A solid sphere rolls down two different inclined planes of the same height but of different inclinations

Options

A. in ot cases the speeds and time of descend will be same
B. the speeds will be same but time of descend will be different
C. the speeds will be different but time of descend will be same
D. speeds and time of descend both will be different.

Explanation

To solve the problem of a solid sphere rolling down two different inclined planes of the same height but with different inclinations, we can follow these steps:

### Step 1: Understanding the Forces Acting on the Sphere
When the solid sphere rolls down the inclined planes, the forces acting on it include:
- Gravitational force (mg) acting downward.
- Normal force (N) acting perpendicular to the surface of the incline.
- Frictional force (f) acting up the incline.

For each incline, we can resolve the gravitational force into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

### Step 2: Setting Up the Equations of Motion
For translational motion along the incline:
\[ mg \sin \theta - f = ma \]
For rotational motion about the center of mass:
\[ f \cdot r = I \cdot \alpha \]
Where \( I \) is the moment of inertia of the sphere, and \( \alpha \) is the angular acceleration. For a solid sphere, \( I = \frac{2}{5}mr^2 \) and \( \alpha = \frac{a}{r} \).

### Step 3: Relating Translational and Rotational Motion
Substituting \( \alpha \) into the rotational motion equation gives:
\[ f \cdot r = \frac{2}{5}mr^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ f = \frac{2}{5}ma \]

### Step 4: Substituting the Friction Force
Substituting \( f \) back into the translational motion equation:
\[ mg \sin \theta - \frac{2}{5}ma = ma \]
Combining terms:
\[ mg \sin \theta = ma + \frac{2}{5}ma \]
\[ mg \sin \theta = \frac{7}{5}ma \]

### Step 5: Finding the Acceleration
From the above equation, we can solve for acceleration \( a \):
\[ a = \frac{5}{7}g \sin \theta \]

### Step 6: Finding the Speed at the Bottom of the Incline
Using the kinematic equation \( v^2 = u^2 + 2as \) (where \( u = 0 \)):
\[ v^2 = 0 + 2aS \]
The distance \( S \) along the incline can be expressed in terms of height \( h \) and angle \( \theta \):
\[ S = \frac{h}{\sin \theta} \]
Thus,
\[ v^2 = 2 \left(\frac{5}{7}g \sin \theta\right) \left(\frac{h}{\sin \theta}\right) \]
This simplifies to:
\[ v^2 = \frac{10h}{7}g \]
Taking the square root gives:
\[ v = \sqrt{\frac{10h}{7}g} \]

### Step 7: Finding the Time of Descent
Using the equation \( s = ut + \frac{1}{2}at^2 \):
\[ \frac{h}{\sin \theta} = 0 + \frac{1}{2} \left(\frac{5}{7}g \sin \theta\right) t^2 \]
This leads to:
\[ t^2 = \frac{2h}{\frac{5}{7}g \sin \theta} \]
Thus,
\[ t = \sqrt{\frac{14h}{5g \sin \theta}} \]

### Conclusion
1. **Speeds**: The speed at the bottom of both inclines is the same, given by \( v = \sqrt{\frac{10h}{7}g} \).
2. **Times**: The time taken to roll down each incline is different, as it depends on the angle \( \theta \).

### Final Answer
The correct option is: **The speeds will be the same but the times of descent will be different.**

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