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].You need to login to perform this action.
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