Two dises having masses in the ratio `1 : 2` and radii in the ratio `1 : 8` roll down without slipping one by one from an inclined plane of height h. The ratio of their linear velocities on reaching the ground is :-

Options

A. `1 : 16`
B. `1:128`
C. `1:8sqrt(2)`
D. `1:1`

Answer

To solve the problem, we need to find the ratio of the linear velocities of two discs rolling down an inclined plane. Let's denote the two discs as Disc 1 and Disc 2.

### Step 1: Define the masses and radii of the discs
Let the mass of Disc 1 be \( m \) and the mass of Disc 2 be \( 2m \). The radius of Disc 1 is \( R \) and the radius of Disc 2 is \( 8R \).

### Step 2: Write the energy conservation equation for Disc 1
The potential energy lost by Disc 1 when it rolls down the height \( h \) is equal to the kinetic energy gained. The total kinetic energy consists of translational kinetic energy and rotational kinetic energy.

\[ mgh = \frac{1}{2} mv_1^2 + \frac{1}{2} I_1 \omega_1^2 \]

For a disc, the moment of inertia \( I \) is given by:

\[ I_1 = \frac{1}{2} m R^2 \]

Since \( \omega_1 = \frac{v_1}{R} \), we substitute \( \omega_1 \) into the equation:

\[ mgh = \frac{1}{2} mv_1^2 + \frac{1}{2} \left(\frac{1}{2} m R^2\right) \left(\frac{v_1^2}{R^2}\right) \]

This simplifies to:

\[ mgh = \frac{1}{2} mv_1^2 + \frac{1}{4} mv_1^2 \]

Combining the terms gives:

\[ mgh = \frac{3}{4} mv_1^2 \]

### Step 3: Solve for \( v_1 \)
Cancelling \( m \) from both sides and solving for \( v_1 \):

\[ gh = \frac{3}{4} v_1^2 \]

\[ v_1^2 = \frac{4gh}{3} \]

\[ v_1 = \sqrt{\frac{4gh}{3}} \]

### Step 4: Write the energy conservation equation for Disc 2
For Disc 2, the potential energy lost is:

\[ 2mgh = \frac{1}{2} (2m) v_2^2 + \frac{1}{2} I_2 \omega_2^2 \]

Where \( I_2 \) is:

\[ I_2 = \frac{1}{2} (2m) (8R)^2 = \frac{1}{2} (2m) (64R^2) = 64mR^2 \]

Substituting \( \omega_2 = \frac{v_2}{8R} \):

\[ 2mgh = mv_2^2 + \frac{1}{2} (64mR^2) \left(\frac{v_2^2}{(8R)^2}\right) \]

This simplifies to:

\[ 2mgh = mv_2^2 + \frac{1}{2} (64mR^2) \left(\frac{v_2^2}{64R^2}\right) \]

Which simplifies further to:

\[ 2mgh = mv_2^2 + mv_2^2 \]

Thus, we have:

\[ 2mgh = 2mv_2^2 \]

### Step 5: Solve for \( v_2 \)
Cancelling \( 2m \) from both sides gives:

\[ gh = v_2^2 \]

\[ v_2 = \sqrt{gh} \]

### Step 6: Find the ratio of the velocities
Now we can find the ratio of the linear velocities \( v_1 \) and \( v_2 \):

\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{4gh}{3}}}{\sqrt{gh}} = \frac{\sqrt{4/3}}{1} = \frac{2}{\sqrt{3}} \]

### Final Answer
The ratio of their linear velocities on reaching the ground is:

\[ \frac{v_1}{v_2} = \frac{2}{\sqrt{3}} \quad \text{or approximately } 1.155 \]