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 Question:
A uniform rod \( AB \) is suspended at a variable distance \( x \) from end \( A \) at a suspension point \( P \). A mass \( m \) is hung from end \( A \) to keep the rod in a horizontal position. We need to find the pair of variables that, when plotted against each other, produce a straight line.

Step 2: Key Formula or Approach:
For the rod to remain in rotational equilibrium (horizontal position), the net torque about the suspension point \( P \) must be zero:
\[ \sum \tau_P = 0 \]
- Let \( L \) be the length of the uniform rod.
- Let \( M_{\text{rod}} \) be the mass of the rod, so its weight acts at its center of mass, which lies at the midpoint \( O \) (at a distance of \( \frac{L}{2} \) from end \( A \)).
- The distance from \( P \) to end \( A \) is \( x \).
- The distance from \( P \) to the center of mass \( O \) is \( \left(\frac{L}{2} - x\right) \).

Step 3: Detailed Explanation:
Equating the counterclockwise torque (due to mass \( m \)) and the clockwise torque (due to the mass of the rod \( M_{\text{rod}} \)) about the suspension point \( P \):
\[ m g \cdot x = M_{\text{rod}} g \cdot \left(\frac{L}{2} - x\right) \]
We can cancel \( g \) on both sides:
\[ m \cdot x = M_{\text{rod}} \cdot \left(\frac{L}{2} - x\right) \]
\[ m \cdot x = \frac{M_{\text{rod}} L}{2} - M_{\text{rod}} \cdot x \]
Divide both sides by \( x \):
\[ m = \left(\frac{M_{\text{rod}} L}{2}\right) \frac{1}{x} - M_{\text{rod}} \]
This equation is of the form:
\[ y = M z + C \]
where:
- \( y = m \)
- \( z = \frac{1}{x} \)
- \( M = \frac{M_{\text{rod}} L}{2} \) (a constant slope)
- \( C = -M_{\text{rod}} \) (a constant intercept)
Since this is a linear relation of the form \( y = M z + C \), plotting \( m \) against \( \frac{1}{x} \) will give a straight line.

Step 4: Final Answer:
The set of variables that will give a straight line is \( m \) and \( \frac{1}{x} \).