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A body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is 0.5 , then its retardation is
( $\mathrm{g}$ - acceleration due to gravity)
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( $\mathrm{g}$ - acceleration due to gravity)
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Verified Answer
The correct answer is:
$\left(\frac{2+\sqrt{3}}{4}\right) \mathrm{g}$
Retardation, $a=g(\sin \theta+\mu \cos \theta)$ $=\mathrm{g}\left[\sin 30^{\circ}+0.5\left(\cos 30^{\circ}\right)\right]$
$=g\left(\frac{1}{2}+\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)=\left(\frac{1}{2}+\frac{\sqrt{3}}{4}\right) \mathrm{g}=\left(\frac{2+\sqrt{3}}{4}\right) \mathrm{g}$
$=g\left(\frac{1}{2}+\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)=\left(\frac{1}{2}+\frac{\sqrt{3}}{4}\right) \mathrm{g}=\left(\frac{2+\sqrt{3}}{4}\right) \mathrm{g}$
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