10,000 small balls, each weighing \(1 \mathrm{~g}\) strike one square \(\mathrm{cm}\) of area per second with a velocity \(100 \mathrm{~ms}^{-1}\) in a normal direction and rebound with the same velocity. The value of pressure on the surface will be (A) \(2 \times 10^{3} \mathrm{Nm}^{-2}\) (B) \(2 \times 10^{5} \mathrm{Nm}^{-2}\) (C) \(10^{7} \mathrm{Nm}^{-2}\) (D) \(2 \times 10^{7} \mathrm{Nm}^{2}\)

To find the force exerted by a single ball, we can use Newton's second law (F = ma). First, find the acceleration of the ball as it rebounds. The acceleration can be found using the formula: \(a = \frac{v^2 - u^2}{2s}\) where v is the final velocity, u is the initial velocity, and s is the distance. Since the balls are rebounding with the same velocity, the initial velocity is -100 m/s, and the final velocity is 100 m/s. \(a = \frac{(100 \mathrm{~ms}^{-1})^2 - (-100 \mathrm{~ms}^{-1})^2}{2 \times 0.01 \mathrm{~m}}\) \(a = \frac{(10000 - 10000) \mathrm{m^2s^{-2}}}{0.02 \mathrm{~m}}\) \(a = 10000 \times 10^3 \mathrm{ms}^{-2}\) Now, find the force exerted by a single ball using F = ma: \(F = m \times a\) \(F = (1 \times 10^{-3} \mathrm{~kg}) \times (10000 \times 10^3 \mathrm{ms}^{-2})\) \(F = 10 \mathrm{N}\)

Since there are 10000 balls hitting the surface per second, we can find the total force by multiplying the force exerted by a single ball by the number of balls: Total Force = Force exerted by a single ball × number of balls Total Force = 10 N × 10000 Total Force = 100000 N

To find the pressure exerted on the surface, we can use the formula for pressure: Pressure = \(\frac{\text{Total Force}}{\text{Area}}\) Pressure = \(\frac{100000 \mathrm{N}}{0.0001 \mathrm{m^2}}\) Pressure = \(2 \times 10^5 \mathrm{Nm}^{-2}\)

Comparing the calculated pressure with the provided options, we find that our answer matches option (B): Pressure = \(2 \times 10^5 \mathrm{Nm}^{-2}\)