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Three block are connected as shown, on a horizontal frictionless table and pulled to the right with a force `T_(3) = 60` N. If `m_(1) = 10 kg`, `m_(2) = 20` kg and `m_(3) = 30` kg, the tension `T_(2)` is-

A

40 N

B

30 N

C

20 N

D

10 N

Text Solution

Verified by Experts

The correct Answer is:
B

`a =F/(M_(1) + M_(2) + M_(3)) = 60/(10 + 20 + 30) = 1 ms^(-2)`
Taking `M_(1)` and `M_(2)` as a system
`T_(2) = (M_(1) + M_(2)) a = (10 + 20) 1 = 30 N`
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