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Step-by-Step Solution
Step 1: Time taken to travel distance on a smooth incline
Consider an incline of angle 30^\circ with the horizontal. When the body slides on a smooth (frictionless) plane, the only component of acceleration is due to gravity along the plane. This acceleration is
g \sin 30^\circ = g \, \frac{1}{2} = \frac{g}{2}.
If the body covers a distance S from rest, the time taken ( T ) can be found from the equation for motion under constant acceleration:
S = \frac{1}{2} \left( \frac{g}{2} \right) T^2.
Solving for T, we get
T = \sqrt{ \frac{2S}{\frac{g}{2}} } = \sqrt{ \frac{4S}{g} } = 2 \,\sqrt{\frac{S}{g}}.
Step 2: Time taken to travel the same distance on a rough incline
When the body slides over the same inclined plane with friction, the net acceleration a_{\text{rough}} is
a_{\text{rough}} = g \sin 30^\circ \;-\; \mu\,g \cos 30^\circ,
where \mu is the coefficient of friction.
Hence,
a_{\text{rough}} = \frac{g}{2} \;-\; \mu\,g \left(\frac{\sqrt{3}}{2}\right) = g \Bigl(\frac{1}{2} - \mu \frac{\sqrt{3}}{2}\Bigr).
For the same distance S from rest, the time taken ( T_{\text{rough}} ) will be
T_{\text{rough}} = \sqrt{ \frac{2S}{a_{\text{rough}}} }
= \sqrt{\frac{2S}{g \bigl(\tfrac{1}{2} - \mu \,\tfrac{\sqrt{3}}{2}\bigr)}}.
Step 3: Relating the two times via the constant \alpha
We are given that T_{\text{rough}} = \alpha T, where \alpha > 1. Thus,
\[
\alpha = \frac{T_{\text{rough}}}{T}
= \sqrt{ \frac{\frac{2S}{g \bigl(\frac{1}{2} - \mu \frac{\sqrt{3}}{2}\bigr)}}{\frac{4S}{g}} }
= \sqrt{ \frac{\frac{1}{2} - \mu \,\frac{\sqrt{3}}{2}}{\frac{1}{2}} } \quad \Bigl(\text{by inverting the fraction correctly}\Bigr).
\]
To be precise, it is often more direct to use the ratio of accelerations:
\[
T_{\text{rough}} = \sqrt{\frac{2S}{g(\sin 30^\circ - \mu \, \cos 30^\circ)}},
\quad
T = \sqrt{\frac{2S}{g\sin 30^\circ}}.
\]
Hence,
\[
\alpha = \frac{T_{\text{rough}}}{T}
= \sqrt{ \frac{\sin 30^\circ}{\sin 30^\circ - \mu \cos 30^\circ} }.
\]
Squaring both sides,
\[
\alpha^2 = \frac{\sin 30^\circ}{\sin 30^\circ - \mu \cos 30^\circ}.
\]
Step 4: Solving for the coefficient of friction \mu
From
\[
\alpha^2 = \frac{\sin 30^\circ}{\sin 30^\circ - \mu \cos 30^\circ},
\]
we rearrange:
\[
\sin 30^\circ - \mu \cos 30^\circ
= \frac{\sin 30^\circ}{\alpha^2}.
\]
Thus,
\[
\mu \cos 30^\circ
= \sin 30^\circ - \frac{\sin 30^\circ}{\alpha^2}
= \sin 30^\circ\Bigl(1 - \frac{1}{\alpha^2}\Bigr).
\]
Therefore,
\[
\mu = \frac{\sin 30^\circ}{\cos 30^\circ} \left(1 - \frac{1}{\alpha^2}\right).
\]
Since \sin 30^\circ = \frac{1}{2} and \cos 30^\circ = \frac{\sqrt{3}}{2},
\[
\mu = \frac{\tfrac{1}{2}}{\tfrac{\sqrt{3}}{2}} \Bigl(1 - \frac{1}{\alpha^2}\Bigr)
= \frac{1}{\sqrt{3}} \Bigl(\frac{\alpha^2 - 1}{\alpha^2}\Bigr).
\]
Step 5: Comparing with the given expression to find x
The problem states \mu = \frac{1}{\sqrt{x}} \Bigl(\frac{\alpha^2 - 1}{\alpha^2}\Bigr).
From our derived expression,
\[
\mu = \frac{1}{\sqrt{3}} \Bigl(\frac{\alpha^2 - 1}{\alpha^2}\Bigr).
\]
By direct comparison,
\[
\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{3}} \quad \Longrightarrow \quad x = 3.
\]
Final Answer
\boxed{x = 3.}
