Search any question & find its solution
Question:
Answered & Verified by Expert
A thin rod of length ' $L$ ' is lying along the $x$-axis with its ends at $x=0$ and $x=L$. Its linear density (mass/length) varies with $x$ ask $\left(\frac{x}{L}\right)^n$, where $n$ can be zero or any positive number. If the position $x_{\mathrm{CM}}$ of the centre of mass of the rod is plotted against ' $n$ ', which of the following graphs best approximates the dependence of $x_{\mathrm{CM}}$ on $n$ ?
Options:
- A
- B
- C
- D
Solution:
2008 Upvotes
Verified Answer
The correct answer is:
$$
\begin{aligned}
& x_{c m}=\frac{\int d m x}{\int d m}=\frac{\int \lambda d x \cdot x}{\int d m}=\frac{\int k\left(\frac{x}{L}\right)^n \cdot x d x}{\int k\left(\frac{x}{L}\right)^n d x}=\left[\frac{\frac{k x^{n+2}}{(n+2) L^n}}{\frac{k x^{n+1}}{(n+1) L^n}}\right]_0^L=\left[\frac{x(n+1)}{n+2}\right]_0^L \\
& x_{c m}=\frac{L}{2}, \frac{2 L}{3}, \frac{3 L}{4}, \frac{4 L}{5}, \frac{5 L}{6}, \ldots
\end{aligned}
$$
\begin{aligned}
& x_{c m}=\frac{\int d m x}{\int d m}=\frac{\int \lambda d x \cdot x}{\int d m}=\frac{\int k\left(\frac{x}{L}\right)^n \cdot x d x}{\int k\left(\frac{x}{L}\right)^n d x}=\left[\frac{\frac{k x^{n+2}}{(n+2) L^n}}{\frac{k x^{n+1}}{(n+1) L^n}}\right]_0^L=\left[\frac{x(n+1)}{n+2}\right]_0^L \\
& x_{c m}=\frac{L}{2}, \frac{2 L}{3}, \frac{3 L}{4}, \frac{4 L}{5}, \frac{5 L}{6}, \ldots
\end{aligned}
$$
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.