A man standing in front of a mountain beats a drum at regular intervals. The drumming rate is gradually increased and he finds that echo is not heard distinctly when the rate becomes `40` per minute. He then moves near to the mountain by `90` metres and finds that echo is again not heard distinctly when the drumming rate becomes `60` per minute. Calculate (a) the distance between the mountain and the initial position of the man and (b) the velocity of sound.

Options

A. i.`330 m`
ii. `330 m//s`
B. i. `300 m`
ii. `720 m//s`
C. i. `240 m`
ii. `300 m//s`
D. i. `270 m`
ii. 270 m//s`

Explanation

To solve the problem step by step, we will break down the information given and apply the principles of sound and echo.

### Step 1: Understand the Problem
The man beats a drum at increasing rates and hears echoes at two different rates. We need to find the distance from the mountain to the man's initial position and the speed of sound.

### Step 2: Define Variables
Let:
- \( d \) = distance from the mountain to the man's initial position (in meters)
- \( V \) = velocity of sound (in meters per second)

### Step 3: Analyze the First Situation
When the drum rate is 40 beats per minute:
- The time for one beat = \( \frac{60}{40} = 1.5 \) seconds.
- The sound travels to the mountain and back, so the total distance traveled by sound is \( 2d \).
- The equation relating speed, distance, and time is:
\[ V = \frac{2d}{1.5} \]
Simplifying gives:
\[ V = \frac{4d}{3} \]

### Step 4: Analyze the Second Situation
After moving 90 meters closer to the mountain, the man beats the drum at 60 beats per minute:
- The time for one beat = \( \frac{60}{60} = 1 \) second.
- The new distance to the mountain is \( d - 90 \).
- The equation for this situation is:
\[ V = \frac{2(d - 90)}{1} \]
Simplifying gives:
\[ V = 2(d - 90) = 2d - 180 \]

### Step 5: Set the Equations Equal
Now we have two expressions for \( V \):
1. \( V = \frac{4d}{3} \)
2. \( V = 2d - 180 \)

Setting them equal to each other:
\[ \frac{4d}{3} = 2d - 180 \]

### Step 6: Solve for \( d \)
To eliminate the fraction, multiply the entire equation by 3:
\[ 4d = 6d - 540 \]
Rearranging gives:
\[ 540 = 6d - 4d \]
\[ 540 = 2d \]
\[ d = 270 \text{ meters} \]

### Step 7: Calculate the Velocity of Sound
Now substitute \( d = 270 \) back into one of the equations for \( V \). Using \( V = \frac{4d}{3} \):
\[ V = \frac{4 \times 270}{3} = \frac{1080}{3} = 360 \text{ meters per second} \]

### Final Answers
(a) The distance between the mountain and the initial position of the man is **270 meters**.
(b) The velocity of sound is **360 meters per second**.

---