Question

Part B A statue of unknown volume and density is suspended from a string. When suspended in air, the tension in the string is Tair however, when submerged in water, the tension is T, < Tair - Derive an expression for the volume of the statue in terms of pn, acceleration due to gravity g, Tair and T. Express your answer in terms of the variables pr: 9, Trir and Tw. ινα ΑΣφ V = Request Answer Submit

Answer 1

we know that when we place a object in a fluid fluid exert a force on object in upward direction. This force is known as buoyant force. Magnitude of buoyant force is equal to weight of displaced fluid by object submerged So, let's come on the question- case I) Statue is in Air Unkทธา -47% ฯ * า ง density and rolume / (volume is r) 8 is density of of Stathe. statue of Tair Tair=mg mis mass of stathe. m = fv weight of stathe is balanced by Tair. so, Tair = ng = prg

Caselt) When Stathe is submerged in water. - t - 7 -W - Images free Body Diagram of stathe T SA FB Now, I buoyant force and Tensions both are directed upward weight of statue is balanced by Twe and FB buoyant force = sw vg = weight of displaced fuiol So, now we can write s displaced fluid volume is) [ equal to volume of stathes Tw t Fog = mg ş mg = sug? Twt for Vg = frg lug is equal to Tair Twt Jw vg = Tair from equation (1), In vg = Tair - Tw = | Tair - Tw I so, this is the required expression for volume of stathe.