d-p)

The maximum depth to which the body sings before returning to float on surface is ph/($\sigma$-p) m

Answer 2

Answer:

h(delta)/(rho-delta)

Explanation:

refer attachment for explanation.

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hope this helps :)