Question

(i)the volume of Holow portion of sphere is 3/4 of trhat of external volume of the sphere.If it floats in a liquid of rd 3/2 half of its external volume immersed,the rd of material of solid is
(ii)A barometer reads 0.76 m.Its torricellian space is 0.09m long.The volume of air measured at atm to be introduced into space to cause Hg drop to 0.57 m is (the cross section of barometer tube is 10^4 sqm)
(iii)A cubical box of wood of side 30 cm weighing 21.6 kg floats on water with 2 faces horizontal.The depth of immersion of box is
please answer all questions with detailed explanations

Answer 1

R = external radius,          radius of hollow portion = r

     Volume of solid part of the sphere =  (1 - 3/4) * 4/3 π R³
      Relative density of fluid = 3/2
     Volume of liquid displaced = 1/2 * 4/3 * π R³

        [ 1/4 * 4/3πR³ ] * rd_solid * g = (2/3 π R³) * 3/2 * g
                 rd_solid = 3
 
       This can be obtained directly as half of the sold is floating and half is immersed, the solid has a density equal to twice of the liquid.

2.   Volume of measured at  what pressure ?  there seems to be a number missing there.

   Let us assume that we want the volume of air measured at P atm to be introduced in to the Torricellian space.

  Area of barometer tube is probably  10^-4 sqm = 1 sq cm.  not  10^4 sqm.     

Height of barometer is 0.76 + 0.09 = 85 cm
  Pressure due to air + pressure due to mercury column = atmospheric pressure
   Pressure of air + density_mercury * 57 cm * g = density_mercury * 76 cm * g

    Pressure of air = 13.6 * 19 * g  units
     Volume of air = at the above pressure = (85 - 57) cm * 1 sq cm = 28 cm³      
             P V = constant

    Volume of air measured at P atm = (13.6 * 19 * g * 28( / (P * 13.6 * 76 * g)  cm³
                     = 7 / P  cm³3)

==============================
      depth of immersion = h
         h * 0.30² * 1000 kg/m³ * g = 21.6 kg * g

             h = 0.24 meter