In the arrangement shown in the figure, accelerations of blocks A,B and C are aA,aB and aC respectively. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then
A
2aA+2aB+aC=0
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B
aA+aB+2aC=0
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C
aA+aB+aC=0
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D
2aA+aB+aC=0
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Solution
The correct option is A2aA+2aB+aC=0 I− Method
Determining relation of accelerations of blocks using constraint of string length i.e total length of the string is constant. X0 is the nearest distance of the pulley from the ceiling xA+xA−x0+xB−x0+xB−x0+xC−x0=L ⇒2xA+2xB+xC=L+3x0=constant Differentiating two times, we have: ⇒2aA+2aB+aC=0 Hence, option (a) is correct
II− Method
By work done method, total work done by the tention on the block is zero and A will have 2T , B will have 2T, C will have T as single string is connected to block C
Total work done by tention =(2T).sA+(2T).sB+(T).sC=0 where sA is displacement of block A