Q168 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
$10{,}000$ 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$10^7\ \text{N/m}^2$
- Correct answerD$2 \times 10^7\ \text{N/m}^2$
Solution
Change in momentum of each ball upon rebounding is $\Delta p = 2mv$.
Force exerted on the surface is the total change in momentum per second, $F = n \times 2mv$.
Substituting the given values, $n = 10^4\ \text{s}^{-1}$, $m = 10^{-3}\ \text{kg}$, and $v = 100\ \text{m/s}$:
$F = 10^4 \times 2 \times 10^{-3} \times 100 = 2 \times 10^3\ \text{N}$
Pressure is force per unit area, $P = \dfrac{F}{A}$.
Given area $A = 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$.
$P = \dfrac{2 \times 10^3}{10^{-4}} = 2 \times 10^7\ \text{N/m}^2$
Answer: $2 \times 10^7\ \text{N/m}^2$
Force exerted on the surface is the total change in momentum per second, $F = n \times 2mv$.
Substituting the given values, $n = 10^4\ \text{s}^{-1}$, $m = 10^{-3}\ \text{kg}$, and $v = 100\ \text{m/s}$:
$F = 10^4 \times 2 \times 10^{-3} \times 100 = 2 \times 10^3\ \text{N}$
Pressure is force per unit area, $P = \dfrac{F}{A}$.
Given area $A = 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$.
$P = \dfrac{2 \times 10^3}{10^{-4}} = 2 \times 10^7\ \text{N/m}^2$
Answer: $2 \times 10^7\ \text{N/m}^2$
Hence, the correct answer is option: 'D'.