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NEET PHYSICSLAWS OF MOTIONMedium

Question

A monkey of mass 20 kg20 \text{ kg} is holding a vertical rope. The rope will not break when a mass of 25 kg25 \text{ kg} is suspended from it but will break if the mass exceeds 25 kg25 \text{ kg}. What is the maximum acceleration with which the monkey can climb up along the rope (g=10 m/s2g = 10 \text{ m/s}^2)?

A

10 m/s210 \text{ m/s}^2

B

25 m/s225 \text{ m/s}^2

C

2.5 m/s22.5 \text{ m/s}^2

D

5 m/s25 \text{ m/s}^2

Step-by-Step Solution

  1. Determine Maximum Tension (TmaxT_{max}): The breaking strength of the rope corresponds to the weight of the maximum mass it can hold (M=25 kgM = 25 \text{ kg}). Thus, the maximum tension is: Tmax=Mg=25 kg×10 m/s2=250 NT_{max} = M \cdot g = 25 \text{ kg} \times 10 \text{ m/s}^2 = 250 \text{ N}

  2. Apply Newton's Second Law: For the monkey (mass m=20 kgm = 20 \text{ kg}) climbing upwards with acceleration aa, the forces are Tension (TT) acting upwards and Weight (mgmg) acting downwards. The equation of motion is: Tmg=maT - mg = ma [Source 118, 126]

  3. Solve for Maximum Acceleration (amaxa_{max}): Substitute TmaxT_{max} into the equation: 250 N(20 kg×10 m/s2)=20 kg×amax250 \text{ N} - (20 \text{ kg} \times 10 \text{ m/s}^2) = 20 \text{ kg} \times a_{max} 250200=20amax250 - 200 = 20 \cdot a_{max} 50=20amax50 = 20 \cdot a_{max} amax=5020=2.5 m/s2a_{max} = \frac{50}{20} = 2.5 \text{ m/s}^2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONmonkeyholdingverticalsuspendedexceeds

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