Question

The linear mass density of a rod of length \( L \) varies from one end to the other as \( \lambda_0 \left( 1 + \frac{x^2}{L^2} \right) \), where \( x \) is the distance from one end with tensions \( T_1 \) and \( T_2 \) in them (see figure), and \( \lambda_0 \) is a constant. The rod is suspended from a ceiling by two massless strings. Then, which of the following statement(s) is (are) correct?

• A. The mass of the rod is \( \frac{2 \lambda_0 L^3}{3} \).
• B. The center of gravity of the rod is located at \( x = \frac{9L}{16} \).
• C. The tension \( T_1 \) in the left string is \( \frac{7\lambda_0 g L^2}{12} \).
• D. The tension \( T_2 \) in the right string is \( \frac{3 \lambda_0 g L^2}{2} \).

Hint:

For objects with non-uniform mass distributions, the center of mass can be found by integrating the mass density over the length of the object. Similarly, tensions in a system can be determined using equilibrium conditions.

Answer 1

The Correct Option is B, C

Step 1: Understanding the mass distribution.
The linear mass density varies along the length of the rod, so the mass distribution is not uniform. To find the mass of the rod, we integrate the linear mass density over the length of the rod. This gives us the total mass and the position of the center of gravity, which is located at \( x = \frac{9L}{16} \). Step 2: Calculating the tension.
By considering the forces acting on the rod and applying the equilibrium conditions, the tensions in the strings are calculated. The tension \( T_1 \) in the left string is \( \frac{7\lambda_0 g L^2}{12} \), and the tension \( T_2 \) in the right string can be derived similarly.
Step 3: Conclusion.
Thus, the correct answers are options (B) and (C).