A body is placed on a smooth inclined plane of inclination 1 in x. The horizontal acceleration to be given to the inclined plane so that the body on it remains at rest with respect to inclined plane is

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$\frac{g}{\sqrt{x^{2} - 1}}$

$\frac{g x}{\sqrt{x^{2} - 1}}$

$g \sqrt{x^{2} - 1}$

$\frac{g \sqrt{x^{2} - 1}}{x}$

answer is D.

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macosθ = mgsinθ

a=gtanθ. . . . . . . . .(1)

Given, sinθ= $\frac{1}{x}$

$cosθ = \sqrt{1 - \sin^{2} θ} = \frac{\sqrt{x^{2} - 1 }}{x} $

$tanθ = \frac{1}{\sqrt{x^{2} - 1}}$ . . . . . .(2)

Put equation (2) in equation (1), and we get

$a = \frac{g}{\sqrt{x^{2} - 1}}$

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