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A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $2^{\frac{5}{4}}$ seconds. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is :

$2^{\frac{1}{4}}$

$2^{\frac{1}{2}}$

$2^{\frac{3}{4}}$

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A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetisation of 5.30 × 10³Amp/m³. What is its magnetic dipole moment ?

$1 \times 10^{- 2} J / T$

$2.08 \times 10^{- 2} J / T$

$3.08 \times 10^{- 2} J / T$

$1.52 \times 10^{- 2} J / T$

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If $ϕ_{1}$ and $ϕ_{2}$ be the angles of dip observed in two vertical planes at right angles to each other and $ϕ$ be the true angle of dip, then

$\cos^{2} ϕ = \cos^{2} ϕ_{1} + \cos^{2} ϕ_{2}$

$\sec^{2} ϕ = \sec^{2} ϕ_{1} + \sec^{2} ϕ_{2}$

$\tan^{2} ϕ = \tan^{2} ϕ_{1} + \tan^{2} ϕ_{2}$

$\cot^{2} ϕ = \cot^{2} ϕ_{1} + \cot^{2} ϕ_{2}$

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A dip needle lies initially in the magnetic meridian when it shows an angle of dip $θ$ at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip $θ'$ . Then $\frac{\tan θ'}{\tan θ}$ is

1/cos x

1/sin x

1/tan x

cos x

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If $θ_{1} a n d θ_{2}$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $θ$ is given by

$c o t^{2} θ = c o t^{2} θ_{1} + c o t^{2} θ_{2}$

$\tan^{2} θ = \tan^{2} θ_{1} - \tan^{2} θ_{2}$

$c o t^{2} θ = c o t^{2} θ_{1} - \tan^{2} θ_{2}$

$\tan^{2} θ = \tan^{2} θ_{1} - \tan^{2} θ_{2}$

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Explanation

If Î¸1 and Î¸2 are the apparent angles of dip observed in two vertical planes at right angles, the true angle of dip Î¸ is given by cot^2 Î¸ = cot^2 Î¸1 + cot^2 Î¸2. This relation is derived from the vector addition of the horizontal and vertical components of the Earth's magnetic field.

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The magnetic susceptibility is negative for

paramagnetic material only

ferromagnetic material only

paramagnetic and ferromagnetic materials

diamagnetic material only

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Explanation

(d) As we know the relation between the magnetic permeability and susceptibility of material i.e.

$μ_{r} = 1 + x_{m}$ ...(i)

so, For diamagnetic substances, $μ_{r} < 1$

so, according to equation (i), the magnetic susceptibility ( $x_{m}$ ) of diamagnetic substance will be negative.

While in case of para and ferromagnetic substances, magnetic susceptibility is positive.

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A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole.It

will become rigid showing no movement

will stay in any position

will stay in north-south direction only

will stay in east-west direction only

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Explanation

It will stay in any position at geomagnetic north and south poles.

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A magnetic needle suspended parallel to a

magnetic field requires $\sqrt{3}$ J of work to turn

it through $60 °$ . The torque needed to maintain

the needle in this position will be

$2 \sqrt{3}$ J

3 J

$\sqrt{3}$ J

$\frac{3}{2}$ J

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Explanation

In this case, work done

$W = MB ({cosθ}_{1} - {cosθ}_{2}) = MB (\cos 0 ° - \cos 60 °) = MB (1 - \frac{1}{2}) = \frac{MB}{2} MB = 2 \sqrt{3} J (so, given W = \sqrt{3} J) τ = MB \sin 60 ° = (2 \sqrt{3}) (\frac{\sqrt{3}}{2}) J = 3 J$

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There are four light-weight-rod samples; A, B, C, D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted:

(i) A is feebly repelled

(ii) B is feebly attracted

(iii) C is strongly attracted

(iv) D remains unaffected

Which one of the following is true?

C is of a diamagnetic material

D is of a ferromagnetic material

A is of a non-magnetic material

B is of a paramagnetic material

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Explanation

Paramagnetic material will be feebly attracted, diamagnetic material will be feebly repelled and Ferromagnetic material will be strongly attracted.

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A short bar magnet of magnetic moment $0.4 {JT}^{- 1}$ is placed in a uniform magnetic field of $0.16 T .$ The magnet is stable equilibrium when the potential energy is

$- 0.64 J$

zero

$- 0.082 J$

$- 0.064 J$

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Explanation

We know that

$U = - M B \cos θ$

For stable equilibrium $θ = 0 °$

So, $U = - M B$

= $- (0.4) (0.16) = - 0.064 J$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.