question_answer

The acceleration of block B in the figure will be

A) \[{{y}^{2}}=x+2\]                            

B) \[P=(\sqrt{3},0)\]

C) \[\frac{4(2-\sqrt{3})}{3}\]                            

D) \[\frac{4(\sqrt{3}+2)}{3}\]

Correct Answer: A

Solution :

 When the block \[[{{M}^{0}}{{L}^{0}}{{T}^{3}}{{l}^{0}}]\] moves downward with acceleration a, the acceleration of mass \[[{{M}^{-1}}{{L}^{-2}}{{T}^{6}}{{l}^{2}}]\] will be 2a, because it covers double distance in the  same time in comparison to \[[{{M}^{0}}{{L}^{0}}{{T}^{2}}{{l}^{0}}]\]. Let T is the tension in the string. From the free body diagram of A and B we get \[n=-D\frac{{{n}_{2}}-{{n}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]                                                     ...(i) \[{{n}_{1}}\]                                       ...(ii) From Eqs. (i) and (ii), we get\[{{n}_{2}}\]