A room (cubical) is made of mirros. An insect is moving along the diagonal on the floor such that the velocity of image of insect on two adjacent wall mirrors is `10(cm)/(sec)`. The velocity of image of insect in ceiling mirror is

Options

A. `10 cm s^(-1)`
B. `20 cm s^(-1)`
C. `10/sqrt2cm s^(-1)`
D. `10sqrt (2) cm s^(-1)`

Answer

To solve the problem, we need to determine the velocity of the image of the insect in the ceiling mirror, given that the velocity of the image on the two adjacent wall mirrors is 10 cm/s.

### Step-by-step Solution:

1. **Understanding the Movement of the Insect:**
The insect is moving along the diagonal of the floor of a cubical room made of mirrors. This means it has a certain velocity \( V_0 \) in the diagonal direction.

2. **Velocity Components:**
Since the insect is moving along the diagonal, we can break its velocity into components. The angle between the diagonal and each wall is 45 degrees. Therefore, the components of the insect's velocity along the two adjacent walls (let's call them wall 1 and wall 2) can be expressed as:
\[ V_{1} = V_{0} \cos(45^\circ) = \frac{V_{0}}{\sqrt{2}} \]
\[ V_{2} = V_{0} \cos(45^\circ) = \frac{V_{0}}{\sqrt{2}} \]

3. **Given Information:**
We are given that the velocity of the image of the insect on the two adjacent walls is 10 cm/s. Therefore:
\[ \frac{V_{0}}{\sqrt{2}} = 10 \text{ cm/s} \]

4. **Finding \( V_0 \):**
To find \( V_0 \), we can rearrange the equation:
\[ V_{0} = 10 \sqrt{2} \text{ cm/s} \]

5. **Velocity of the Image in the Ceiling Mirror:**
The velocity of the image in the ceiling mirror is equal to the actual velocity of the insect, which is \( V_0 \). Thus:
\[ V_{\text{ceiling}} = V_0 = 10 \sqrt{2} \text{ cm/s} \]

### Final Answer:
The velocity of the image of the insect in the ceiling mirror is \( 10\sqrt{2} \) cm/s.

---