A cubicla room is formed with 6 plane mirrors. An insect moves along the diagonal of the floor with uniform speed. The speed of its image in two adjacent walls is `20 sqrt2 cm//s`.
Then the speed of the image formed on the roof is

Options

A. `20 cm//s`
B. 40 cm/s
C. `20 sqrt2 cm//s`
D. `10sqrt2 cm//s`

Answer

To solve the problem step by step, we need to analyze the motion of the insect and how its image is formed in the mirrors of the cubic room.

### Step 1: Understand the setup
The insect moves along the diagonal of the floor of a cubic room formed by six plane mirrors. The speed of the insect is uniform, and we denote this speed as \( V \).

### Step 2: Analyze the speed of the image in adjacent walls
According to the problem, the speed of the insect's image in two adjacent walls is given as \( 20\sqrt{2} \, \text{cm/s} \). Since the walls are perpendicular to each other, we can denote the speed of the image in one wall as \( V_x = 20\sqrt{2} \, \text{cm/s} \) and in the adjacent wall as \( V_y = 20\sqrt{2} \, \text{cm/s} \).

### Step 3: Calculate the resultant speed of the insect
The resultant speed \( V \) of the insect can be calculated using the Pythagorean theorem, since the two speeds are perpendicular:

\[ V = \sqrt{V_x^2 + V_y^2} \]

Substituting the values:

\[ V = \sqrt{(20\sqrt{2})^2 + (20\sqrt{2})^2} \]
\[ V = \sqrt{400 \cdot 2 + 400 \cdot 2} \]
\[ V = \sqrt{800 + 800} = \sqrt{1600} = 40 \, \text{cm/s} \]

### Step 4: Determine the speed of the image on the roof
In a cubic room, the speed of the image formed on the roof is equal to the component of the insect's speed that is perpendicular to the mirror surface (the roof). Since the insect is moving along the diagonal of the floor, the speed of the image on the roof will be the same as the resultant speed of the insect.

Thus, the speed of the image formed on the roof is:

\[ \text{Speed of image on roof} = V = 40 \, \text{cm/s} \]

### Final Answer
The speed of the image formed on the roof is \( 40 \, \text{cm/s} \).

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