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Step-by-Step Solution
Step 1: Understanding the two motions (slipping vs. rolling)
When a circular disc is released from the top of an inclined plane of length L , it can either slip down without rolling or roll down without slipping. The time taken in each scenario is different because the accelerations are different.
Step 2: Time taken when the disc slips
For slipping (pure translation with no rolling), the acceleration down the plane is:
a_1 = g \sin \theta.
Since the disc starts from rest and travels a distance s = L (the length of the inclined plane), we use the equation of motion:
s = \frac{1}{2} a_1 t_1^2 \quad \Longrightarrow \quad L = \frac{1}{2} g \sin \theta \, t_1^2.
Hence,
t_1 = \sqrt{\frac{2L}{g \sin \theta}}.
Step 3: Time taken when the disc rolls
For rolling without slipping, the acceleration a_2 of the disc down the plane is given by:
a_2 = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}},
where K is the radius of gyration and R is the radius of the disc. For a solid disc, \frac{K^2}{R^2} = \frac{1}{2}. Therefore,
a_2 = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2}{3} g \sin \theta.
Again, using s = \frac{1}{2} a_2 t_2^2 with s = L , we get:
L = \frac{1}{2} \left(\frac{2}{3} g \sin \theta\right) t_2^2 = \frac{1}{3} g \sin \theta \, t_2^2.
Hence,
t_2 = \sqrt{\frac{3L}{g \sin \theta}}.
Step 4: Forming the ratio of times
Divide t_2 by t_1 :
\frac{t_2}{t_1} = \frac{\sqrt{\frac{3L}{g \sin \theta}}}{\sqrt{\frac{2L}{g \sin \theta}}}
= \sqrt{\frac{\frac{3L}{g \sin \theta}}{\frac{2L}{g \sin \theta}}}
= \sqrt{\frac{3}{2}}.
Step 5: Identifying the value of x
According to the question, \frac{t_2}{t_1} = \sqrt{\frac{3}{x}}. From our calculation, \frac{t_2}{t_1} = \sqrt{\frac{3}{2}}. By comparison:
\sqrt{\frac{3}{x}} = \sqrt{\frac{3}{2}}
\quad \Longrightarrow \quad
\frac{3}{x} = \frac{3}{2}
\quad \Longrightarrow \quad
x = 2.
