Question

\(10,000\) small balls each weighing \(1\) gram strike \(1\ \text{cm}^{2}\) 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

• A. \(2\times 10^{3}\ \text{N/m}^{2}\)
• B. \(2\times 10^{5}\ \text{N/m}^{2}\)
• C. \(10^{7}\ \text{N/m}^{2}\)
• D. \(2\times 10^{7}\ \text{N/m}^{2}\)

Hint:

Force equals momentum change per second; pressure is force divided by area.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Each ball rebounds with the same speed, so its momentum changes by \(2mv\). The force on the surface is the total momentum change per second.

Step 2: Force
Mass of a ball \(=10^{-3}\) kg, so \(2mv=2\times10^{-3}\times100=0.2\) kg m/s.
\(F=10^{4}\times0.2=2\times10^{3}\) N

Step 3: Pressure
Area \(=1\ \text{cm}^{2}=10^{-4}\ \text{m}^{2}\).
\(P=\dfrac{2\times10^{3}}{10^{-4}}=2\times10^{7}\ \text{N/m}^{2}\)

Final Answer:
The pressure on the surface is 2 times 10 to the power 7 N per square metre. \[ \boxed{\text{(D) }2\times10^{7}\ \text{N/m}^{2}} \]