10,000 small balls, each weighing 1 gm , strike one square cm of area per second with a velocity 100 m/s in a normal direction and rebound with the same velocity. The value of pressure on the surface will be

Explanation

To solve the problem, we need to calculate the pressure exerted on a surface by the small balls striking it and rebounding back. Here's a step-by-step solution:

### Step 1: Understand the Given Data
- Number of balls (N) = 10,000
- Mass of each ball (m) = 1 gram = \(1 \times 10^{-3}\) kg
- Velocity of the balls (v) = 100 m/s
- Area (A) = 1 cm² = \(1 \times 10^{-4}\) m²

### Step 2: Calculate the Total Momentum Change
When the balls strike the surface and rebound, they undergo a change in momentum. The change in momentum (Δp) for one ball is given by:
\[ \Delta p = mv - (-mv) = mv + mv = 2mv \]
For 10,000 balls, the total change in momentum (ΔP) is:
\[ \Delta P = N \times \Delta p = N \times 2mv = 10,000 \times 2 \times (1 \times 10^{-3}) \times 100 \]

### Step 3: Calculate the Total Change in Momentum
Substituting the values:
\[ \Delta P = 10,000 \times 2 \times (1 \times 10^{-3}) \times 100 = 10,000 \times 2 \times 0.001 \times 100 \]
\[ \Delta P = 10,000 \times 0.2 = 2000 \text{ kg m/s} \]

### Step 4: Calculate the Force
The force (F) exerted by the balls on the surface is equal to the rate of change of momentum. Since the balls strike the surface once every second, the force can be calculated as:
\[ F = \frac{\Delta P}{\Delta t} = \frac{2000 \text{ kg m/s}}{1 \text{ s}} = 2000 \text{ N} \]

### Step 5: Calculate the Pressure
Pressure (P) is defined as force per unit area:
\[ P = \frac{F}{A} \]
Substituting the values:
\[ P = \frac{2000 \text{ N}}{1 \times 10^{-4} \text{ m}^2} = 2000 \times 10^{4} \text{ Pa} = 2 \times 10^{7} \text{ Pa} \]

### Final Answer
The pressure exerted on the surface by the balls is:
\[ P = 2 \times 10^{7} \text{ Pa} \]

---