AP EAMCET · PHYSICS · Laws of Motion
If the breaking strength of a rope is \(\frac{4}{3}\) times the weight of a person, then the maximum acceleration with which the person can safely climb up the rope is (g - accelereation due to gravity)
- A \(\frac{\mathrm{g}}{2}\)
- B g
- C \(\frac{\mathrm{g}}{3}\)
- D \(\frac{2 \mathrm{~g}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{g}}{3}\)
Step-by-step Solution
Detailed explanation
\(T_{max} = \frac{4}{3} mg\) \(T_{max} - mg = ma\) \(\frac{4}{3} mg - mg = ma\) \(\frac{1}{3} mg = ma\) \(a = \frac{g}{3}\)
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