A rod of length 'L' and mass 'M' is acted on by two unequal forces F1 and F2 (<F1) as shown in the following figure. The Tension in the rod at a distance 'y' from the end 'A' is given by :
A
F1(1−yL)+F2(yL)
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B
F2(1−yL)+F1(yL)
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C
(F1−F2)yL
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D
none of these
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Solution
The correct option is AF1(1−yL)+F2(yL) acceleration =F1−F2m