Question

P-12.15. An object has a mass of 35 kg when measured in air. It is then tied to a string and immersed in water. The ten- sion in the string is now 280 N. Later, when immersed in oil the tension becomes 300 N. Find (a) the density of the object and (b) the density of the oil.

Answer 1

@u Consider the free-body diagram of the inmersed object in water. Ta tension in the string Fg Buoyant force my weight of the object In equilibrium, the force on the object is zero T + Fo - mg=0 - Foamg-T Punter IV = mg-T - 0 here v Es the volume of the object v=mg-T Pwater g Using m= 35 kg, g=9.81 mls?, T=280N, Pwater = 1000 kg/m3 we get V = 6.458 x10-3 m

The density of the object is Poject = massa Volume [ Perfect = 5420 kg/m3 (6) "Using similar analysis for in oil, we will arrive the object innersed at o Por gv = mg - T. Cil - (35 kg) x 19.81 misa) – 300N (9.81 m/s-)x(6:458x6m) Tlum = 684 kg / m3