153975
A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 $T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet ?
1 $0.36 \mathrm{~J} \mathrm{~T}^{-1}$
2 $0.036 \mathrm{~J} \mathrm{~T}^{-1}$
3 $3.6 \mathrm{~J} \mathrm{~T}^{-1}$
4 $36 \mathrm{~J} \mathrm{~T}^{-1}$
Explanation:
A Given, magnetic field (B) $=0.25 \mathrm{~T}$ Torque $(\tau)=4.5 \times 10^{-2} \mathrm{~J}$ Angle $(\theta)=30^{\circ}$ We know that, $\tau=M B \sin \theta$ $\therefore \quad \mathrm{M} =\frac{\tau}{\mathrm{B} \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}$ $\mathrm{M} =0.36 \mathrm{JT}^{-1}$
GUJCET 18.04.2022
Magnetism and Matter
153976
A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of original magnet is ' $T_{0}$ ' in a uniform magnetic field $B$. Then, the time period of each part in the same magnetic field is-
1 $\frac{T_{0}}{2}$
2 $\frac{T_{0}}{3}$
3 $\frac{T_{0}}{4}$
4 $\mathrm{T}_{0}$
Explanation:
B Initial time period $\mathrm{T}_{\mathrm{o}}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}} \quad(\mathrm{I}=\text { Moment of inertia })$ $\mathrm{I}=\frac{\mathrm{mL}^{2}}{12}$ When magnet is cut in three parts then new inertia $I^{\prime}=\frac{\frac{\mathrm{m}}{3} \times\left(\frac{\mathrm{L}}{3}\right)^{2}}{12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{mL}^{2}}{27 \times 12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{27} \quad\left(\because \mathrm{I}=\frac{\mathrm{mL}^{2}}{12}\right)$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\frac{\mathrm{M}}{3}$ New time period $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{\mathrm{I}^{\prime}}{\mathrm{M}^{\prime} \mathrm{B}}}$ $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\frac{\mathrm{I}}{27}}{\frac{\mathrm{M}}{3} \times \mathrm{B}}}=2 \pi \sqrt{\frac{1}{9} \frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{1}{3} \times 2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{\mathrm{T}_{\mathrm{o}}}{3}$
CG PET-22.05.2022
Magnetism and Matter
153980
Figure shows a straight wire placed between the pole pieces of a magnet. Induced emf will be developed across the ends of the wire when it is moved towards
1 $\mathrm{N}$
2 $\mathrm{S}$
3 $\mathrm{P}$
4 $\mathrm{Q}$
Explanation:
C Magnetic field lines moves from $\mathrm{N}$ to $\mathrm{S}$ in order to induced emf there should be change in electric flux $e=\frac{d \phi}{d t}$ So, when wire moves towards $\mathrm{P}$ direction it will cut the magnetic field lines result in change in magnetic field.
AP EAMCET-25.08.2021
Magnetism and Matter
153982
The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_{1}$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_{2}$. The declination $\theta$ at the place is
C $\because \tan \delta_{1}=\frac{\tan \theta}{\cos \alpha}$ $\cos \alpha=\frac{\tan \theta}{\tan \delta_{1}}$ Again, $\tan \delta_{2}=\frac{\tan \theta}{\sin \alpha}$ $\sin \alpha=\frac{\tan \theta}{\tan \delta_{2}}$ From equation (ii) dividing by equation (i), we get- $\frac{\sin \alpha}{\cos \alpha}=\frac{\tan \theta / \tan \delta_{2}}{\tan \theta / \tan \delta_{1}}$ $\tan \alpha=\frac{\tan \delta_{1}}{\tan \delta_{2}}$ $\alpha=\tan ^{-1}\left(\frac{\tan \delta_{1}}{\tan \delta_{2}}\right)$
AP EAMCET-19.08.2021
Magnetism and Matter
153983
A magnet of magnetic 2J.T ${ }^{-1}$ is aligned in the direction of magnetic field of 0.1T. What is the net work done to bring the magnet normal to the magnetic field?
1 $0.1 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $1.0 \mathrm{~J}$
4 $2.0 \mathrm{~J}$
Explanation:
B Magnetic moment (M) $=2 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.1 \mathrm{~T}$ Work done, $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ $\theta_{1}=0 \text { (along the direction of magnetic field) }$ $\theta_{2}=90^{\circ}$ $\mathrm{W}=2 \times 0.1\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=0.2(1-0)$ $\mathrm{W}=0.2 \mathrm{~J}$
153975
A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 $T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet ?
1 $0.36 \mathrm{~J} \mathrm{~T}^{-1}$
2 $0.036 \mathrm{~J} \mathrm{~T}^{-1}$
3 $3.6 \mathrm{~J} \mathrm{~T}^{-1}$
4 $36 \mathrm{~J} \mathrm{~T}^{-1}$
Explanation:
A Given, magnetic field (B) $=0.25 \mathrm{~T}$ Torque $(\tau)=4.5 \times 10^{-2} \mathrm{~J}$ Angle $(\theta)=30^{\circ}$ We know that, $\tau=M B \sin \theta$ $\therefore \quad \mathrm{M} =\frac{\tau}{\mathrm{B} \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}$ $\mathrm{M} =0.36 \mathrm{JT}^{-1}$
GUJCET 18.04.2022
Magnetism and Matter
153976
A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of original magnet is ' $T_{0}$ ' in a uniform magnetic field $B$. Then, the time period of each part in the same magnetic field is-
1 $\frac{T_{0}}{2}$
2 $\frac{T_{0}}{3}$
3 $\frac{T_{0}}{4}$
4 $\mathrm{T}_{0}$
Explanation:
B Initial time period $\mathrm{T}_{\mathrm{o}}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}} \quad(\mathrm{I}=\text { Moment of inertia })$ $\mathrm{I}=\frac{\mathrm{mL}^{2}}{12}$ When magnet is cut in three parts then new inertia $I^{\prime}=\frac{\frac{\mathrm{m}}{3} \times\left(\frac{\mathrm{L}}{3}\right)^{2}}{12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{mL}^{2}}{27 \times 12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{27} \quad\left(\because \mathrm{I}=\frac{\mathrm{mL}^{2}}{12}\right)$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\frac{\mathrm{M}}{3}$ New time period $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{\mathrm{I}^{\prime}}{\mathrm{M}^{\prime} \mathrm{B}}}$ $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\frac{\mathrm{I}}{27}}{\frac{\mathrm{M}}{3} \times \mathrm{B}}}=2 \pi \sqrt{\frac{1}{9} \frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{1}{3} \times 2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{\mathrm{T}_{\mathrm{o}}}{3}$
CG PET-22.05.2022
Magnetism and Matter
153980
Figure shows a straight wire placed between the pole pieces of a magnet. Induced emf will be developed across the ends of the wire when it is moved towards
1 $\mathrm{N}$
2 $\mathrm{S}$
3 $\mathrm{P}$
4 $\mathrm{Q}$
Explanation:
C Magnetic field lines moves from $\mathrm{N}$ to $\mathrm{S}$ in order to induced emf there should be change in electric flux $e=\frac{d \phi}{d t}$ So, when wire moves towards $\mathrm{P}$ direction it will cut the magnetic field lines result in change in magnetic field.
AP EAMCET-25.08.2021
Magnetism and Matter
153982
The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_{1}$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_{2}$. The declination $\theta$ at the place is
C $\because \tan \delta_{1}=\frac{\tan \theta}{\cos \alpha}$ $\cos \alpha=\frac{\tan \theta}{\tan \delta_{1}}$ Again, $\tan \delta_{2}=\frac{\tan \theta}{\sin \alpha}$ $\sin \alpha=\frac{\tan \theta}{\tan \delta_{2}}$ From equation (ii) dividing by equation (i), we get- $\frac{\sin \alpha}{\cos \alpha}=\frac{\tan \theta / \tan \delta_{2}}{\tan \theta / \tan \delta_{1}}$ $\tan \alpha=\frac{\tan \delta_{1}}{\tan \delta_{2}}$ $\alpha=\tan ^{-1}\left(\frac{\tan \delta_{1}}{\tan \delta_{2}}\right)$
AP EAMCET-19.08.2021
Magnetism and Matter
153983
A magnet of magnetic 2J.T ${ }^{-1}$ is aligned in the direction of magnetic field of 0.1T. What is the net work done to bring the magnet normal to the magnetic field?
1 $0.1 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $1.0 \mathrm{~J}$
4 $2.0 \mathrm{~J}$
Explanation:
B Magnetic moment (M) $=2 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.1 \mathrm{~T}$ Work done, $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ $\theta_{1}=0 \text { (along the direction of magnetic field) }$ $\theta_{2}=90^{\circ}$ $\mathrm{W}=2 \times 0.1\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=0.2(1-0)$ $\mathrm{W}=0.2 \mathrm{~J}$
153975
A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 $T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet ?
1 $0.36 \mathrm{~J} \mathrm{~T}^{-1}$
2 $0.036 \mathrm{~J} \mathrm{~T}^{-1}$
3 $3.6 \mathrm{~J} \mathrm{~T}^{-1}$
4 $36 \mathrm{~J} \mathrm{~T}^{-1}$
Explanation:
A Given, magnetic field (B) $=0.25 \mathrm{~T}$ Torque $(\tau)=4.5 \times 10^{-2} \mathrm{~J}$ Angle $(\theta)=30^{\circ}$ We know that, $\tau=M B \sin \theta$ $\therefore \quad \mathrm{M} =\frac{\tau}{\mathrm{B} \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}$ $\mathrm{M} =0.36 \mathrm{JT}^{-1}$
GUJCET 18.04.2022
Magnetism and Matter
153976
A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of original magnet is ' $T_{0}$ ' in a uniform magnetic field $B$. Then, the time period of each part in the same magnetic field is-
1 $\frac{T_{0}}{2}$
2 $\frac{T_{0}}{3}$
3 $\frac{T_{0}}{4}$
4 $\mathrm{T}_{0}$
Explanation:
B Initial time period $\mathrm{T}_{\mathrm{o}}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}} \quad(\mathrm{I}=\text { Moment of inertia })$ $\mathrm{I}=\frac{\mathrm{mL}^{2}}{12}$ When magnet is cut in three parts then new inertia $I^{\prime}=\frac{\frac{\mathrm{m}}{3} \times\left(\frac{\mathrm{L}}{3}\right)^{2}}{12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{mL}^{2}}{27 \times 12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{27} \quad\left(\because \mathrm{I}=\frac{\mathrm{mL}^{2}}{12}\right)$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\frac{\mathrm{M}}{3}$ New time period $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{\mathrm{I}^{\prime}}{\mathrm{M}^{\prime} \mathrm{B}}}$ $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\frac{\mathrm{I}}{27}}{\frac{\mathrm{M}}{3} \times \mathrm{B}}}=2 \pi \sqrt{\frac{1}{9} \frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{1}{3} \times 2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{\mathrm{T}_{\mathrm{o}}}{3}$
CG PET-22.05.2022
Magnetism and Matter
153980
Figure shows a straight wire placed between the pole pieces of a magnet. Induced emf will be developed across the ends of the wire when it is moved towards
1 $\mathrm{N}$
2 $\mathrm{S}$
3 $\mathrm{P}$
4 $\mathrm{Q}$
Explanation:
C Magnetic field lines moves from $\mathrm{N}$ to $\mathrm{S}$ in order to induced emf there should be change in electric flux $e=\frac{d \phi}{d t}$ So, when wire moves towards $\mathrm{P}$ direction it will cut the magnetic field lines result in change in magnetic field.
AP EAMCET-25.08.2021
Magnetism and Matter
153982
The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_{1}$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_{2}$. The declination $\theta$ at the place is
C $\because \tan \delta_{1}=\frac{\tan \theta}{\cos \alpha}$ $\cos \alpha=\frac{\tan \theta}{\tan \delta_{1}}$ Again, $\tan \delta_{2}=\frac{\tan \theta}{\sin \alpha}$ $\sin \alpha=\frac{\tan \theta}{\tan \delta_{2}}$ From equation (ii) dividing by equation (i), we get- $\frac{\sin \alpha}{\cos \alpha}=\frac{\tan \theta / \tan \delta_{2}}{\tan \theta / \tan \delta_{1}}$ $\tan \alpha=\frac{\tan \delta_{1}}{\tan \delta_{2}}$ $\alpha=\tan ^{-1}\left(\frac{\tan \delta_{1}}{\tan \delta_{2}}\right)$
AP EAMCET-19.08.2021
Magnetism and Matter
153983
A magnet of magnetic 2J.T ${ }^{-1}$ is aligned in the direction of magnetic field of 0.1T. What is the net work done to bring the magnet normal to the magnetic field?
1 $0.1 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $1.0 \mathrm{~J}$
4 $2.0 \mathrm{~J}$
Explanation:
B Magnetic moment (M) $=2 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.1 \mathrm{~T}$ Work done, $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ $\theta_{1}=0 \text { (along the direction of magnetic field) }$ $\theta_{2}=90^{\circ}$ $\mathrm{W}=2 \times 0.1\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=0.2(1-0)$ $\mathrm{W}=0.2 \mathrm{~J}$
153975
A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 $T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet ?
1 $0.36 \mathrm{~J} \mathrm{~T}^{-1}$
2 $0.036 \mathrm{~J} \mathrm{~T}^{-1}$
3 $3.6 \mathrm{~J} \mathrm{~T}^{-1}$
4 $36 \mathrm{~J} \mathrm{~T}^{-1}$
Explanation:
A Given, magnetic field (B) $=0.25 \mathrm{~T}$ Torque $(\tau)=4.5 \times 10^{-2} \mathrm{~J}$ Angle $(\theta)=30^{\circ}$ We know that, $\tau=M B \sin \theta$ $\therefore \quad \mathrm{M} =\frac{\tau}{\mathrm{B} \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}$ $\mathrm{M} =0.36 \mathrm{JT}^{-1}$
GUJCET 18.04.2022
Magnetism and Matter
153976
A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of original magnet is ' $T_{0}$ ' in a uniform magnetic field $B$. Then, the time period of each part in the same magnetic field is-
1 $\frac{T_{0}}{2}$
2 $\frac{T_{0}}{3}$
3 $\frac{T_{0}}{4}$
4 $\mathrm{T}_{0}$
Explanation:
B Initial time period $\mathrm{T}_{\mathrm{o}}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}} \quad(\mathrm{I}=\text { Moment of inertia })$ $\mathrm{I}=\frac{\mathrm{mL}^{2}}{12}$ When magnet is cut in three parts then new inertia $I^{\prime}=\frac{\frac{\mathrm{m}}{3} \times\left(\frac{\mathrm{L}}{3}\right)^{2}}{12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{mL}^{2}}{27 \times 12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{27} \quad\left(\because \mathrm{I}=\frac{\mathrm{mL}^{2}}{12}\right)$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\frac{\mathrm{M}}{3}$ New time period $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{\mathrm{I}^{\prime}}{\mathrm{M}^{\prime} \mathrm{B}}}$ $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\frac{\mathrm{I}}{27}}{\frac{\mathrm{M}}{3} \times \mathrm{B}}}=2 \pi \sqrt{\frac{1}{9} \frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{1}{3} \times 2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{\mathrm{T}_{\mathrm{o}}}{3}$
CG PET-22.05.2022
Magnetism and Matter
153980
Figure shows a straight wire placed between the pole pieces of a magnet. Induced emf will be developed across the ends of the wire when it is moved towards
1 $\mathrm{N}$
2 $\mathrm{S}$
3 $\mathrm{P}$
4 $\mathrm{Q}$
Explanation:
C Magnetic field lines moves from $\mathrm{N}$ to $\mathrm{S}$ in order to induced emf there should be change in electric flux $e=\frac{d \phi}{d t}$ So, when wire moves towards $\mathrm{P}$ direction it will cut the magnetic field lines result in change in magnetic field.
AP EAMCET-25.08.2021
Magnetism and Matter
153982
The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_{1}$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_{2}$. The declination $\theta$ at the place is
C $\because \tan \delta_{1}=\frac{\tan \theta}{\cos \alpha}$ $\cos \alpha=\frac{\tan \theta}{\tan \delta_{1}}$ Again, $\tan \delta_{2}=\frac{\tan \theta}{\sin \alpha}$ $\sin \alpha=\frac{\tan \theta}{\tan \delta_{2}}$ From equation (ii) dividing by equation (i), we get- $\frac{\sin \alpha}{\cos \alpha}=\frac{\tan \theta / \tan \delta_{2}}{\tan \theta / \tan \delta_{1}}$ $\tan \alpha=\frac{\tan \delta_{1}}{\tan \delta_{2}}$ $\alpha=\tan ^{-1}\left(\frac{\tan \delta_{1}}{\tan \delta_{2}}\right)$
AP EAMCET-19.08.2021
Magnetism and Matter
153983
A magnet of magnetic 2J.T ${ }^{-1}$ is aligned in the direction of magnetic field of 0.1T. What is the net work done to bring the magnet normal to the magnetic field?
1 $0.1 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $1.0 \mathrm{~J}$
4 $2.0 \mathrm{~J}$
Explanation:
B Magnetic moment (M) $=2 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.1 \mathrm{~T}$ Work done, $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ $\theta_{1}=0 \text { (along the direction of magnetic field) }$ $\theta_{2}=90^{\circ}$ $\mathrm{W}=2 \times 0.1\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=0.2(1-0)$ $\mathrm{W}=0.2 \mathrm{~J}$
153975
A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 $T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet ?
1 $0.36 \mathrm{~J} \mathrm{~T}^{-1}$
2 $0.036 \mathrm{~J} \mathrm{~T}^{-1}$
3 $3.6 \mathrm{~J} \mathrm{~T}^{-1}$
4 $36 \mathrm{~J} \mathrm{~T}^{-1}$
Explanation:
A Given, magnetic field (B) $=0.25 \mathrm{~T}$ Torque $(\tau)=4.5 \times 10^{-2} \mathrm{~J}$ Angle $(\theta)=30^{\circ}$ We know that, $\tau=M B \sin \theta$ $\therefore \quad \mathrm{M} =\frac{\tau}{\mathrm{B} \sin \theta}=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}$ $\mathrm{M} =0.36 \mathrm{JT}^{-1}$
GUJCET 18.04.2022
Magnetism and Matter
153976
A magnet is cut in three equal parts by cutting it perpendicular to its length. The time period of original magnet is ' $T_{0}$ ' in a uniform magnetic field $B$. Then, the time period of each part in the same magnetic field is-
1 $\frac{T_{0}}{2}$
2 $\frac{T_{0}}{3}$
3 $\frac{T_{0}}{4}$
4 $\mathrm{T}_{0}$
Explanation:
B Initial time period $\mathrm{T}_{\mathrm{o}}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}} \quad(\mathrm{I}=\text { Moment of inertia })$ $\mathrm{I}=\frac{\mathrm{mL}^{2}}{12}$ When magnet is cut in three parts then new inertia $I^{\prime}=\frac{\frac{\mathrm{m}}{3} \times\left(\frac{\mathrm{L}}{3}\right)^{2}}{12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{mL}^{2}}{27 \times 12}$ $\mathrm{I}^{\prime}=\frac{\mathrm{I}}{27} \quad\left(\because \mathrm{I}=\frac{\mathrm{mL}^{2}}{12}\right)$ New magnetic moment $\left(\mathrm{M}^{\prime}\right)=\frac{\mathrm{M}}{3}$ New time period $\left(\mathrm{T}^{\prime}\right)=2 \pi \sqrt{\frac{\mathrm{I}^{\prime}}{\mathrm{M}^{\prime} \mathrm{B}}}$ $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\frac{\mathrm{I}}{27}}{\frac{\mathrm{M}}{3} \times \mathrm{B}}}=2 \pi \sqrt{\frac{1}{9} \frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{1}{3} \times 2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$ $\mathrm{T}^{\prime}=\frac{\mathrm{T}_{\mathrm{o}}}{3}$
CG PET-22.05.2022
Magnetism and Matter
153980
Figure shows a straight wire placed between the pole pieces of a magnet. Induced emf will be developed across the ends of the wire when it is moved towards
1 $\mathrm{N}$
2 $\mathrm{S}$
3 $\mathrm{P}$
4 $\mathrm{Q}$
Explanation:
C Magnetic field lines moves from $\mathrm{N}$ to $\mathrm{S}$ in order to induced emf there should be change in electric flux $e=\frac{d \phi}{d t}$ So, when wire moves towards $\mathrm{P}$ direction it will cut the magnetic field lines result in change in magnetic field.
AP EAMCET-25.08.2021
Magnetism and Matter
153982
The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_{1}$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_{2}$. The declination $\theta$ at the place is
C $\because \tan \delta_{1}=\frac{\tan \theta}{\cos \alpha}$ $\cos \alpha=\frac{\tan \theta}{\tan \delta_{1}}$ Again, $\tan \delta_{2}=\frac{\tan \theta}{\sin \alpha}$ $\sin \alpha=\frac{\tan \theta}{\tan \delta_{2}}$ From equation (ii) dividing by equation (i), we get- $\frac{\sin \alpha}{\cos \alpha}=\frac{\tan \theta / \tan \delta_{2}}{\tan \theta / \tan \delta_{1}}$ $\tan \alpha=\frac{\tan \delta_{1}}{\tan \delta_{2}}$ $\alpha=\tan ^{-1}\left(\frac{\tan \delta_{1}}{\tan \delta_{2}}\right)$
AP EAMCET-19.08.2021
Magnetism and Matter
153983
A magnet of magnetic 2J.T ${ }^{-1}$ is aligned in the direction of magnetic field of 0.1T. What is the net work done to bring the magnet normal to the magnetic field?
1 $0.1 \mathrm{~J}$
2 $0.2 \mathrm{~J}$
3 $1.0 \mathrm{~J}$
4 $2.0 \mathrm{~J}$
Explanation:
B Magnetic moment (M) $=2 \mathrm{~J} / \mathrm{T}$ Magnetic field $(\mathrm{B})=0.1 \mathrm{~T}$ Work done, $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ $\theta_{1}=0 \text { (along the direction of magnetic field) }$ $\theta_{2}=90^{\circ}$ $\mathrm{W}=2 \times 0.1\left(\cos 0^{\circ}-\cos 90^{\circ}\right)$ $\mathrm{W}=0.2(1-0)$ $\mathrm{W}=0.2 \mathrm{~J}$
