P is a solid sphere and Q is a hollow sphere both having the same mass and radius .If they roll down from the top of an inclined plane , on reaching the bottom .

Options

A. Velocity of P is more
B. Velocity of Q is more
C. Velocity of P = velocity of Q
D. None

Answer

To solve the problem of comparing the velocities of a solid sphere (P) and a hollow sphere (Q) when they roll down an inclined plane, we can follow these steps:

### Step 1: Understand the Moment of Inertia
The moment of inertia (I) is a measure of an object's resistance to changes in its rotation. For the two spheres:
- For the solid sphere (P), the moment of inertia is given by:
\[ I_P = \frac{2}{5} MR^2 \]
- For the hollow sphere (Q), the moment of inertia is:
\[ I_Q = \frac{2}{3} MR^2 \]

### Step 2: Apply Newton's Second Law
When the spheres roll down the incline, we can apply Newton's second law. The forces acting on the spheres are:
- The gravitational force component along the incline: \( mg \sin \theta \)
- The frictional force (which provides the torque for rotation)

For the solid sphere (P):
\[ mg \sin \theta - f = Ma \]
For the hollow sphere (Q):
\[ mg \sin \theta - f = Ma \]

### Step 3: Relate Linear Acceleration and Angular Acceleration
In rolling motion, the linear acceleration (a) is related to the angular acceleration (\(\alpha\)) by:
\[ \alpha = \frac{a}{R} \]
The torque (\(\tau\)) due to friction (f) can be expressed as:
\[ \tau = fR = I \alpha \]

### Step 4: Substitute Moment of Inertia
Substituting the moment of inertia into the torque equation for both spheres:
- For the solid sphere:
\[ fR = \frac{2}{5} MR^2 \cdot \frac{a}{R} \]
Simplifying gives:
\[ f = \frac{2}{5} Ma \]

- For the hollow sphere:
\[ fR = \frac{2}{3} MR^2 \cdot \frac{a}{R} \]
Simplifying gives:
\[ f = \frac{2}{3} Ma \]

### Step 5: Substitute Friction into the Force Equation
Now substitute the expression for friction back into the force equations:
- For the solid sphere:
\[ mg \sin \theta - \frac{2}{5} Ma = Ma \]
Rearranging gives:
\[ mg \sin \theta = \left(1 + \frac{2}{5}\right) Ma = \frac{7}{5} Ma \]
Thus, the acceleration \(a_P\) for the solid sphere is:
\[ a_P = \frac{5g \sin \theta}{7} \]

- For the hollow sphere:
\[ mg \sin \theta - \frac{2}{3} Ma = Ma \]
Rearranging gives:
\[ mg \sin \theta = \left(1 + \frac{2}{3}\right) Ma = \frac{5}{3} Ma \]
Thus, the acceleration \(a_Q\) for the hollow sphere is:
\[ a_Q = \frac{3g \sin \theta}{5} \]

### Step 6: Compare Accelerations
Now we can compare the accelerations:
- \(a_P = \frac{5g \sin \theta}{7}\)
- \(a_Q = \frac{3g \sin \theta}{5}\)

To compare, we can find a common denominator:
- \(a_P = \frac{5g \sin \theta}{7}\)
- \(a_Q = \frac{21g \sin \theta}{35}\)

Since \(5/7 < 21/35\), we find that:
\[ a_P > a_Q \]

### Step 7: Conclusion on Velocities
Since the solid sphere has greater acceleration than the hollow sphere, it will reach the bottom of the incline with a higher velocity. Therefore, we conclude:
\[ V_P > V_Q \]

### Final Answer
The velocity of the solid sphere (P) is greater than the velocity of the hollow sphere (Q) when they reach the bottom of the incline.

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