question_answer

Two small balls of same size and masses \[{{m}_{1}}\] and \[{{m}_{2}}\] \[\left( {{m}_{1}}\,\,>\,\,{{m}_{2}} \right)\] are tied by a thin weightless thread and dropped from a certain height. Taking upward buoyancy force F into account the tension T of the thread during the flight after the motion of the balls becomes uniform will be-

A) \[\left( {{m}_{1}}-{{m}_{2}} \right)g~\]                        

B) \[\left( {{m}_{1}}-\text{ }{{m}_{2}} \right)\,\frac{g}{2}\]

C) \[\left( {{m}_{1}}+{{m}_{2}} \right)g\]             

D) \[\left( {{m}_{1}}+{{m}_{2}} \right)\,\frac{g}{2}\]

Correct Answer: B

Solution :

  \[\,{{F}_{b}}\] = Buoyancy force of air \[{{m}_{1}}:\text{ }{{m}_{1}}g\text{ }=T\,\,+\,\,{{F}_{b}}~~~~~~~~....\left( i \right)\] \[{{m}_{2}}:{{m}_{2}}g\,\,+\,\,T\,\,=\,\,{{F}_{b}}~~~~~~~~....\left( ii \right)\] From eq. (i) & (ii) \[{{m}_{1}}g-T={{m}_{2}}\,g+T\] \[\Rightarrow \,\,\,T\,\,=\,\,\frac{\left( {{m}_{1}}-{{m}_{2}} \right)g}{2}\]