Physics Rotational & Circular Motion UHS 2024
PMDC Verified Question 12 of 95
A particle of mass 'm' is moving on a circular path of radius 'r' with velocity 'v', then centripetal force acting on it is F. If the velocity of particle increases by 2 time and radius of circular path increases by 4 times then new centripetal force F' will be:
A
\( F' = 2F \)
B
\( F' = \frac{1}{2}F \)
C
\( F' = 4F \)
D
\( F' = F \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( F' = F \)
Concept:

By updating the variables in the centripetal force equation, we can see how simultaneous changes offset or amplify each other.

Formula:

$$ F = \frac{mv^2}{r} $$

Solution:

  • Let the new velocity be \( v' = 2v \).
  • Let the new radius be \( r' = 4r \).
  • Substitute these into the force equation: \( F' = \frac{m(2v)^2}{4r} \).
  • Square the velocity term: \( F' = \frac{m(4v^2)}{4r} \).
  • The 4 in the numerator and the 4 in the denominator perfectly cancel out: \( F' = \frac{mv^2}{r} \).
  • This is strictly equal to the original force \( F \).


Why other options are incorrect:

  • Option A, Option B, Option C fail to correctly square the velocity change or miss the cancellation algebra.

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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