Physics 1 Solved Past Papers 2000 – 2000 Archives

Fluid Dynamics Past Papers

Solved past paper MCQs for Fluid Dynamics from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.

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#1 of 1 MDCAT 2000

The flow speed of a fluid through a horizontal pipe increases from $ 1.0 \text{ m/s} $ to $ 3.0 \text{ m/s} $. If the density of the fluid is $ 1000 \text{ kg/m}^3 $, what is the pressure difference between these two points?

$ 1000 \text{ N/m}^2 $

$ 4000 \text{ N/m}^2 $

$ 8000 \text{ N/m}^2 $

$ 9000 \text{ N/m}^2 $

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Correct Key: Option B Diagnostic Explanation

Concept:

For a horizontal pipe, the potential energy term ($ \rho g h $) in Bernoulli's equation is zero, leaving only pressure and kinetic energy.

Formula:

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

Solution:

We have $ \rho = 1000 \text{ kg/m}^3 $, $ v_1 = 1.0 \text{ m/s} $, $ v_2 = 3.0 \text{ m/s} $.
Calculate the difference in the squares of the speeds: $ (3.0)^2 - (1.0)^2 = 9 - 1 = 8 \text{ m}^2/\text{s}^2 $.
Multiply by half the density: $ \Delta P = \frac{1}{2} \times 1000 \times 8 $.
$ \Delta P = 500 \times 8 = 4000 \text{ N/m}^2 $.

Why other options are incorrect:

Calculating $ (v_2 - v_1)^2 $ instead of $ (v_2^2 - v_1^2) $ gives $ (3-1)^2 = 4 $, leading to 2000. Forgetting the $ 1/2 $ multiplier leads to 8000.

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