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(A) For a smooth inclined plane:The acceleration is $a_1 = g \sin 30^{\circ} = \frac{g}{2}$.Using the equation of motion $S = ut + \frac{1}{2}at^2$,wi

Vedclass · Accepted Answer

$3$

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(A) For a smooth inclined plane:The acceleration is $a_1 = g \sin 30^{\circ} = \frac{g}{2}$.Using the equation of motion $S = ut + \frac{1}{2}at^2$,with $u = 0$:$S = \frac{1}{2} \left(\frac{g}{2}\right) T^2 = \frac{g T^2}{4} \quad \dots (i)$For a rough inclined plane:The acceleration is $a_2 = g \sin 30^{\circ} - \mu g \cos 30^{\circ} = g(\sin 30^{\circ} - \mu \cos 30^{\circ}) = g\left(\frac{1}{2} - \mu \frac{\sqrt{3}}{2}\right) = \frac{g}{2}(1 - \sqrt{3}\mu)$.Using the same distance $S$ and time $\alpha T$:$S = \frac{1}{2} \left[\frac{g}{2}(1 - \sqrt{3}\mu)\right] (\alpha T)^2 = \frac{g}{4}(1 - \sqrt{3}\mu) \alpha^2 T^2 \quad \dots (ii)$Equating $(i)$ and $(ii)$:$\frac{g T^2}{4} = \frac{g}{4}(1 - \sqrt{3}\mu) \alpha^2 T^2$$1 = (1 - \sqrt{3}\mu) \alpha^2$$\frac{1}{\alpha^2} = 1 - \sqrt{3}\mu$$\sqrt{3}\mu = 1 - \frac{1}{\alpha^2} = \frac{\alpha^2 - 1}{\alpha^2}$$\mu = \frac{1}{\sqrt{3}} \left(\frac{\alpha^2 - 1}{\alpha^2}\right)$Comparing this with the given expression $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$,we get $x = 3$.

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