154377
Assertion : The ferromagnetic substance do not obey Curie's law . Reason: At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B The susceptibility of ferromagnetic substance decreases with the rise of temperature. After curies point the susceptibility of ferromagnetic substance varies inversely with its absolute temperature ferromagnetic substance obey's curies law only above its curie point.
AIIMS-2014
Magnetism and Matter
154379
A magnet makes 40 oscillations per minute at a place having magnetic intensity of $0.1 \times 10^{-5}$ tesla. At another place it takes $2.5 \mathrm{sec}$ to complete one oscillation. The value of earth's horizontal field at that place is
1 $0.76 \times 10^{-6}$ tesla
2 $0.18 \times 10^{-6}$ tesla
3 $0.09 \times 10^{-6}$ tesla
4 $0.36 \times 10^{-6}$ tesla
Explanation:
D Given, Number of oscillation $=40$ oscillation per minute. Magnetic field intensity $\left(\mathrm{B}_{\mathrm{H}_{1}}\right)=0.1 \times 10^{-5} \mathrm{~T}$ Calculate earth's horizontal field, $\mathrm{B}_{\mathrm{H}_{2}}=$ ? We know that, $\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{mB}_{\mathrm{H}}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}_{\mathrm{H}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{\mathrm{~B}_{\mathrm{H}_{1}}}}$ Where $\mathrm{T}_{1}=\frac{60}{40}, \mathrm{~T}_{2}=2.5$ $\frac{60 / 40}{2.5}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ $0.6=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ Squares on both side $0.36 =\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 0.1 \times 10^{-5}$ $=0.036 \times 10^{-5}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 10^{-6} \mathrm{~T}$
AIIMS-2011
Magnetism and Matter
154381
Of the following Fig., the lines of magnetic induction due to a magnet $\mathrm{SN}$, are given by
1 1
2 2
3 3
4 4
Explanation:
A Magnetic field lines emerge from north pole of a magnet and enter the south pole. Thus the field lines move from south pole to north pole inside the bar magnet. As lines of magnetic induction B are continuous curves, they run continuously through the bar and outside. magnetic field lines does not cut anywhere.
AIIMS-2012
Magnetism and Matter
154389
The magnetic moment produced in a substance of $1 \mathrm{~g}$ is $6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$. If its density is $5 \mathrm{~g} \mathrm{~cm}^{-3}$, then the intensity of magnetization in $\mathrm{A} / \mathrm{m}$ will be
1 3.0
2 $3 \times 10^{-6}$
3 $8.3 \times 10^{6}$
4 $1.2 \times 10^{-7}$
Explanation:
A Given that, Magnetic moment $=6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$ Mass $=1 \mathrm{gm}=10^{-3} \mathrm{~kg}$ Density $=5 \mathrm{gcm}^{-3}=\frac{5 \times 10^{-3} \mathrm{~kg}}{\left(10^{-2}\right)^{3} \mathrm{~m}^{3}}=5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Volume, $\mathrm{V}=\frac{\text { Mass }}{\text { density }}=\frac{10^{-3}}{5 \times 10^{3}}$ $=0.2 \times 10^{-6} \mathrm{~m}^{3}$ Magnetization $=\frac{\text { Magnetic moment }}{\text { volume }}$ $=\frac{6 \times 10^{-7}}{0.2 \times 10^{-6}}=\frac{6 \times 10^{-7}}{2 \times 10^{-7}}$ $=3.0 \mathrm{~A} / \mathrm{m}$
BCECE-2012
Magnetism and Matter
154397
A magnetic needle lying parallel to the magnetic field requires $W$ units of work to turn it through an angle $45^{\circ}$. The torque required to maintain the needle in this position will be
154377
Assertion : The ferromagnetic substance do not obey Curie's law . Reason: At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B The susceptibility of ferromagnetic substance decreases with the rise of temperature. After curies point the susceptibility of ferromagnetic substance varies inversely with its absolute temperature ferromagnetic substance obey's curies law only above its curie point.
AIIMS-2014
Magnetism and Matter
154379
A magnet makes 40 oscillations per minute at a place having magnetic intensity of $0.1 \times 10^{-5}$ tesla. At another place it takes $2.5 \mathrm{sec}$ to complete one oscillation. The value of earth's horizontal field at that place is
1 $0.76 \times 10^{-6}$ tesla
2 $0.18 \times 10^{-6}$ tesla
3 $0.09 \times 10^{-6}$ tesla
4 $0.36 \times 10^{-6}$ tesla
Explanation:
D Given, Number of oscillation $=40$ oscillation per minute. Magnetic field intensity $\left(\mathrm{B}_{\mathrm{H}_{1}}\right)=0.1 \times 10^{-5} \mathrm{~T}$ Calculate earth's horizontal field, $\mathrm{B}_{\mathrm{H}_{2}}=$ ? We know that, $\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{mB}_{\mathrm{H}}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}_{\mathrm{H}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{\mathrm{~B}_{\mathrm{H}_{1}}}}$ Where $\mathrm{T}_{1}=\frac{60}{40}, \mathrm{~T}_{2}=2.5$ $\frac{60 / 40}{2.5}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ $0.6=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ Squares on both side $0.36 =\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 0.1 \times 10^{-5}$ $=0.036 \times 10^{-5}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 10^{-6} \mathrm{~T}$
AIIMS-2011
Magnetism and Matter
154381
Of the following Fig., the lines of magnetic induction due to a magnet $\mathrm{SN}$, are given by
1 1
2 2
3 3
4 4
Explanation:
A Magnetic field lines emerge from north pole of a magnet and enter the south pole. Thus the field lines move from south pole to north pole inside the bar magnet. As lines of magnetic induction B are continuous curves, they run continuously through the bar and outside. magnetic field lines does not cut anywhere.
AIIMS-2012
Magnetism and Matter
154389
The magnetic moment produced in a substance of $1 \mathrm{~g}$ is $6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$. If its density is $5 \mathrm{~g} \mathrm{~cm}^{-3}$, then the intensity of magnetization in $\mathrm{A} / \mathrm{m}$ will be
1 3.0
2 $3 \times 10^{-6}$
3 $8.3 \times 10^{6}$
4 $1.2 \times 10^{-7}$
Explanation:
A Given that, Magnetic moment $=6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$ Mass $=1 \mathrm{gm}=10^{-3} \mathrm{~kg}$ Density $=5 \mathrm{gcm}^{-3}=\frac{5 \times 10^{-3} \mathrm{~kg}}{\left(10^{-2}\right)^{3} \mathrm{~m}^{3}}=5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Volume, $\mathrm{V}=\frac{\text { Mass }}{\text { density }}=\frac{10^{-3}}{5 \times 10^{3}}$ $=0.2 \times 10^{-6} \mathrm{~m}^{3}$ Magnetization $=\frac{\text { Magnetic moment }}{\text { volume }}$ $=\frac{6 \times 10^{-7}}{0.2 \times 10^{-6}}=\frac{6 \times 10^{-7}}{2 \times 10^{-7}}$ $=3.0 \mathrm{~A} / \mathrm{m}$
BCECE-2012
Magnetism and Matter
154397
A magnetic needle lying parallel to the magnetic field requires $W$ units of work to turn it through an angle $45^{\circ}$. The torque required to maintain the needle in this position will be
154377
Assertion : The ferromagnetic substance do not obey Curie's law . Reason: At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B The susceptibility of ferromagnetic substance decreases with the rise of temperature. After curies point the susceptibility of ferromagnetic substance varies inversely with its absolute temperature ferromagnetic substance obey's curies law only above its curie point.
AIIMS-2014
Magnetism and Matter
154379
A magnet makes 40 oscillations per minute at a place having magnetic intensity of $0.1 \times 10^{-5}$ tesla. At another place it takes $2.5 \mathrm{sec}$ to complete one oscillation. The value of earth's horizontal field at that place is
1 $0.76 \times 10^{-6}$ tesla
2 $0.18 \times 10^{-6}$ tesla
3 $0.09 \times 10^{-6}$ tesla
4 $0.36 \times 10^{-6}$ tesla
Explanation:
D Given, Number of oscillation $=40$ oscillation per minute. Magnetic field intensity $\left(\mathrm{B}_{\mathrm{H}_{1}}\right)=0.1 \times 10^{-5} \mathrm{~T}$ Calculate earth's horizontal field, $\mathrm{B}_{\mathrm{H}_{2}}=$ ? We know that, $\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{mB}_{\mathrm{H}}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}_{\mathrm{H}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{\mathrm{~B}_{\mathrm{H}_{1}}}}$ Where $\mathrm{T}_{1}=\frac{60}{40}, \mathrm{~T}_{2}=2.5$ $\frac{60 / 40}{2.5}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ $0.6=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ Squares on both side $0.36 =\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 0.1 \times 10^{-5}$ $=0.036 \times 10^{-5}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 10^{-6} \mathrm{~T}$
AIIMS-2011
Magnetism and Matter
154381
Of the following Fig., the lines of magnetic induction due to a magnet $\mathrm{SN}$, are given by
1 1
2 2
3 3
4 4
Explanation:
A Magnetic field lines emerge from north pole of a magnet and enter the south pole. Thus the field lines move from south pole to north pole inside the bar magnet. As lines of magnetic induction B are continuous curves, they run continuously through the bar and outside. magnetic field lines does not cut anywhere.
AIIMS-2012
Magnetism and Matter
154389
The magnetic moment produced in a substance of $1 \mathrm{~g}$ is $6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$. If its density is $5 \mathrm{~g} \mathrm{~cm}^{-3}$, then the intensity of magnetization in $\mathrm{A} / \mathrm{m}$ will be
1 3.0
2 $3 \times 10^{-6}$
3 $8.3 \times 10^{6}$
4 $1.2 \times 10^{-7}$
Explanation:
A Given that, Magnetic moment $=6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$ Mass $=1 \mathrm{gm}=10^{-3} \mathrm{~kg}$ Density $=5 \mathrm{gcm}^{-3}=\frac{5 \times 10^{-3} \mathrm{~kg}}{\left(10^{-2}\right)^{3} \mathrm{~m}^{3}}=5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Volume, $\mathrm{V}=\frac{\text { Mass }}{\text { density }}=\frac{10^{-3}}{5 \times 10^{3}}$ $=0.2 \times 10^{-6} \mathrm{~m}^{3}$ Magnetization $=\frac{\text { Magnetic moment }}{\text { volume }}$ $=\frac{6 \times 10^{-7}}{0.2 \times 10^{-6}}=\frac{6 \times 10^{-7}}{2 \times 10^{-7}}$ $=3.0 \mathrm{~A} / \mathrm{m}$
BCECE-2012
Magnetism and Matter
154397
A magnetic needle lying parallel to the magnetic field requires $W$ units of work to turn it through an angle $45^{\circ}$. The torque required to maintain the needle in this position will be
154377
Assertion : The ferromagnetic substance do not obey Curie's law . Reason: At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B The susceptibility of ferromagnetic substance decreases with the rise of temperature. After curies point the susceptibility of ferromagnetic substance varies inversely with its absolute temperature ferromagnetic substance obey's curies law only above its curie point.
AIIMS-2014
Magnetism and Matter
154379
A magnet makes 40 oscillations per minute at a place having magnetic intensity of $0.1 \times 10^{-5}$ tesla. At another place it takes $2.5 \mathrm{sec}$ to complete one oscillation. The value of earth's horizontal field at that place is
1 $0.76 \times 10^{-6}$ tesla
2 $0.18 \times 10^{-6}$ tesla
3 $0.09 \times 10^{-6}$ tesla
4 $0.36 \times 10^{-6}$ tesla
Explanation:
D Given, Number of oscillation $=40$ oscillation per minute. Magnetic field intensity $\left(\mathrm{B}_{\mathrm{H}_{1}}\right)=0.1 \times 10^{-5} \mathrm{~T}$ Calculate earth's horizontal field, $\mathrm{B}_{\mathrm{H}_{2}}=$ ? We know that, $\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{mB}_{\mathrm{H}}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}_{\mathrm{H}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{\mathrm{~B}_{\mathrm{H}_{1}}}}$ Where $\mathrm{T}_{1}=\frac{60}{40}, \mathrm{~T}_{2}=2.5$ $\frac{60 / 40}{2.5}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ $0.6=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ Squares on both side $0.36 =\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 0.1 \times 10^{-5}$ $=0.036 \times 10^{-5}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 10^{-6} \mathrm{~T}$
AIIMS-2011
Magnetism and Matter
154381
Of the following Fig., the lines of magnetic induction due to a magnet $\mathrm{SN}$, are given by
1 1
2 2
3 3
4 4
Explanation:
A Magnetic field lines emerge from north pole of a magnet and enter the south pole. Thus the field lines move from south pole to north pole inside the bar magnet. As lines of magnetic induction B are continuous curves, they run continuously through the bar and outside. magnetic field lines does not cut anywhere.
AIIMS-2012
Magnetism and Matter
154389
The magnetic moment produced in a substance of $1 \mathrm{~g}$ is $6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$. If its density is $5 \mathrm{~g} \mathrm{~cm}^{-3}$, then the intensity of magnetization in $\mathrm{A} / \mathrm{m}$ will be
1 3.0
2 $3 \times 10^{-6}$
3 $8.3 \times 10^{6}$
4 $1.2 \times 10^{-7}$
Explanation:
A Given that, Magnetic moment $=6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$ Mass $=1 \mathrm{gm}=10^{-3} \mathrm{~kg}$ Density $=5 \mathrm{gcm}^{-3}=\frac{5 \times 10^{-3} \mathrm{~kg}}{\left(10^{-2}\right)^{3} \mathrm{~m}^{3}}=5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Volume, $\mathrm{V}=\frac{\text { Mass }}{\text { density }}=\frac{10^{-3}}{5 \times 10^{3}}$ $=0.2 \times 10^{-6} \mathrm{~m}^{3}$ Magnetization $=\frac{\text { Magnetic moment }}{\text { volume }}$ $=\frac{6 \times 10^{-7}}{0.2 \times 10^{-6}}=\frac{6 \times 10^{-7}}{2 \times 10^{-7}}$ $=3.0 \mathrm{~A} / \mathrm{m}$
BCECE-2012
Magnetism and Matter
154397
A magnetic needle lying parallel to the magnetic field requires $W$ units of work to turn it through an angle $45^{\circ}$. The torque required to maintain the needle in this position will be
154377
Assertion : The ferromagnetic substance do not obey Curie's law . Reason: At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
B The susceptibility of ferromagnetic substance decreases with the rise of temperature. After curies point the susceptibility of ferromagnetic substance varies inversely with its absolute temperature ferromagnetic substance obey's curies law only above its curie point.
AIIMS-2014
Magnetism and Matter
154379
A magnet makes 40 oscillations per minute at a place having magnetic intensity of $0.1 \times 10^{-5}$ tesla. At another place it takes $2.5 \mathrm{sec}$ to complete one oscillation. The value of earth's horizontal field at that place is
1 $0.76 \times 10^{-6}$ tesla
2 $0.18 \times 10^{-6}$ tesla
3 $0.09 \times 10^{-6}$ tesla
4 $0.36 \times 10^{-6}$ tesla
Explanation:
D Given, Number of oscillation $=40$ oscillation per minute. Magnetic field intensity $\left(\mathrm{B}_{\mathrm{H}_{1}}\right)=0.1 \times 10^{-5} \mathrm{~T}$ Calculate earth's horizontal field, $\mathrm{B}_{\mathrm{H}_{2}}=$ ? We know that, $\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{mB}_{\mathrm{H}}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}_{\mathrm{H}}}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{\mathrm{~B}_{\mathrm{H}_{1}}}}$ Where $\mathrm{T}_{1}=\frac{60}{40}, \mathrm{~T}_{2}=2.5$ $\frac{60 / 40}{2.5}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ $0.6=\sqrt{\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}}$ Squares on both side $0.36 =\frac{\mathrm{B}_{\mathrm{H}_{2}}}{0.1 \times 10^{-5}}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 0.1 \times 10^{-5}$ $=0.036 \times 10^{-5}$ $\mathrm{~B}_{\mathrm{H}_{2}} =0.36 \times 10^{-6} \mathrm{~T}$
AIIMS-2011
Magnetism and Matter
154381
Of the following Fig., the lines of magnetic induction due to a magnet $\mathrm{SN}$, are given by
1 1
2 2
3 3
4 4
Explanation:
A Magnetic field lines emerge from north pole of a magnet and enter the south pole. Thus the field lines move from south pole to north pole inside the bar magnet. As lines of magnetic induction B are continuous curves, they run continuously through the bar and outside. magnetic field lines does not cut anywhere.
AIIMS-2012
Magnetism and Matter
154389
The magnetic moment produced in a substance of $1 \mathrm{~g}$ is $6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$. If its density is $5 \mathrm{~g} \mathrm{~cm}^{-3}$, then the intensity of magnetization in $\mathrm{A} / \mathrm{m}$ will be
1 3.0
2 $3 \times 10^{-6}$
3 $8.3 \times 10^{6}$
4 $1.2 \times 10^{-7}$
Explanation:
A Given that, Magnetic moment $=6 \times 10^{-7} \mathrm{~A}-\mathrm{m}^{2}$ Mass $=1 \mathrm{gm}=10^{-3} \mathrm{~kg}$ Density $=5 \mathrm{gcm}^{-3}=\frac{5 \times 10^{-3} \mathrm{~kg}}{\left(10^{-2}\right)^{3} \mathrm{~m}^{3}}=5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Volume, $\mathrm{V}=\frac{\text { Mass }}{\text { density }}=\frac{10^{-3}}{5 \times 10^{3}}$ $=0.2 \times 10^{-6} \mathrm{~m}^{3}$ Magnetization $=\frac{\text { Magnetic moment }}{\text { volume }}$ $=\frac{6 \times 10^{-7}}{0.2 \times 10^{-6}}=\frac{6 \times 10^{-7}}{2 \times 10^{-7}}$ $=3.0 \mathrm{~A} / \mathrm{m}$
BCECE-2012
Magnetism and Matter
154397
A magnetic needle lying parallel to the magnetic field requires $W$ units of work to turn it through an angle $45^{\circ}$. The torque required to maintain the needle in this position will be
