Physics Fluid Dynamics PMDC Conceptual Practice
PMDC Verified Question 15 of 37
A machine worker places a cylinder with a diameter of \( 0.2 \text{ m} \) between the plates of a hydraulic press. If an applied force of \( 3.14 \times 10^5 \text{ N} \) is exerted, what is the approximate stress acting on the end of the cylinder?
A
\( 1.0 \times 10^6 \text{ Pa} \)
B
\( 3.14 \times 10^6 \text{ Pa} \)
C
\( 1.0 \times 10^7 \text{ Pa} \)
D
\( 1.5 \times 10^7 \text{ Pa} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( 1.0 \times 10^7 \text{ Pa} \)
Concept:

Stress is defined as the internal restoring force per unit cross-sectional area.

Formula:

$$ \text{Stress} = \frac{F}{A} $$

Solution:

  • The diameter is \( d = 0.2 \text{ m} \), so the radius \( r = 0.1 \text{ m} \).
  • Calculate the cross-sectional area: \( A = \pi r^2 = 3.14 \times (0.1)^2 = 3.14 \times 0.01 = 0.0314 \text{ m}^2 \).
  • Calculate stress: \( \text{Stress} = \frac{3.14 \times 10^5}{0.0314} \).
  • Simplify: \( \frac{3.14 \times 10^5}{3.14 \times 10^{-2}} = 10^5 \times 10^2 = 1.0 \times 10^7 \text{ Pa} \).


Why other options are incorrect:

Using diameter instead of radius to calculate the area yields an incorrect factor of 4. Omitting the area calculation entirely yields Option B.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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