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Question

A chain of 10 links each of mass 0.1kg is lifted vertically with a constant acceleration of 2ms2.The force of interaction between the top link and the one immediately below it is (g=10ms−2):

A
0.2N
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B
2N
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C
10.8N
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D
20N
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Solution

The correct option is C 10.8N
Mass of each link m=0.1kg
Mass of the system included in dotted box M=9m=0.9kg
Let the force of interaction between 1st link and 2nd link be T.
Writing equation of motion for the system ;
9ma=T9mg
T=9m(a+g)
Or T=0.9(2+10)=10.8 N

692354_297753_ans_b0c95b192ad74523a1df5262b511068c.png

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