A solid sphere and a solid cylinder of same mass are rolled down on two inclined planes of heights `h_(1)` and `h_(2)` respectively. If at the bottom of the plane the two objects have same linear velocities, then the ratio of `h_(1):h_(2)` is

Options

A. `2:3`
B. `7:5`
C. `14:15`
D. `15:14`

Answer

To solve the problem, we need to find the ratio of the heights \( h_1 \) and \( h_2 \) for a solid sphere and a solid cylinder, respectively, when they are rolled down inclined planes and reach the bottom with the same linear velocities.

### Step-by-Step Solution:

1. **Identify the formulas for linear velocity**:
For a rolling object, the linear velocity \( v \) can be expressed as:
\[ v = \sqrt{\frac{2gh}{1 + \frac{k^2}{r^2}}} \]
where \( h \) is the height of the incline, \( k \) is the radius of gyration, and \( r \) is the radius of the object.

2. **Determine the radius of gyration for each object**:
- For a solid sphere, the radius of gyration \( k \) is given by:
\[ k^2 = \frac{2}{5} r^2 \]
- For a solid cylinder, the radius of gyration \( k \) is given by:
\[ k^2 = \frac{1}{2} r^2 \]

3. **Calculate the linear velocity for the solid sphere**:
Substituting \( k^2 \) for the sphere into the velocity formula:
\[ v_{\text{sphere}} = \sqrt{\frac{2gh_1}{1 + \frac{2/5 r^2}{r^2}}} = \sqrt{\frac{2gh_1}{1 + \frac{2}{5}}} = \sqrt{\frac{2gh_1}{\frac{7}{5}}} = \sqrt{\frac{10gh_1}{7}} \]

4. **Calculate the linear velocity for the solid cylinder**:
Substituting \( k^2 \) for the cylinder into the velocity formula:
\[ v_{\text{cylinder}} = \sqrt{\frac{2gh_2}{1 + \frac{1/2 r^2}{r^2}}} = \sqrt{\frac{2gh_2}{1 + \frac{1}{2}}} = \sqrt{\frac{2gh_2}{\frac{3}{2}}} = \sqrt{\frac{4gh_2}{3}} \]

5. **Set the velocities equal to each other**:
Since both objects have the same linear velocity at the bottom of the incline:
\[ \sqrt{\frac{10gh_1}{7}} = \sqrt{\frac{4gh_2}{3}} \]

6. **Square both sides to eliminate the square roots**:
\[ \frac{10gh_1}{7} = \frac{4gh_2}{3} \]

7. **Cancel \( g \) from both sides**:
\[ \frac{10h_1}{7} = \frac{4h_2}{3} \]

8. **Cross-multiply to solve for the ratio \( h_1:h_2 \)**:
\[ 10h_1 \cdot 3 = 4h_2 \cdot 7 \]
\[ 30h_1 = 28h_2 \]

9. **Divide both sides by \( h_2 \) and rearrange**:
\[ \frac{h_1}{h_2} = \frac{28}{30} = \frac{14}{15} \]

### Final Answer:
The ratio of the heights \( h_1:h_2 \) is \( 14:15 \).