A24. A 0.500 kg solid mass is totally submerged in water while suspended from a

Question

A24. A 0.500 kg solid mass is totally submerged in water while suspended from a spring balance. The density of the material from which the mass is composed is 5 times the density of water. The density of water = 1.00 x 10^3 kgm-3. (i). Calculate the volume of the solid mass. (ii). Calculate the reading on the spring balance when the object is fully immersed. A25. An oil tanker fully loaded has a mass of 6 10^8 kg . Assume the ship has a shape like a shoebox: 300 m long, 80 m wide and 40 m high. Calculate how far in meters the bottom of the ship is below sea level when the density of seawater is 1020 kg m3 .

Explanation / Answer

A24 (!) density    of the solid mass '   = mass /Volume    = 5*10^3 kg / m^3 Volume of the solid mass   V =   (0.5kg ) / ( 5*10^3 kg / m^3  ) = 1*10^-4 m^3 According to archimede's principle weight of the solid mass   in water W'   =   weight of the solid mass in air (w) - weight of the water displaced W'    =   mg   -   Vg Here  g = accealration due to gravity =   9.8 m/s^2 m = mass = 0.5 kg = density of water   = 1000 kg / m^3 (!!)   wiight mesured by the scale W'   = (0.5)(9.8 m/s^2)    -   ( 1000 kg / m^3 ) ( 1*10^-4 m^3)(9.8)   W' = 3.92 N

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