Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force F applied on the upper string produces an acceleration of 2 $m / s^{2}$ in the upward direction in both the blocks. If T and T′ be the tensions in the two parts of the string, then

T=70.8 N and T′=47.2 N

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T=58.8 N and T′=47.2 N

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T=70.8 N and T′=58.8 N

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T=70.8 N and T′=0

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Solution

The correct option is A
T=70.8 N and T′=47.2 N

FBD of mass 2 kg FBD of mass 4kg

$T - T^{'} - 20 = 4 . . . (1) T^{'} - 40 = 8 (i i)$
By solving (i) and (ii) T'=47.23 N and T=70.8 N

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Q. Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force F applied on the upper string produces an acceleration of

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