question_answer

In the system in figure \[{{m}_{B}}=4\,kg\]and \[{{m}_{A}}=2\,kg.\]The pulleys are massless and friction is absent every- where. The acceleration of block A is\[(g=10\,m{{s}^{-2}})\]

A) \[2\,m/{{s}^{2}}\]               

B) \[4\,m/{{s}^{2}}\]               

C) \[\frac{10}{3}m/{{s}^{2}}\]                      

D) \[\frac{20}{3}\,m/{{s}^{2}}\]   

Correct Answer: C

Solution :

As is clear from figure acceleration of B must be half the acceleration of A, i.e., (i)                                 (ii)             If \[{{a}_{B}}=a,\]then,\[{{a}_{A}}=2a.\] From free body diagram of B in the figure [a], \[4g-T=4a\] or \[40-T=4a\]                                         (i) From free body diagram of A in the figure [b], \[2\times 2a=\frac{T}{2}-2\,g\,\sin {{30}^{o}}\]                               (ii) From Eq. (i), \[T=40-4a\] Put in Eq. (ii), \[4a=\frac{40-4a}{2}-2\times 10\times \frac{1}{2}\] \[4a+2a=20-10=10,\,a=\frac{10}{6}=\frac{5}{3}\,m/{{s}^{2}}\] \[{{a}_{A}}=2\alpha =\frac{10}{3}\,m/{{s}^{2}}\] Hence, the correction option is [c].