**Problem:**
A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of (m,x) values is recorded. The appropriate variables that give a straight line, when plotted, are
(JEE Mains 2018)

- m, x
- m, 1/x
- m, 1/x
^{2} - m, x
^{2}

**Solution:**
Let $M$ be the mass and $L$ be the length of the rod AB. The centre of mass C of uniform rod lies at its geometrical centre i.e., at a distance $L/2$ from A.

The forces on the rod are $mg$ at A, $Mg$ at C and the string tension $T$ at X. To keep the rod horizontal, the torque about any point of the rod should be zero. Take torque about the suspension point X to get \begin{align} \tau=mgx-Mg(L/2-x)=0,\nonumber \end{align} which gives \begin{align} m=\frac{ML}{2}\cdot \frac{1}{x}-M.\nonumber \end{align} Thus, $(m,1/x)$ graph is a straight line with slope $ML/2$ and intercept $-M$.

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