Search any question & find its solution
Question:
Answered & Verified by Expert
The velocity $(v)$ of a particle (under a force $F$ ) depends on its distance $(x)$ from the origin (with $x>0) v \propto\frac{1}{\sqrt{x}}$. Find how the magnitude
of the force $(F)$ on the particle depends on $x ?$
Options:
of the force $(F)$ on the particle depends on $x ?$
Solution:
1460 Upvotes
Verified Answer
The correct answer is:
$F \propto \frac{1}{x^{2}}$
According to the question.
$$
v \propto \frac{1}{\sqrt{x}}
$$
or $v=\frac{k}{\sqrt{x}}$
$$
\frac{d v}{d t}=\frac{d}{d t} \cdot \frac{K}{\sqrt{x}}=k \cdot \frac{-1}{2} \cdot x^{-\frac{1}{2}-1} \cdot \frac{d x}{d t}
$$
$a=-\frac{k}{2} \cdot x^{-\frac{3}{2}} \cdot \frac{k}{\sqrt{x}}\left[\because \frac{d x}{d t}=v\right.$ and $v=\frac{k}{\sqrt{x}}$ from $\left.\mathrm{Eq}(i)\right]$
$$
=\frac{-k^{2}}{2} \cdot \frac{1}{x^{2}}
$$
$=m \frac{k^{2}}{2} \cdot \frac{1}{x^{2}}$
(magnitue
$F=\frac{k^{2}}{2} \cdot \frac{m}{x^{2}}$
$\therefore$
$F x \frac{1}{x^{2}}$
$$
v \propto \frac{1}{\sqrt{x}}
$$
or $v=\frac{k}{\sqrt{x}}$
$$
\frac{d v}{d t}=\frac{d}{d t} \cdot \frac{K}{\sqrt{x}}=k \cdot \frac{-1}{2} \cdot x^{-\frac{1}{2}-1} \cdot \frac{d x}{d t}
$$
$a=-\frac{k}{2} \cdot x^{-\frac{3}{2}} \cdot \frac{k}{\sqrt{x}}\left[\because \frac{d x}{d t}=v\right.$ and $v=\frac{k}{\sqrt{x}}$ from $\left.\mathrm{Eq}(i)\right]$
$$
=\frac{-k^{2}}{2} \cdot \frac{1}{x^{2}}
$$
$=m \frac{k^{2}}{2} \cdot \frac{1}{x^{2}}$
(magnitue
$F=\frac{k^{2}}{2} \cdot \frac{m}{x^{2}}$
$\therefore$
$F x \frac{1}{x^{2}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.