Question

A jet water issues from a nozzel with a velocity of $20m/s$ and it impinges normally on a flat plate moving away from it at $10m/s$. If the cross-sectional area of the jet is $0.02m^2$ and the density of water is taken as $1000kg/m^3$, then the force developed on the plate will be

• A. $10N$
• B. $100N$
• C. $1000N$
• D. $2000N$

Answer 1

The correct option is A $10N$

A jet water issues from a nozzel with a velocity$=20m/s$

It impinges normally on a flat plate moving away from it$=10m/s$

Cross-sectional area of the jet $=0.02m^2$

The density of water is taken$=1000kg/m^3$

The force developed on the plate will be

Hydrostatic force o the bottom surface of the tank will be

$F_b=\rho g(l\times b)\times h{F}_b\rho g\times (1\times 2)\times 2F_b=4\rho g\left(lbh\right)\to \left(1\right)$

Hydrostatic force on vertical surface will be

$F_v=\rho g(l\times h)\times \frac{h}{2}{F}_v=\rho g(2\times 2)\frac{l{h}^2}{2}{F}_v=2\rho g\left(l{h}^2\right)\to \left(2\right)$

Ratio will be

$\frac{F_b}{F_V}=\frac{4\rho g\left(lbh\right)}{2\rho g\left(l{h}^2\right)}\frac{F_b}{F_V}=1F_b=1\times 10=10N$