BSc Nursing GNM 2019 Solved Paper

Time period of a thin bar magnet of vibration magnetometer is given by $T=2\pi\sqrt{\frac{I}{MH}}$ where, $I$ is moment of inertia of a bar magnet of mass $m_b$, length $l$ about an axis passing through its centre and perpendicular to its length,C $I=\frac{m_b l^2}{12}$ $M$ is magnetic moment of bar magnet, i.e. $M=m\times l$, where $m$ is magnetic strength of bar magnet and $H$ is the uniform magnetic field in which the magnet is oscillating. Case I When magnet is cut parallel to its length $I' = \frac{m_b}{2} \cdot \frac{l^2}{12} = \frac{m_b l^2}{2 \times 12} = \frac{I}{2}$ $\text{ and } M' = \frac{m}{2} \times l = \frac{M}{2}$ $\therefore$ Time period of each part, when bar magnet is cut parallel to its length is $T'=2\pi\sqrt{\frac{I'}{M' H}}=2\pi\sqrt{\frac{I/2}{(M/2)H}}$ $\Rightarrow T'=2\pi\sqrt{\frac{I}{MH}}=T$ Case II When magnet is cut perpendicular to its length. $I'' = \frac{m_b}{2} \cdot \frac{(l/2)^2}{12} =$ and $M'' = m \times \frac{l}{2} = \frac{M}{2}$ $\therefore$ Time period of each part, when bar magnet is cut perpendicular to its length is $T'' = 2\pi\sqrt{\frac{I''}{M'' H}}=2\pi\sqrt{\frac{I/8}{(M/2)H}}$ $T'' = 2\pi\sqrt{\frac{I}{4 M H}} = \frac{1}{2} \times 2\pi\sqrt{\frac{I}{MH}}$ $\Rightarrow T'' = \frac{T}{2}$