Question

A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of the U is more than 10 cm deep. Then, oil with a density of 950 kg/m3 is poured into one side, until the column of oil is 6.0 cm tall. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?

Answer 1

Answer:

The value is  $\Delta h = 0.003 \ m$

Explanation:

From the question we are told that

   The  height of the water is  $h_1 = 10 \ cm = 0.10 \ m$

    The  density of  oil is $\rho_o = 950 \ kg/m^3$

  The  height of  oil  is  $h_2 = 6 \ cm = 0.06 \ m$

Given that both arms of the tube are open then the pressure on both side is the same

So  

      $P_a = P_b$

=>   Here  

             $P_a = P_z + \rho_w * g * h$

where  $\rho_w$ is the density of water with value  $\rho_w = 1000 \ kg/m^3$

and  $P_z$ is the atmospheric pressure

and  

        $P_b = P_z + \rho_o * g * h_2$

=>   $P_z + \rho_w * g * h = P_z + \rho_o * g * h_2$

=>    $\rho_w * h = \rho_o * h_2$

=>      $h = \frac{950 * 0.06 }{1000}$

=>      $h = 0.057 \ m$

The  difference in height is evaluated as    

           $\Delta h = 0.06 - 0.057$

          $\Delta h = 0.003 \ m$