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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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In the arrangement shown by what acceleration the boy must go up so that 100 kg block remains stationary on the wedge. The wedge is fixed and friction is absent, every where $(g=10 m/s^2)$


\[(a)\;2 m/s^2 \quad (b)\; 6m/s^2 \quad (c)\;4 m/s^2 \quad(d)\;8 m/s^2 \]

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For the block to be stationary
$T=mg \sin \theta$
$\quad=100 \times 10 \sin 53^{\circ}$
$\quad=800 N$
$800-500=50 a$
$a=6 m/s^2$
hence b is the correct answer.


answered Jul 9, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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