Two solid spheres P and Q are released from rest from the same point on an inclined plane. Plane is sufficiently rough to provide pure rolling Mass of P is greater than the mass of Q whereas radius of Q is greater than the radius of P

Options

A. Both the spheres will reach the bottom simultaneously
B. Sphere Preaches the bottom first
C. Sphere Q reaches the bottom first
D. K.E. of P will be more than K.E. of Q when they reach the bottom

Explanation

To solve the problem, we need to analyze the motion of two solid spheres, P and Q, released from the same height on an inclined plane. The plane is rough enough to allow for pure rolling. We are given that the mass of sphere P is greater than that of sphere Q, and the radius of sphere Q is greater than that of sphere P. We need to determine two things: which sphere reaches the bottom first and which sphere has greater kinetic energy at the bottom.

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Spheres:**
- Both spheres experience gravitational force acting down the incline, which can be expressed as \( F_{\text{gravity}} = mg \sin \theta \).
- There is also a frictional force acting up the incline, which is necessary for rolling without slipping.

2. **Equations of Motion:**
- For pure rolling motion, the acceleration \( a \) of the center of mass of the sphere is related to the angular acceleration \( \alpha \) by the equation:
\[ a = \alpha r \]
- The net force acting on the sphere along the incline can be expressed as:
\[ mg \sin \theta - F_{\text{friction}} = ma \]

3. **Torque Equation:**
- The torque due to friction about the center of mass is given by:
\[ \tau = F_{\text{friction}} \cdot r = I \cdot \alpha \]
- Where \( I \) is the moment of inertia of the sphere. For a solid sphere, \( I = \frac{2}{5} m r^2 \).

4. **Substituting for Angular Acceleration:**
- We can substitute \( \alpha \) in terms of \( a \):
\[ F_{\text{friction}} \cdot r = \frac{2}{5} m r^2 \cdot \frac{a}{r} \]
- This simplifies to:
\[ F_{\text{friction}} = \frac{2}{5} ma \]

5. **Combining the Equations:**
- Substitute \( F_{\text{friction}} \) back into the net force equation:
\[ mg \sin \theta - \frac{2}{5} ma = ma \]
- Rearranging gives:
\[ mg \sin \theta = ma + \frac{2}{5} ma = \frac{7}{5} ma \]
- Thus, the acceleration \( a \) can be expressed as:
\[ a = \frac{5g \sin \theta}{7} \]

6. **Conclusion on Time to Reach the Bottom:**
- Since the acceleration \( a \) is the same for both spheres, they will reach the bottom of the incline simultaneously.

7. **Kinetic Energy at the Bottom:**
- The potential energy lost by each sphere when descending a height \( h \) is converted into kinetic energy:
\[ mgh = \Delta KE \]
- The final kinetic energy for each sphere is:
\[ KE = mgh \]
- Since mass of sphere P is greater than that of sphere Q, the kinetic energy of sphere P will be greater than that of sphere Q:
\[ KE_P > KE_Q \]

### Final Answers:
- Both spheres will reach the bottom simultaneously.
- The kinetic energy of sphere P will be greater than the kinetic energy of sphere Q when they reach the bottom.