NEET Magnetic Properties of Matter MCQs & PYQs

Q: 18. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is given by (1995)

$W_{90^{\circ}} = MB(1 - \cos 90^{\circ}) = MB(1 - 0) = MB$. $W_{60^{\circ}} = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB$. Since $W_{90^{\circ}} = n W_{60^{\circ}}$, we have $MB = n(0.5MB) \Rightarrow n = 2$.

Q: 19. An iron rod of susceptibility $599$ is subjected to a magnetising field of $1200$ A m$^{-1}$. The permeability of the material of the rod is : (2020) ($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)

Relative permeability $\mu_r = 1 + \chi = 1 + 599 = 600$. Permeability $\mu = \mu_r \mu_0 = 600 \times 4\pi \times 10^{-7} = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4}$ T m A$^{-1}$.

Q: 7. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now the torque required to keep the magnet in this new position is: (2016 - II)

Work done, $W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} \Rightarrow MB = 2W$. Torque required, $\tau = MB \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}W$.

Q: 8. A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be: (2012 Mains)

$W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} = \sqrt{3} \Rightarrow MB = 2\sqrt{3}$ J. Torque, $\tau = MB \sin 60^{\circ} = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$ J.

Q: 9. A short bar magnet of magnetic moment $0.4$ J T$^{-1}$ is placed in a uniform magnetic field of $0.16$ T. The magnet is in stable equilibrium when the potential energy is: (2011 Mains)

For stable equilibrium, the angle between $M$ and $B$ is $0^{\circ}$. Potential energy $U = -MB \cos 0^{\circ} = - (0.4)(0.16)(1) = -0.064$ J.

Q: 12. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It: (2012 Pre)

At the geomagnetic poles, the Earth's magnetic field is entirely vertical, meaning the horizontal component $B_H$ is zero. A compass free to rotate only horizontally experiences no directing torque and will stay in any position.

Practice NEET Magnetic Properties of Matter Questions

Question 41:

easy

6. A 250 turn rectangular coil of length $2.1\text{ cm}$ and width $1.25\text{ cm}$ carries a current of $85\text{ }\mu\text{A}$ and subjected to a magnetic field of strength $0.85\text{ T}$. Work done for rotating the coil by $180^\circ$ against the torque is: (2017-Delhi)

Magnetic moment $M = NIA = 250 \times (85 \times 10^{-6}\text{ A}) \times (2.1 \times 1.25 \times 10^{-4}\text{ m}^2) \approx 5.58 \times 10^{-6}\text{ A m}^2$.
Work done $W = MB(1 - \cos 180^\circ) = 2MB = 2 \times (5.58 \times 10^{-6}) \times 0.85 \approx 9.1\text{ }\mu\text{J}$.

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Question 42:

easy

15. A bar magnet having a magnetic moment of $2 \times 10^4$ J T$^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times 10^{-4}$ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is: (2009)

Work done, $W = MB(\cos \theta_1 - \cos \theta_2) = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - \frac{1}{2}) = \frac{MB}{2}$.
$W = \frac{1}{2} \times (2 \times 10^4) \times (6 \times 10^{-4}) = 6$ J.

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Question 43:

easy

16. A bar magnet is oscillating in the Earth’s magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled? (2003)

Time period is given by $T = 2\pi \sqrt{\frac{I}{MB}}$. Moment of inertia $I$ is directly proportional to mass $m$. If mass is quadrupled, $I$ becomes $4I$.
New period $T' = 2\pi \sqrt{\frac{4I}{MB}} = 2T$. The restoring torque is still $-MB \sin \theta$, so motion remains S.H.M.

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Question 44:

easy

17. Two bar magnets having same geometry with magnetic moments $M$ and $2M$, are firstly placed in such a way that their similar poles are same side then its time period of oscillation is $T_1$. Now the polarity of one of the magnet is reversed then time period of oscillation is $T_2$, then: (2002)

$T = 2\pi \sqrt{\frac{I}{M_{net} B_H}}$. Initially $M_{net} = 2M + M = 3M$, so $T_1 \propto \frac{1}{\sqrt{3M}}$.
After reversing polarity, $M_{net} = 2M - M = M$, so $T_2 \propto \frac{1}{\sqrt{M}}$. Therefore, $T_1 < T_2$.

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Question 45:

easy

18. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is given by (1995)

$W_{90^{\circ}} = MB(1 - \cos 90^{\circ}) = MB(1 - 0) = MB$.
$W_{60^{\circ}} = MB(1 - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5MB$.
Since $W_{90^{\circ}} = n W_{60^{\circ}}$, we have $MB = n(0.5MB) \Rightarrow n = 2$.

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Question 46:

easy

19. An iron rod of susceptibility $599$ is subjected to a magnetising field of $1200$ A m$^{-1}$. The permeability of the material of the rod is : (2020)
($\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$)

Relative permeability $\mu_r = 1 + \chi = 1 + 599 = 600$.
Permeability $\mu = \mu_r \mu_0 = 600 \times 4\pi \times 10^{-7} = 2400\pi \times 10^{-7} = 2.4\pi \times 10^{-4}$ T m A$^{-1}$.

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Question 47:

easy

7. A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now the torque required to keep the magnet in this new position is: (2016 – II)

Work done, $W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} \Rightarrow MB = 2W$.
Torque required, $\tau = MB \sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2} = \sqrt{3}W$.

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Question 48:

easy

8. A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be: (2012 Mains)

$W = MB(1 - \cos 60^{\circ}) = \frac{MB}{2} = \sqrt{3} \Rightarrow MB = 2\sqrt{3}$ J.
Torque, $\tau = MB \sin 60^{\circ} = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$ J.

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Question 49:

easy

9. A short bar magnet of magnetic moment $0.4$ J T$^{-1}$ is placed in a uniform magnetic field of $0.16$ T. The magnet is in stable equilibrium when the potential energy is: (2011 Mains)

For stable equilibrium, the angle between $M$ and $B$ is $0^{\circ}$.
Potential energy $U = -MB \cos 0^{\circ} = - (0.4)(0.16)(1) = -0.064$ J.

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Question 50:

easy

12. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It: (2012 Pre)

At the geomagnetic poles, the Earth's magnetic field is entirely vertical, meaning the horizontal component $B_H$ is zero. A compass free to rotate only horizontally experiences no directing torque and will stay in any position.

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