• # 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) $\frac{20}{3}\,m/{{s}^{2}}$

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].