Two blocks, each having mass M rest on frictionless surfaces as shown in figure. If pulleys are ligh

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4-1.Newton's Laws of Motion

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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

A

$\frac{2}{3}\,M g \sin \theta$

B

$\frac{3}{2}\,M g \sin \theta$

C

$\frac{M g \sin \theta}{2}$

D

$2\,M g \sin \theta$

Solution

$M g \sin \theta-T=M a$

$T = Ma$

Now eq.$(1)$ – eq.$(2)$

$M g \sin \theta-2 T=0$

$T =\frac{ Mg \sin \theta}{2}$

Standard 11

Physics

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