Question

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?

• A. \( m, x^2 \)
• B. \( m, \frac{1}{x^2} \)
• C. \( m, \frac{1}{x} \)
• D. \( m, x \)

Hint:

To quickly find straight-line relations in experimental physics questions, rearrange the equilibrium equation to isolate the dependent variable (e.g., \( m \)) on one side. If the other side contains a term like \( \frac{\text{constant}}{x} \), the straight line will always be obtained by plotting \( y \) versus \( \frac{1}{x} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a rod to remain horizontal in equilibrium, the net torque about the point of suspension \(P\) must be zero.
Torque is the product of the force and its perpendicular distance from the pivot.
The weights acting are the suspended mass \(m\) at end \(A\) and the weight of the uniform rod (\(M\)) acting at its center of mass.
Step 2: Key Formula or Approach:
Torque balance: \(\sum \tau_P = 0\).
Weight of rod acts at center \(L/2\).
Step 3: Detailed Explanation:
Let \(L\) be the total length of the uniform rod and \(M\) be its mass.
Pivot is at point \(P\), at distance \(x\) from end \(A\).
Distance from \(A\) to \(P = x\).
Distance from \(P\) to the center of mass (midpoint) of the rod \(= (L/2) - x\).
Taking torques about \(P\):
Anti-clockwise torque due to \(m\): \(\tau_1 = (mg) \cdot x\).
Clockwise torque due to rod's weight \(M\): \(\tau_2 = (Mg) \cdot (L/2 - x)\).
Equating torques:
\[mx = M(L/2 - x)\]
\[mx = \frac{ML}{2} - Mx\]
\[x(m + M) = \frac{ML}{2}\]
We want to see the relationship between \(m\) and \(x\).
\[m + M = \frac{ML}{2} \cdot \frac{1}{x}\]
\[m = \left(\frac{ML}{2}\right) \frac{1}{x} - M\]
This is of the form \(y = k \cdot Z + C\), where \(y = m\) and \(Z = 1/x\).
Thus, a plot of \(m\) versus \(1/x\) will yield a straight line with a negative intercept.
This matches option (C).
Step 4: Final Answer:
From the torque balance equation, \(m\) is linearly proportional to the reciprocal of the distance \(x\). Thus, plotting \(m\) against \(1/x\) gives a straight line.