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Step-by-Step Solution
Step 1: Identify the Physical Principle
A disc and a solid sphere rolling down an inclined plane experience both translational motion and rotational motion. The time taken by each to roll down depends on their moments of inertia about the axis of rotation.
Step 2: Write Down the Time of Rolling Formula
The time taken by a rolling object (without slipping) to descend an inclined plane of length $l$ and angle of inclination $\theta$ can be expressed as:
$$
t = \sqrt{\frac{2l\left(1 + \frac{k^2}{R^2}\right)}{g \sin \theta}}
$$
where:
$R$ is the radius of the rolling object,
$k$ is the radius of gyration (related to the moment of inertia),
$g$ is the acceleration due to gravity,
$\theta$ is the angle of the incline.
Step 3: Compare the Moments of Inertia of Disc and Sphere
1. Solid Disc: Its moment of inertia about the center is
$$ I_d = \frac{1}{2} M R^2. $$
Hence, its radius of gyration $k_d$ satisfies
$$ k_d^2 = \frac{1}{2} R^2. $$
2. Solid Sphere: Its moment of inertia about the center is
$$ I_s = \frac{2}{5} M R^2. $$
Hence, its radius of gyration $k_s$ satisfies
$$ k_s^2 = \frac{2}{5} R^2. $$
Step 4: Form the Ratio of Times
Since $g$, $l$, $R$, and $\sin \theta$ are the same for both objects, we compare their times by examining the factor
$$\sqrt{1 + \frac{k^2}{R^2}}.$$
We define the ratio of the time taken by the disc ($t_d$) to that taken by the sphere ($t_s$) as follows:
$$
\frac{t_d}{t_s}
= \frac{\sqrt{1 + \frac{k_d^2}{R^2}}}{\sqrt{1 + \frac{k_s^2}{R^2}}}
= \sqrt{\frac{1 + \frac{1}{2}}{1 + \frac{2}{5}}}
= \sqrt{\frac{\frac{3}{2}}{\frac{7}{5}}}
= \sqrt{\frac{3 \times 5}{2 \times 7}}
= \sqrt{\frac{15}{14}}
> 1.
$$
Step 5: Conclude Which Object Arrives First
Since
$$ \frac{t_d}{t_s} > 1, $$
it follows that $t_d > t_s,$ meaning the disc takes longer to reach the bottom than the sphere. Therefore, the sphere reaches the bottom of the plane first.
