Question

an object of mass 500g and a relative density of 2,4 is suspended in air by rope. if the object is held suspended by the rope but totally submerged in water, the tension in the rope will be

Answer 1

From the problem, given that

mass of the body = 500 grams = 0.5 kg

the relative density of the body is = 2.4

from the figure,

• T - The tension in the string when it is hanged from the regid support in air....
• 

*when the body is hanged in air because of equilibrium, tension in the body is equal to weight of the body in air

hence

         T = mg

            = 0.5 X 10

            = 5 newton; (here g = 10 m/s² - the acceleration due to gravity)

Calculating the buoyant force: F_B= D_o x V_b x g

      (here D_o - is density of water = 1000 kg /m³;

               V - volume of the body = mass of the body / density of the body

                                                = 0.5 / 2400 = 5 / 24000 = 0.0002083 m³

              g - acceleration due to gravity = 10 m/s² ;

• again here density of the body = relative density x density of water = 2.4 x 1000 = 2400 kg / m³ )

Now the force of buoyancy is F_B = D_o x V_b x g

                                                = 1000 x 0.0002083 x 10

                                                = 2.083 newton;

from the figure,

• T' - The tension in the string when it is hanged from the regid support and immersed in water
• F_B - the buoyant force act on the hanged body in water

*when the body is hanged in water the equilibrium occurs because of tension force and buoyant force with weight of the body

..

hence   

                              T' + F_B = mg

The tension in the string when it is immersed in water is T' is

                              T' = mg - F_B

                                  = 5 - 2.083

                                  = 2.917 newton;

and now the tensin in the string when it is place in water is T' = 2.917 newton

Hence the solution to the given problem