153987
A bar magnet is hanging freely in a magnetic field, If work done to rotate it from equilibrium position to $60^{\circ}$ is $W_{1}$ and then $60^{\circ}$ to $90^{\circ}$ is $W_{2}$, then ratio of $W_{1}: W_{2}$ will be
1 $1: 1$
2 $2: 1$
3 $\sqrt{3: 2}$
4 $2: \sqrt{3}$
Explanation:
A We know that, Work done $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ When it rotate from equilibrium position to $60^{\circ}$, it means $\theta_{1}=0 \& \theta_{2}=60^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{1}=\frac{\mathrm{MB}}{2}$ Similarly, when it rotate from $60^{\circ}$ to $90^{\circ}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\frac{1}{2}-0\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 60^{\circ}-\cos 90^{\circ}\right)$ Dividing equation (i) and (ii), we get- $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\frac{\mathrm{MB}}{2}}{\frac{\mathrm{MB}}{2}}=\frac{1}{1}$ $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{1}{1}$ So, $\mathrm{W}_{1}: \mathrm{W}_{2}=1: 1$
Tripura-2020
Magnetism and Matter
153988
An electron revolving in a circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{2}$
2 $\frac{M}{4}$
3 $2 \mathrm{M}$
4 $\mathrm{M}$
Explanation:
C We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And, $I=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{e} \times \mathrm{f}$ $\mathrm{M}=\mathrm{IA}=\mathrm{ef} \times \mathrm{A}$ $\because \quad \mathrm{M} \propto \mathrm{f}$ $\therefore \quad \frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f_{1}}{f_{2}}$ $\frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f}{2 f}=\frac{1}{2}$ $M_{\text {final }}=2 M$
MHT-CET 2020
Magnetism and Matter
153989
An electron of charge ' $\mathrm{e}$ ' is revolving in a fixed orbit of radius ' $r$ ' with frequency ' $f$ '. Its magnetic dipole moment is
1 $\pi^{2}$ efr
2 $\pi$ efr ${ }^{2}$
3 $\pi$ efr
4 $\pi^{2}$ efr ${ }^{2}$
Explanation:
B Given, radius of orbit $=r$, area $(A)=\pi r^{2}$, frequency $=\mathrm{f}$ For a revolving electron- $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=\mathrm{ef}$ Where, $\mathrm{e}=$ Charge of electron $\mathrm{t}=\text { Time period of revolution of the electron }$ Magnet dipole moment $(\mathrm{M})=\mathrm{IA}$ Putting the value in above equation, we get- $\mathrm{M}=\mathrm{ef} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\pi \mathrm{efr} \mathrm{r}^{2}$
MHT-CET 2020
Magnetism and Matter
153990
A cylindrical magnetic rod has length $5 \mathrm{~cm}$ and diameter $1 \mathrm{~cm}$. It has uniform magnetization $5.3 \times 10^{3} \frac{\mathrm{A}}{\mathrm{m}}$. Its net magnetic dipole moment is nearly $\left(\pi=\frac{22}{7}\right)$
153987
A bar magnet is hanging freely in a magnetic field, If work done to rotate it from equilibrium position to $60^{\circ}$ is $W_{1}$ and then $60^{\circ}$ to $90^{\circ}$ is $W_{2}$, then ratio of $W_{1}: W_{2}$ will be
1 $1: 1$
2 $2: 1$
3 $\sqrt{3: 2}$
4 $2: \sqrt{3}$
Explanation:
A We know that, Work done $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ When it rotate from equilibrium position to $60^{\circ}$, it means $\theta_{1}=0 \& \theta_{2}=60^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{1}=\frac{\mathrm{MB}}{2}$ Similarly, when it rotate from $60^{\circ}$ to $90^{\circ}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\frac{1}{2}-0\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 60^{\circ}-\cos 90^{\circ}\right)$ Dividing equation (i) and (ii), we get- $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\frac{\mathrm{MB}}{2}}{\frac{\mathrm{MB}}{2}}=\frac{1}{1}$ $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{1}{1}$ So, $\mathrm{W}_{1}: \mathrm{W}_{2}=1: 1$
Tripura-2020
Magnetism and Matter
153988
An electron revolving in a circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{2}$
2 $\frac{M}{4}$
3 $2 \mathrm{M}$
4 $\mathrm{M}$
Explanation:
C We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And, $I=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{e} \times \mathrm{f}$ $\mathrm{M}=\mathrm{IA}=\mathrm{ef} \times \mathrm{A}$ $\because \quad \mathrm{M} \propto \mathrm{f}$ $\therefore \quad \frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f_{1}}{f_{2}}$ $\frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f}{2 f}=\frac{1}{2}$ $M_{\text {final }}=2 M$
MHT-CET 2020
Magnetism and Matter
153989
An electron of charge ' $\mathrm{e}$ ' is revolving in a fixed orbit of radius ' $r$ ' with frequency ' $f$ '. Its magnetic dipole moment is
1 $\pi^{2}$ efr
2 $\pi$ efr ${ }^{2}$
3 $\pi$ efr
4 $\pi^{2}$ efr ${ }^{2}$
Explanation:
B Given, radius of orbit $=r$, area $(A)=\pi r^{2}$, frequency $=\mathrm{f}$ For a revolving electron- $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=\mathrm{ef}$ Where, $\mathrm{e}=$ Charge of electron $\mathrm{t}=\text { Time period of revolution of the electron }$ Magnet dipole moment $(\mathrm{M})=\mathrm{IA}$ Putting the value in above equation, we get- $\mathrm{M}=\mathrm{ef} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\pi \mathrm{efr} \mathrm{r}^{2}$
MHT-CET 2020
Magnetism and Matter
153990
A cylindrical magnetic rod has length $5 \mathrm{~cm}$ and diameter $1 \mathrm{~cm}$. It has uniform magnetization $5.3 \times 10^{3} \frac{\mathrm{A}}{\mathrm{m}}$. Its net magnetic dipole moment is nearly $\left(\pi=\frac{22}{7}\right)$
153987
A bar magnet is hanging freely in a magnetic field, If work done to rotate it from equilibrium position to $60^{\circ}$ is $W_{1}$ and then $60^{\circ}$ to $90^{\circ}$ is $W_{2}$, then ratio of $W_{1}: W_{2}$ will be
1 $1: 1$
2 $2: 1$
3 $\sqrt{3: 2}$
4 $2: \sqrt{3}$
Explanation:
A We know that, Work done $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ When it rotate from equilibrium position to $60^{\circ}$, it means $\theta_{1}=0 \& \theta_{2}=60^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{1}=\frac{\mathrm{MB}}{2}$ Similarly, when it rotate from $60^{\circ}$ to $90^{\circ}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\frac{1}{2}-0\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 60^{\circ}-\cos 90^{\circ}\right)$ Dividing equation (i) and (ii), we get- $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\frac{\mathrm{MB}}{2}}{\frac{\mathrm{MB}}{2}}=\frac{1}{1}$ $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{1}{1}$ So, $\mathrm{W}_{1}: \mathrm{W}_{2}=1: 1$
Tripura-2020
Magnetism and Matter
153988
An electron revolving in a circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{2}$
2 $\frac{M}{4}$
3 $2 \mathrm{M}$
4 $\mathrm{M}$
Explanation:
C We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And, $I=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{e} \times \mathrm{f}$ $\mathrm{M}=\mathrm{IA}=\mathrm{ef} \times \mathrm{A}$ $\because \quad \mathrm{M} \propto \mathrm{f}$ $\therefore \quad \frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f_{1}}{f_{2}}$ $\frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f}{2 f}=\frac{1}{2}$ $M_{\text {final }}=2 M$
MHT-CET 2020
Magnetism and Matter
153989
An electron of charge ' $\mathrm{e}$ ' is revolving in a fixed orbit of radius ' $r$ ' with frequency ' $f$ '. Its magnetic dipole moment is
1 $\pi^{2}$ efr
2 $\pi$ efr ${ }^{2}$
3 $\pi$ efr
4 $\pi^{2}$ efr ${ }^{2}$
Explanation:
B Given, radius of orbit $=r$, area $(A)=\pi r^{2}$, frequency $=\mathrm{f}$ For a revolving electron- $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=\mathrm{ef}$ Where, $\mathrm{e}=$ Charge of electron $\mathrm{t}=\text { Time period of revolution of the electron }$ Magnet dipole moment $(\mathrm{M})=\mathrm{IA}$ Putting the value in above equation, we get- $\mathrm{M}=\mathrm{ef} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\pi \mathrm{efr} \mathrm{r}^{2}$
MHT-CET 2020
Magnetism and Matter
153990
A cylindrical magnetic rod has length $5 \mathrm{~cm}$ and diameter $1 \mathrm{~cm}$. It has uniform magnetization $5.3 \times 10^{3} \frac{\mathrm{A}}{\mathrm{m}}$. Its net magnetic dipole moment is nearly $\left(\pi=\frac{22}{7}\right)$
153987
A bar magnet is hanging freely in a magnetic field, If work done to rotate it from equilibrium position to $60^{\circ}$ is $W_{1}$ and then $60^{\circ}$ to $90^{\circ}$ is $W_{2}$, then ratio of $W_{1}: W_{2}$ will be
1 $1: 1$
2 $2: 1$
3 $\sqrt{3: 2}$
4 $2: \sqrt{3}$
Explanation:
A We know that, Work done $(\mathrm{W})=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ When it rotate from equilibrium position to $60^{\circ}$, it means $\theta_{1}=0 \& \theta_{2}=60^{\circ}$ $\mathrm{W}_{1}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$ $\mathrm{W}_{1}=\mathrm{MB}\left(1-\frac{1}{2}\right)$ $\mathrm{W}_{1}=\frac{\mathrm{MB}}{2}$ Similarly, when it rotate from $60^{\circ}$ to $90^{\circ}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\frac{1}{2}-0\right)$ $\mathrm{W}_{2}=\frac{\mathrm{MB}}{2}$ $\mathrm{W}_{2}=\mathrm{MB}\left(\cos 60^{\circ}-\cos 90^{\circ}\right)$ Dividing equation (i) and (ii), we get- $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\frac{\mathrm{MB}}{2}}{\frac{\mathrm{MB}}{2}}=\frac{1}{1}$ $\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{1}{1}$ So, $\mathrm{W}_{1}: \mathrm{W}_{2}=1: 1$
Tripura-2020
Magnetism and Matter
153988
An electron revolving in a circular orbit of radius ' $r$ ' with velocity ' $V$ ' and frequency ' $v$ ' has orbital magnetic moment ' $M$ '. If the frequency of revolution is doubled then the new magnetic moment will be
1 $\frac{M}{2}$
2 $\frac{M}{4}$
3 $2 \mathrm{M}$
4 $\mathrm{M}$
Explanation:
C We know that, Magnetic moment $(\mathrm{M})=\mathrm{IA}$ And, $I=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{e} \times \mathrm{f}$ $\mathrm{M}=\mathrm{IA}=\mathrm{ef} \times \mathrm{A}$ $\because \quad \mathrm{M} \propto \mathrm{f}$ $\therefore \quad \frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f_{1}}{f_{2}}$ $\frac{M_{\text {initial }}}{M_{\text {final }}}=\frac{f}{2 f}=\frac{1}{2}$ $M_{\text {final }}=2 M$
MHT-CET 2020
Magnetism and Matter
153989
An electron of charge ' $\mathrm{e}$ ' is revolving in a fixed orbit of radius ' $r$ ' with frequency ' $f$ '. Its magnetic dipole moment is
1 $\pi^{2}$ efr
2 $\pi$ efr ${ }^{2}$
3 $\pi$ efr
4 $\pi^{2}$ efr ${ }^{2}$
Explanation:
B Given, radius of orbit $=r$, area $(A)=\pi r^{2}$, frequency $=\mathrm{f}$ For a revolving electron- $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{t}}=\mathrm{ef}$ Where, $\mathrm{e}=$ Charge of electron $\mathrm{t}=\text { Time period of revolution of the electron }$ Magnet dipole moment $(\mathrm{M})=\mathrm{IA}$ Putting the value in above equation, we get- $\mathrm{M}=\mathrm{ef} \times \pi \mathrm{r}^{2}$ $\mathrm{M}=\pi \mathrm{efr} \mathrm{r}^{2}$
MHT-CET 2020
Magnetism and Matter
153990
A cylindrical magnetic rod has length $5 \mathrm{~cm}$ and diameter $1 \mathrm{~cm}$. It has uniform magnetization $5.3 \times 10^{3} \frac{\mathrm{A}}{\mathrm{m}}$. Its net magnetic dipole moment is nearly $\left(\pi=\frac{22}{7}\right)$
