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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A block is placed on the top of a smooth inclined plane of inclination $\theta$ kept on the floor of a lift.When the lift is descending with a retardation $a$,the block is released.The acceleration of the block relative to the incline is

$\begin{array}{1 1}g\sin \theta\\a\sin \theta\\(g-a)\sin \theta\\(g+a)\sin \theta\end{array} $

Can you answer this question?
 
 

1 Answer

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Answer : $(g+a)\sin \theta$
When the lift is descending with a retardation (negative acceleration) $a$,the effective value of $g$ is $g_{eff}=g+a$.
The component of this acceleration along the inclined plane is
$g_{eff}\sin \theta=(g+a)\sin \theta$.
Hence the correct choice is $(g+a)\sin \theta$
answered Aug 27, 2014 by sreemathi.v
 

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