$A$ magnet is suspended horizontally in the Earth's magnetic field. When it is made to oscillate in a horizontal plane,its time period is $T$. If a wooden block of the same moment of inertia as the magnet is attached to it,what will be the new time period of the system?

Before using the tangent galvanometer,its coil is set in

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \, s$. The magnet is cut along its length into three equal parts and these three parts are then placed on each other with their like poles together. The time period of this combination will be

$A$ certain amount of current when flowing in a properly set tangent galvanometer,produces a deflection of $45^{\circ}$. If the current is reduced by a factor of $\sqrt{3}$,the deflection would:

In a uniform magnetic field,a magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, A \cdot m^{2}$ and a moment of inertia $5 \times 10^{-6} \, kg \cdot m^{2}$. If it performs $10$ complete oscillations in $5 \, s$,then the magnitude of the magnetic field is $.... \, mT$. [ Take $\pi^{2} = 9.85$ ]

$A$ thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T'$,the ratio $T'/T$ is :-