An object is placed on a smooth inclined plane of 1 in $l$ . The horizontal acceleration to be imparted to inclined plane so that, the object is stationary relative to the inclined is given by :

\frac{g}{l^{2} - 1}

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\frac{g}{\sqrt{l^{2} - 1}}

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g \sqrt{l^{2} - 1}

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g (l^{2} - 1)

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Solution

The correct option is D $\frac{g}{\sqrt{l^{2} - 1}}$
Consider the diagram
Here, $sin θ = \frac{1}{l} \Rightarrow tan θ = \frac{1}{\sqrt{l^{2} - 1}}$
From free body diagram, we have
$m a cos θ = m g sin θ$
$a = g tan θ = g \frac{1}{\sqrt{l^{2} - 1}}$

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An object is placed on a smooth inclined plane of 1 in l. The horizontal acceleration to be imparted to inclined plane so that, the object is stationary relative to the inclined is given by :

The correct option is D g√l2−1Consider the diagramHere, sinθ=1l⇒tanθ=1√l2−1From free body diagram, we havemacosθ=mgsinθa=gtanθ=g1√l2−1

An object is placed on a smooth inclined plane of 1 in $l$ . The horizontal acceleration to be imparted to inclined plane so that, the object is stationary relative to the inclined is given by :

The correct option is D $\frac{g}{\sqrt{l^{2} - 1}}$
Consider the diagram
Here, $sin θ = \frac{1}{l} \Rightarrow tan θ = \frac{1}{\sqrt{l^{2} - 1}}$
From free body diagram, we have
$m a cos θ = m g sin θ$
$a = g tan θ = g \frac{1}{\sqrt{l^{2} - 1}}$