(d) $I = \frac{L}{r^{2}}$ $\Rightarrow \frac{dI}{I} = - \frac{2dr}{r}\ \ \ \ \ \ \ \ \ \ \ \lbrack\because L = constant\rbrack$ $\Rightarrow \frac{dI}{I} \times 100 = - \frac{2 \times dr}{r} \times 100 = - 2 \times 1 = - 2\%$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.