An experiment is conducted to measure the pressure drop Δ P across...

Let the cross-sectional area of the wider part of the pipe be $A_1$ and that of the constriction be $A_2$.

From the equation of continuity, the volume flow rate is $Q = A_1 v_1 = A_2 v_2$.

The velocities at the two sections are $v_1 = \dfrac{Q}{A_1}$ and $v_2 = \dfrac{Q}{A_2}$.

Applying Bernoulli's principle for a horizontal pipe:

$P_1 + \dfrac{1}{2} \rho v_1^2 = P_2 + \dfrac{1}{2} \rho v_2^2$

The pressure drop $\Delta P$ is given by:

$\Delta P = P_1 - P_2 = \dfrac{1}{2} \rho (v_2^2 - v_1^2)$

Substituting the expressions for velocities:

$\Delta P = \dfrac{1}{2} \rho \left[ \left(\dfrac{Q}{A_2}\right)^2 - \left(\dfrac{Q}{A_1}\right)^2 \right]$

$\Delta P = \left[ \dfrac{1}{2} \rho \left( \dfrac{1}{A_2^2} - \dfrac{1}{A_1^2} \right) \right] Q^2$

Since the density $\rho$ and the areas $A_1$ and $A_2$ are constant for a given setup, the term in the square brackets is a positive constant $k$.

Thus, $\Delta P = k Q^2$.

This equation is of the form $y = k x^2$, which represents a parabola opening towards the positive y-axis (the $\Delta P$-axis).

Answer: A parabola opening towards the positive $\Delta P$-axis