fLUIDS O13,14

Solution for Question 13

We apply Bernoulli’s principle between a point inside the syringe (just before the orifice) and the point just outside the orifice.

Let $P_{in}$ be the pressure inside the syringe and $v$ be the velocity of efflux. Assuming the cross-sectional area of the syringe $A$ is much larger than the orifice, the velocity of the fluid inside is negligible ($v_{in} \approx 0$).

$$ P_{in} = P_{atm} + \frac{F}{A} $$

According to Bernoulli’s equation:

$$ P_{in} + \frac{1}{2}\rho v_{in}^2 = P_{out} + \frac{1}{2}\rho v^2 $$ $$ \left(P_{atm} + \frac{F}{A}\right) + 0 = P_{atm} + \frac{1}{2}\rho v^2 $$ $$ \frac{F}{A} = \frac{1}{2}\rho v^2 $$ $$ v^2 = \frac{2F}{\rho A} \implies v = \sqrt{\frac{2F}{\rho A}} $$

---

Solution for Question 14

The liquid exits horizontally and undergoes projectile motion. We need to find the resultant velocity when it strikes the ground.

Horizontal component ($v_x$): Remains constant.

$$ v_x = \sqrt{\frac{2F}{\rho A}} $$

Vertical component ($v_y$): Acquired due to gravity over height $h$.

$$ v_y^2 = u_y^2 + 2gh = 0 + 2gh = 2gh $$

Resultant Velocity ($V$):

$$ V = \sqrt{v_x^2 + v_y^2} $$ $$ V = \sqrt{\left(\frac{2F}{\rho A}\right) + 2gh} $$ $$ V = \sqrt{\frac{2F + 2\rho A g h}{\rho A}} = \sqrt{\frac{2(F + \rho g h A)}{\rho A}} $$