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A chain consisting of 6 links each of mass $0.2 k g$ is lifted vertically up with a constant acceleration of $1.2 m s^{- 2}$ .The force of interaction between the top link and the link immediately below it is

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5.5N

7.6N

11N

answer is D.

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Detailed Solution

F – 6mg = 6ma $\Rightarrow F = 6 m (g + a)$

Force of interaction b/w top link and its next link is T. Then ma = F– mg – T

$T = F - m (g + a) = 6 m (g + a) - m (g + a)$

$= 5 m (g + a) = 5 \times 0.2 (9.8 + 1.2) = 11 N$

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