When a bar magnet is freely suspended for oscillation,its time period is $T$. If it is cut into $3$ equal parts perpendicular to its length,then the time period of oscillation for each part will be

The deflection produced in a tangent galvanometer,whose coil has a resistance of $9 \Omega$,is $30^{\circ}$. The potential difference across the coil is $4.5 \ V$. If the number of turns in the coil is $10$,the radius of the coil is (Given,$B_{H} = 3.14 \times 10^{-5} \ T$)

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 $\pi^{2} = 9.85$ ]

$A$ magnetic needle of negligible breadth and thickness compared to its length oscillates in a horizontal plane with a period $T$. What is the period of oscillation of each part obtained after breaking the magnet into $n$ equal parts perpendicular to its length?

$A$ magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of Earth's magnetic field is $20 \mu T$. If the neutral point is found at a distance of $20 \ cm$ in the plane bisecting the dipole, then the magnetic moment of the dipole is (Assume $\mu_0 = 4 \pi \times 10^{-7} \text{ S.I. units}$) (in $\text{ A m}^2$)

$A$ tangent galvanometer shows a deflection of $45^{\circ}$ when a $10 \, mA$ current passes through it. If the horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \, T$ and the radius of the coil is $10 \, cm$,find the number of turns in the coil.