When a freely suspended magnet is heated,its magnetic moment decreases by $19\%$. What is the percentage change in its time period?

$A$ deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling of the deflection is $T$. When the magnet is removed,the period of oscillation of the needle is $T_0$ before settling to $0^{\circ}-0^{\circ}$. If the earth's horizontal magnetic field is $B_H$,the relation between $T$ and $T_0$ is

$A$ 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:

$A$ magnet makes $40$ oscillations per minute at a place having magnetic field intensity of $0.1 \times 10^{-5} \,T$. At another place,it takes $2.5 \,s$ to complete one vibration. The value of earth's horizontal field at that place is:

$A$ dip circle lies initially in the magnetic meridian,it shows an angle of dip $\delta$ at a place. The dip circle is rotated through an angle $\alpha$ in the horizontal plane and then it shows an angle of dip $\delta^{\prime}$. Hence $\frac{\tan \delta^{\prime}}{\tan \delta}$ is

Two bar magnets having the same geometry with magnetic moments $M$ and $2M$ are first placed in such a way that their similar poles are on the same side,and their time period of oscillation is $T_{1}$. Now,the polarity of one of the magnets is reversed,and the time period of oscillation becomes $T_{2}$. Then: