Variable Mass And Impulsive Force

The force on the wall equals the rate at which the water's momentum changes as it is stopped, a direct application of Newton's Second Law in its momentum form $F=\frac{dp}{dt}$. Each second a mass $\frac{dm}{dt}=10\text{ kg}$ arrives carrying speed $v=15\text{ m/s}$, and since the water does not rebound, its forward velocity drops from 15 m/s to zero. The momentum destroyed per second is $\frac{dm}{dt}\cdot v=10\times 15=150\text{ N}$, which is the average force the wall exerts on the water and, by the Third Law, the water exerts on the wall. The 75 N value halves this, perhaps by averaging the velocity incorrectly. The 300 N value would apply if the water rebounded elastically, doubling the momentum change, but the problem states it does not bounce. The 100 N value has no consistent basis. The deeper idea is that Newton's Second Law in its original momentum form $F=\frac{dp}{dt}$ handles a continuous stream more naturally than the special case $F=ma$, since here it is the mass flow rate, not a single accelerating body, that carries the momentum. This same reasoning underlies rocket thrust and the force on turbine blades. This is the NCERT thrust-and-impulse application. As a check, the units of mass-rate times velocity, kilograms per second times metres per second, give newtons, confirming a force, and an elastic rebound would double this to 300 N.