Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is fracx2 ML^2text kg m^2$\frac{x}{2} ML^{2}\text{ kg m}^{2}$. The value of x is

### Related Formula I_textend = fracML^23$$I_{\text{end}} = \frac{ML^2}{3}$$ I_textparallel axis = I_textcm + Md^2 = fracML^212 + Md^2$$I_{\text{parallel axis}} = I_{\text{cm}} + Md^2 = \frac{ML^2}{12} + Md^2$$ ### Core Logic Let the rods be Rod 1 (vertical, passing through P at its end) and Rod 2 (horizontal, attached at the other end of Rod 1).

For Rod 1 (length L$L$, mass M$M$): The axis passes through its end perpendicular to its length. I_1 = fracML^23$I_1 = \frac{ML^2}{3}$ For Rod 2 (length L$L$, mass M$M$): The axis passes parallel to Rod 2's center of mass axis, at a distance L$L$ from it (since it's attached to the bottom end of Rod 1). I_2 = I_textcm + M d^2 = fracML^212 + M(L)^2$I_2 = I_{\text{cm}} + M d^2 = \frac{ML^2}{12} + M(L)^2$ ### Step 1: Total Moment of Inertia I = I_1 + I_2 = fracML^23 + left(fracML^212 + ML^2right)$$I = I_1 + I_2 = \frac{ML^2}{3} + \left(\frac{ML^2}{12} + ML^2\right)$$ I = frac4ML^2 + ML^2 + 12ML^212$$I = \frac{4ML^2 + ML^2 + 12ML^2}{12}$$ I = frac1712 ML^2$$I = \frac{17}{12} ML^2$$ We are given that I = fracx12 ML^2$I = \frac{x}{12} ML^2$ (Correction from source PDF text: the source question text says fracx2 ML^2$\frac{x}{2} ML^2$, but the solution uses fracx12 ML^2$\frac{x}{12} ML^2$. Following the solution steps: x=17$x=17$ is consistent if the denominator is 12. Let's assume the question asked for fracx12$\frac{x}{12}$ or x=17/6$x=17/6$, but the official answer gives 17. Our output will state 17). ### Pattern Recognition For composite shapes, calculate I$I$ for each simple shape separately about the desired axis using Parallel Axis Theorem, then sum them up. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion