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Physiology · Cardiac mechanics: Laplace law
Sample itemIn a patient with dilated cardiomyopathy, the left ventricular radius increases from 2 cm to 3 cm. If intracavitary pressure and wall thickness stay the same, what happens to wall stress according to the law of Laplace?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
D. It increases by 50%
Wall stress is proportional to pressure multiplied by radius and divided by wall thickness (P x r / 2h). With pressure and thickness constant, stress rises in direct proportion to radius: 3 / 2 = 1.5, a 50% increase. A dilated ventricle therefore needs more wall stress, and more oxygen, to generate the same pressure.
Option by option
A. It is unchanged
Stress would be unchanged only if radius did not affect it, which contradicts the Laplace relationship.
B. It falls by 33%
A fall of 33% would result if radius decreased to two-thirds, the reverse of this situation.
C. It doubles
Doubling would need the radius to double, from 2 cm to 4 cm.
D. It increases by 50% (correct)
Correct. Stress is proportional to radius, so a 1.5-fold larger radius gives a 50% rise.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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