Laws of MotionHard
Question
Two blocks, each having mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then the tension in the string will be :
Options
A.2/3 Mg sin θ
B.3/2 Mg sin θ
C.
D.2 Mg sin θ
Solution
Mg sinθ − T = Ma [Newton′s II law for block 1]
T = Ma [Newton′s II law for block 2]
By dividing both equations
2 T = Mg sinθ T =
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