A uniform rod $AB$ is suspended from a point $P$, at a variable distance $x$, from $A$, as shown in figure. To make the rod horizontal, a mass '$m$' is suspended from its end $A$. Which set of variables will give a straight line when they are plotted?
To solve this problem, we need to understand the concept of equilibrium in a rotational system.
The uniform rod \(AB\) is suspended at a point \(P\), and we want to keep it horizontal by attaching a mass \(m\) at a variable distance \(x\) from \(A\). The key is to ensure the torques about point \(P\) sum to zero for rotational equilibrium.
The torque due to the weight of the mass \(m\) is given by:
\(\tau = m \cdot g \cdot x\)
where \(g\) is the acceleration due to gravity and \(x\) is the distance from \(P\) to \(A\).
The torque due to the weight of the rod acts at its center, which is at distance \(\frac{L}{2}\) from \(B\) (assuming length \(L\)). The rod's weight acting at the center causes a torque that must be balanced by the torque due to \(m\).
The condition for equilibrium (sum of torques about \(P\) = 0) gives us:
\(m \cdot g \cdot x = W_{rod} \cdot \left(\frac{L}{2} - x_{P}\right)\)
Simplifying, and assuming potential other forces at equilibrium, we find:
\(m = \frac{1}{x}\)
Thus, plotting mass \(m\) against \(\frac{1}{x}\) will give us a straight line, matching option: '\(m, \frac{1}{x}\)'.
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