1. Set up the torque balance

Let the rod have:

• length $L$
• mass $M$

Since the rod is uniform, its weight $Mg$ acts at its center, i.e. at a distance $L/2$ from end $A$.

The rod is suspended from point $X$, which is at a distance $x$ from $A$. A mass $m$ is hung at end $A$ to keep the rod horizontal.

For rotational equilibrium about point $X$, net torque must be zero.

2. Identify lever arms about $X$

• The suspended mass $m$ at $A$ produces torque with arm $x$.
• The rod’s own weight acts at its center, so its lever arm is $\left(\frac{L}{2}-x\right)$.

Assuming $X$ lies between $A$ and the center of mass, the torques balance as:

$mg\,x = Mg\left(\frac{L}{2}-x\right)$

3. Solve for $m$ in terms of $x$

Cancel $g$:

$mx = M\left(\frac{L}{2}-x\right)$

$mx = \frac{ML}{2} - Mx$

$m = \frac{ML}{2x} - M$

So,

$m = \left(\frac{ML}{2}\right)\left(\frac{1}{x}\right) - M$

4. Interpret the result

This is of the form

$m = a\left(\frac{1}{x}\right) + b$

which is a straight-line equation in $m$ versus $\dfrac{1}{x}$.

Hence, plotting:

• vertical axis: $m$
• horizontal axis: $\dfrac{1}{x}$

will give a straight line.

5. Check options

• A: $m, x$ → not linear
• B: $m, \dfrac{1}{x}$ → linear ✔
• C: $m, \dfrac{1}{x^2}$ → not linear
• D: $m, x^2$ → not linear

Therefore, the correct option is B.