A soap bubble is blown with the help of mechanical pump at the mouth of a tube. The pump produces a cartain increase per minute in the volume of the bubble, irrespective of its internal pressure. The graph between the pressure inside the soap bubble and time `t` will be

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Explanation

To solve the problem, we need to analyze the relationship between the pressure inside a soap bubble and time as the bubble is inflated by a mechanical pump. Here’s a step-by-step solution:

### Step 1: Understand the Pressure Inside the Soap Bubble
The pressure inside a soap bubble can be expressed using the formula:
\[ P = P_0 + \frac{4T}{R} \]
where:
- \( P \) is the pressure inside the bubble,
- \( P_0 \) is the atmospheric pressure,
- \( T \) is the surface tension of the soap solution,
- \( R \) is the radius of the bubble.

### Step 2: Relate Volume and Time
The volume \( V \) of a soap bubble is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
According to the problem, the volume of the bubble increases at a constant rate per minute, which means:
\[ V \propto t \]
Thus, we can write:
\[ V = k \cdot t \]
for some constant \( k \).

### Step 3: Express Radius in Terms of Time
From the volume equation, we can express the radius \( R \) in terms of time \( t \):
\[ \frac{4}{3} \pi R^3 = k \cdot t \]
This implies:
\[ R^3 = \frac{3k}{4\pi} t \]
Taking the cube root:
\[ R = \left( \frac{3k}{4\pi} \right)^{1/3} t^{1/3} \]
Let’s denote the constant \( \left( \frac{3k}{4\pi} \right)^{1/3} \) as \( C \):
\[ R = C \cdot t^{1/3} \]

### Step 4: Substitute Radius Back into the Pressure Equation
Now, we substitute \( R \) back into the pressure equation:
\[ P = P_0 + \frac{4T}{C \cdot t^{1/3}} \]
This can be rewritten as:
\[ P = P_0 + \frac{4T}{C} t^{-1/3} \]

### Step 5: Analyze the Relationship
From the equation \( P = P_0 + \frac{4T}{C} t^{-1/3} \), we can see that as time \( t \) increases, the term \( t^{-1/3} \) decreases. Therefore, the pressure \( P \) will approach \( P_0 \) as \( t \) increases.

### Step 6: Graphical Representation
The graph of \( P \) versus \( t \) will start from \( P_0 \) and will decrease as \( t \) increases. The relationship is inversely proportional to \( t^{1/3} \), which means the graph will have a decreasing curve.

### Conclusion
The correct graph between the pressure inside the soap bubble and time \( t \) will show a decreasing trend starting from \( P_0 \) and approaching \( P_0 \) as time increases.