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Two blocks, each having a 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

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$\frac{2}{3} M g \sin θ$

$\frac{3}{2} M g \sin θ$

$\frac{M g \sin θ}{2}$

$2 M g \sin θ$

answer is C.

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Detailed Solution

Acceleration of system, $a = \frac{Net pulling force}{Total mass}$

$= \frac{M g \sin θ}{2 M} a = \frac{1}{2} g \sin θ$

Now, the block on ground is moving due to tension.

Hence, $T = M a = \frac{M g \sin θ}{2}$

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