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Two blocks, each having a mass $M$ , rest on frictionless surface 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:

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$\frac{M g}{2} \sin θ$

$\frac{3}{2} M g \sin θ$

$2.3 M g \sin θ$

$2 M g \sin θ$

answer is C.

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Detailed Solution

$a = \frac{M g \sin θ}{(M + M)} = \frac{g}{2} \sin θ ∴ T = M a = \frac{M g}{2} \sin θ$

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