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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

A particle of mass MM starting from rest undergoes uniform acceleration. If the speed acquired in time TT is vv, the power delivered to the particle is:

A

Mv2T\frac{Mv^2}{T}

B

12Mv2T2\frac{1}{2}\frac{Mv^2}{T^2}

C

Mv2T2\frac{Mv^2}{T^2}

D

12Mv2T\frac{1}{2}\frac{Mv^2}{T}

Step-by-Step Solution

  1. Identify the Goal: The problem asks for the power delivered to the particle. Given the probable answer provided, this refers to the Average Power delivered over the time interval TT to achieve speed vv.
  2. Apply Work-Energy Theorem: The work done (WW) on the particle is equal to the change in its kinetic energy (ΔK\Delta K).
  • Initial Kinetic Energy (KiK_i) = 0 (starts from rest)
  • Final Kinetic Energy (KfK_f) = 12Mv2\frac{1}{2}Mv^2 W=ΔK=12Mv20=12Mv2W = \Delta K = \frac{1}{2}Mv^2 - 0 = \frac{1}{2}Mv^2 [Class 11 Physics, Ch 5, Sec 5.2, Eq 5.2a]
  1. Calculate Average Power: Average power (PavP_{av}) is defined as the work done divided by the time taken. Pav=WtP_{av} = \frac{W}{t} Substituting the values: Pav=12Mv2T=12Mv2TP_{av} = \frac{\frac{1}{2}Mv^2}{T} = \frac{1}{2}\frac{Mv^2}{T} [Class 11 Physics, Ch 5, Sec 5.10, Eq 5.20]

(Note: If the question asked for Instantaneous Power at time TT, the answer would be P=Fv=(Ma)v=M(vT)v=Mv2TP = F \cdot v = (Ma)v = M(\frac{v}{T})v = \frac{Mv^2}{T}).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. 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.

PHYSICSWORK, ENERGY AND POWERparticlestartingundergoesuniformacceleration

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