question_answer

Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force F applied on the upper string produces an acceleration of \[2m/{{s}^{2}}\] in the upward direction in both the blocks. If T and \[{T}'\]  be the tensions in the two parts of the string, then                 [AMU (Engg.) 2000]

A)             \[T=70.8N\] and \[{T}'=47.2N\]

B)             \[T=58.8N\] and \[{T}'=47.2N\]

C)             \[T=70.8N\] and \[{T}'=58.8N\]

D)             \[T=70.8N\] and \[{T}'=0\]

Correct Answer: A

Solution :

                FBD of mass 2 kg FBD of mass 4kg                                           \[T-{T}'-20=4\]  ?.(i) \[{T}'-40=8\]              ?(ii)             By solving (i) and (ii) \[{T}'=47.23\,N\] and \[T=70.8\,N\]