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 a non-uniform mass distribution, the center of mass can be calculated by integrating the mass density over the length of the object.

Answer 1

The Correct Option is B

Step 1: Understanding the mass distribution.
The linear mass density varies along the length of the rod, so the mass distribution is not uniform. The mass of the rod is obtained by integrating the linear mass density over the length of the rod. By integrating \( \lambda(x) \), we can find the mass and the location of the center of gravity. The correct location is \( x = \frac{9L}{16} \).
Step 2: Conclusion.
Thus, the correct answer is option (B).