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In the above diagram all surface friction less what horizontal force has to be applied on wedge such that in equilibrium steady state spring is compressed by $\frac{m g \sin θ}{K}$

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$2 m g \sin θ$

$2 m g \tan θ$

$4 m g \tan θ$

answer is C.

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Detailed Solution

applying newtons law among incline

$kx + mgsin θ = m a \cos θ$
$2 mgsin θ = \frac{F}{2} \cos θ; 4 mgtan θ = F$

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