Two spheres of same mass and radius are in contact with each other. If the moment of inertia of a sphere about its diameter is I, then the moment of inertia of both the spheres about the tangent at their common point would be

Options

A. 3I
B. 7I
C. 4I
D. 5I

Explanation

To solve the problem of finding the moment of inertia of two spheres about the tangent at their common point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Moment of Inertia of a Sphere**:
The moment of inertia of a single sphere about its diameter is given as \( I = \frac{2}{5} m r^2 \).

2. **Apply the Parallel Axis Theorem**:
To find the moment of inertia about a tangent at the common point of the two spheres, we will use the parallel axis theorem. The theorem states that if you know the moment of inertia about an axis through the center of mass, you can find the moment of inertia about any parallel axis by adding \( md^2 \), where \( d \) is the distance between the two axes.

3. **Calculate the Distance**:
The distance \( d \) from the center of the sphere to the tangent at the point of contact is equal to the radius \( r \) of the sphere.

4. **Moment of Inertia for One Sphere about the Tangent**:
Using the parallel axis theorem, the moment of inertia of one sphere about the tangent is:
\[ I_{\text{tangent}} = I + m r^2 = \frac{2}{5} m r^2 + m r^2 = \frac{2}{5} m r^2 + \frac{5}{5} m r^2 = \frac{7}{5} m r^2 \]

5. **Calculate for Two Spheres**:
Since there are two identical spheres, the total moment of inertia about the tangent at their common point will be:
\[ I_{\text{total}} = 2 \times I_{\text{tangent}} = 2 \times \frac{7}{5} m r^2 = \frac{14}{5} m r^2 \]

6. **Express in Terms of \( I \)**:
We know that \( I = \frac{2}{5} m r^2 \). Therefore, we can express \( \frac{14}{5} m r^2 \) in terms of \( I \):
\[ I_{\text{total}} = 7 \times \frac{2}{5} m r^2 = 7I \]

### Final Answer:
The moment of inertia of both spheres about the tangent at their common point is \( 7I \).