Question

when a stone is in the air its weight is for an and in the water weight is its weight is 18 and the loss of weight is 1.6 and what is its buoyancy​

Answer 1

So when a body is weighed in water its weight is lossed due to upward force (Buoyant Force).

So,

Loss in weight = buoyant force on the immersed body

Weight (in air) - weight (in liquid) = buoyant force —(equation-1)

Weight (in air) = weight (in liquid) + buoyant force

Let σ & ρ be the density of body and liquid and 'V ' be the volume of body and we know that buoyant force F=ρVg

σ Vg = W(in liquid)+ ρ Vg

σVg=W(liquid)×ρVgρVg+ρVg

Taking ρ Vg as common

σVg=ρVg(W(liquid)ρVg+1)

σρ=W(liquid)+ρVgρVg

σρ=Weight(air)ρVg

Here, ρ Vg is buoyant force and σρ is Relative Density (R.D)

σρ=Weight(air)W(air)−W(liquid) — (using equation-1)

So, R.D=Weight(air)W(air)−W(liquid)

Remember this relation.

Now , we have given

W(in air)= 64

W(in water)=58

We know density of water is (ρ) = 1000Kg/m³

σ be the density of body

then, By using above relation we get

σρ=Weight(air)W(air)−W(liquid)

σ1000=6464–58

σ=640006

σ=10666.66Kg/m³

Density of body is 10666.66Kg/m³

I hope you are understand my solution