Question

There are two identical small holes on the opposite side of a tank containing full of a liquid. The tank is open at the top. The difference in height between the two holes is ' $h$ '. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

• A. $h^{3/2}$
• B. $h^2$
• C. $\sqrt{h}$
• D. $h$

Hint:

For liquid jets from holes, use Torricelli’s theorem: \[ v=\sqrt{2gh} \] and momentum flux arguments to relate force with depth.

Answer 1

The Correct Option is C

Concept:
Efflux speed from a hole at depth \(y\) below the liquid surface is: \[ v=\sqrt{2gy} \] The reaction force due to the jet is proportional to the rate of momentum flow: \[ F \propto \dot{m}v \] Since \(\dot{m}\propto v\), we get: \[ F\propto v^2 \propto y \] ip

Step 1: Write force due to each jet.
If the depths of the two holes are \(y_1\) and \(y_2\), then: \[ F_1\propto y_1,\qquad F_2\propto y_2 \] These forces act in opposite horizontal directions. ip

Step 2: Find net horizontal force.
\[ F_{\text{net}} \propto |y_2-y_1| \] If the difference in depths is \(h\), then: \[ F_{\text{net}}\propto h \] This direct derivation gives proportionality to \(h\), which corresponds to option (D), not the keyed option sequence implied in the source. ip The direct physics result gives:
\[ \boxed{F_{\text{net}}\propto h} \]