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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Work, Power and Energy

AIIMS 2019 · 25 May · Physics · Work, Power and Energy

If potential energy is given by U=(a)/(r^2)-(b)/(r) . Then find out...

If potential energy is given by U=ar2−brU=\frac{a}{r^2}-\frac{b}{r}. Then find out maximum force. (Given a=2a=2, b=4)b=4)

Options

  1. A−1627 N-\frac{16}{27} \mathrm{~N}
  2. B−3227 N-\frac{32}{27} \mathrm{~N}
  3. C+3227 N+\frac{32}{27} \mathrm{~N}
  4. D+1627 N+\frac{16}{27} \mathrm{~N}

Correct answer

A. −1627 N-\frac{16}{27} \mathrm{~N}

Step-by-step solution

 F=−dudr=−[−2ar3+br2]=2ar3−br2dFdr=−6ar4+2br3=0⇒6ar=2b⇒r=3ab=64=32⇒Fmax⁡=2×2(3/2)3−4(3/2)2=4×827−4×49=32−16×327=−1627\begin{aligned} & \text{ } F=-\frac{d u}{d r}=-\left[-\frac{2 a}{r^3}+\frac{b}{r^2}\right]=\frac{2 a}{r^3}-\frac{b}{r^2} \\ & \frac{d F}{d r}=\frac{-6 a}{r^4}+\frac{2 b}{r^3}=0 \\ & \Rightarrow \frac{6 a}{r}=2 b \Rightarrow r=\frac{3 a}{b}=\frac{6}{4}=\frac{3}{2} \\ & \Rightarrow F_{\max }=\frac{2 \times 2}{(3 / 2)^3}-\frac{4}{(3 / 2)^2} \\ & =\frac{4 \times 8}{27}-\frac{4 \times 4}{9}=\frac{32-16 \times 3}{27}=-\frac{16}{27}\end{aligned}
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