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\[{{n}_{2}}\]You need to login to perform this action.
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