NEET (UG) 2024 — Question 7

Time period $T = \frac{5\ \text{s}}{20} = 0.25\ \text{s}$. For a magnetic dipole in a uniform field, $T = 2\pi\sqrt{\frac{I}{MB}}$. Squaring: $T^2 = 4\pi^2 \frac{I}{MB}$ $\Rightarrow M = \frac{4\pi^2 I}{T^2 B}$. Substitute: $I = 9.8 \times 10^{-6}\ \mathrm{kg\ m^2}$, $T^2 = (0.25)^2 = 0.0625$, $B = 0.049\ \mathrm{T}$. $M = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{0.0625 \times 0.049} = \frac{39.2\pi^2 \times 10^{-6}}{0.0030625} = 12.8\pi^2 \times 10^{-3}\ \mathrm{A\ m^2}$. Since the desired form is $x \times 10^{-5}\ \mathrm{A\ m^2}$, rewrite: $12.8\pi^2 \times 10^{-3} = 1280\pi^2 \times 10^{-5}$. Thus $x = 1280\pi^2$, matching option D.