(d) $I = I_{CM} + Mx^{2}$ $= \frac{ML^{2}}{12} + M\left\lbrack \frac{L}{4} \right\rbrack^{2}$ $= \frac{ML^{2}}{12} + \frac{ML^{2}}{16} = \frac{7ML^{2}}{48}$ [image:7_image99.emf]
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.