Question

An object of weight W and density p is submerged in a fluid of density p1 .its apparent weight will be ??

Answer 1

Answer:

The apparent weight will be $W(1-\dfrac{\rho_{1}}{\rho})$

Explanation:

Given that,

Weight of object = W

Density of object = $\rho$

Density of liquid =$\rho_{1}$

We know that,

Apparent weight = Real weight - Buoyant force

$W'=W-F$.....(I)

Apparent weight $W'=mg'=\rho V g'$

Real weight $W= \rho V g$

Buoyant force = mass of liquid x g

$F= mg=\rho_{1} V g$

Now put the all values in equation (I)

$\rho V g=\rho V g-\rho_{1} V g$

$g'=g(\dfrac{\rho-\rho_{1}}{\rho})$....(II)

Now, multiply by m both sides in equation (II)

$mg'=mg(\dfrac{\rho-\rho_{1}}{\rho})$

$W'=W(1-\dfrac{\rho_{1}}{\rho})$

Hence, The apparent weight will be $W(1-\dfrac{\rho_{1}}{\rho})$

Answer 2

Answer:

hope this helps you :)

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 ρ₁

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

Apparent loss = ρ₁V = ρ₁ ₓ W/ρ

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

Weight in water = W − ρ₁ ₓ W/ρ

                           = W(1−ρ₁/ρ)