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A particle slides down from rest on an inclined plane of angle ' $θ$ ' with horizontal and the distances are as shown. The particle slides down from top to the position ' $A$ '. Where it’s velocity is ' $V$ '. Then below correct options are.

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$(V^{2} - 2 g H)$ will remain constant

$(V^{2} - 2 g S \sin θ)$ will remain constant

$[V^{2} - \frac{2 g S (H - h)}{P - S}]$ will remain constant

$[V^{2} - \frac{2 g S H}{P}]$ will remain constant

answer is B, C, D.

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Detailed Solution

from fig, $h = S (\sin θ)$

$\begin{array}{l} \frac{1}{2} {mV}^{2} + mg (H - h) = mgH \\ \frac{1}{2} {mV}^{2} - mgh = 0 \\ V^{2} = 2 gh, V^{2} = 2 gSsin θ \end{array}$

From fig, $\sin θ = \frac{H}{P} = \frac{(H - h)}{(P - S)}$

$\begin{aligned} V^{2} = 2 gS [\frac{H - h}{P - S}] \\ V^{2} - 2 gS (\frac{H}{P}) = 0 \end{aligned}$

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