Question

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the fraction of the legth tof the rod above water

Answer 1

Answer:

    $\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          $\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          $\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

           5/9 = $\frac{h_{liquid} }{ h_{body} }$