A disc and a solid sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?

Options

A. Sphere
B. Both reach at the same time
C. Depends on their masses
D. Disc

Answer

To solve the problem of determining which object, a disc or a solid sphere, reaches the bottom of an inclined plane first, we can follow these steps:

### Step 1: Understand the Problem
We have two objects: a disc and a solid sphere, both with the same radius but different masses. They roll down two inclined planes of the same height and length. We need to determine which one reaches the bottom first.

### Step 2: Identify the Relevant Formula
The time \( t \) taken to reach the bottom of the inclined plane for a rolling body can be expressed as:
\[ t = \sqrt{\frac{2s \cdot k}{g \sin \theta}} \]
where:
- \( s \) is the distance along the incline,
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of the incline,
- \( k \) is a constant that depends on the moment of inertia of the object.

### Step 3: Determine the Moment of Inertia
The moment of inertia \( I \) for each object is needed to calculate \( k \):
- For the solid sphere:
\[ I = \frac{2}{5} m r^2 \]
- For the disc:
\[ I = \frac{1}{2} m r^2 \]

### Step 4: Calculate the Value of \( k \)
The constant \( k \) is given by:
\[ k = 1 + \frac{I}{mr^2} \]
Calculating \( k \) for both objects:
- For the solid sphere:
\[ k_{\text{sphere}} = 1 + \frac{\frac{2}{5} m_2 r^2}{m_2 r^2} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4 \]
- For the disc:
\[ k_{\text{disc}} = 1 + \frac{\frac{1}{2} m_1 r^2}{m_1 r^2} = 1 + \frac{1}{2} = \frac{3}{2} = 1.5 \]

### Step 5: Compare the Values of \( k \)
From the calculations:
- \( k_{\text{sphere}} = 1.4 \)
- \( k_{\text{disc}} = 1.5 \)

Since \( k_{\text{sphere}} < k_{\text{disc}} \), it implies that the time taken to reach the bottom is less for the solid sphere.

### Step 6: Conclusion
Since the time taken to reach the bottom is directly proportional to \( \sqrt{k} \), the solid sphere, having a smaller \( k \), will reach the bottom of the inclined plane first.

### Final Answer
The solid sphere will reach the bottom of the inclined plane first.

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