Question

An object of weight W and density p is
dipped in a fluid of density Pi. Its
apparent weight will be​

Answer 1

Answer:

Here is your answer buddy

The apparent weight will be W(1-\dfrac{\rho_{1}}{\rho. Explanation: Given that,. Weight of object = W. Density of objectwill be V = W/ρ Now, this object is immersed in a fluid of density v=W/p.

Now this object is immersed in a fluid of density.

Thanks...

Answer 2

Answer:

W(1−ρ  1  /ρ)

Explanation:

Weight of object in vacuum = W ; It's density = ρ

Therefore, the volume of the object will be V = W/ρ

Now, this object is immersed in a fluid of density ρ  1

​  

Its apparent loss in weight would be equal to the weight of fluid displaced (as is evident from Archimedes' principle)

Apparent loss = ρ  1  V = ρ  1  W/ρ

Therefore, the weight in water = weight in vacuum - the loss in weight

Weight in water = W−ρ  1  W/ρ

= W(1−ρ  1  /ρ)