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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Two blocks of masses $m_1=m$ and $m_2=2m$ are connected by a light string passing over a frictionless pulley.The mass $m_1$ is placed on a smooth inclined plane of inclination $\theta=30^{\large\circ}$ and mass $m_2$ hangs vertically as shown in figure below.If the system is released the tension in the string is

$\begin{array}{1 1}mg\\\large\frac{3mg}{2}\\2mg\\\large\frac{2mg}{3}\end{array} $

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Answer : $mg$
Since the inclined plane is smooth and $m_2 > m_1$,block $m_1$ will up the plane and block $m_2$ will move vertically with a common acceleration $a$.If $T$ is the tension in the string,the free-body diagrams of masses $m_1$ and $m_2$ are as shown in figure.
The equations of motion of the blocks are
$T-m_1g\sin \theta=m_1a$-------(1)
$m_2g-T=m_2a$--------(2)
From equations (1) and (2),we get
$T= m_2(g-a)=2m\times (g-\large\frac{g}{2})$$=mg$
answered Aug 28, 2014 by sreemathi.v
 

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