Step 1: Recall the Formula for Magnetic Moment
The spin-only magnetic moment (\( \mu \)) is given by:
$$ \mu = \sqrt{n(n+2)} $$
where \( n \) is the number of unpaired electrons. The magnetic moment depends only on the number of unpaired electrons in the ion.
Step 2: Analyze the Given Ions
- Has 1 unpaired electron.
- Spin-only magnetic moment:
$$ \mu = \sqrt{1(1+2)} = \sqrt{3} $$
- Has 4 unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = \sqrt{4(4+2)} = \sqrt{24} $$
- Has 5 unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = \sqrt{5(5+2)} = \sqrt{35} $$
- Has 4 unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = \sqrt{4(4+2)} = \sqrt{24} $$
- Has no unpaired electrons.
- Spin-only magnetic moment:
$$ \mu = 0 $$
Step 3: Compare the Magnetic Moments
Step 4: Conclusion
The correct answer is: Option (1) - B and D only, since they have the same spin-only magnetic moment.
Given below are two statements:
Statement I: Ferromagnetism is considered as an extreme form of paramagnetism.
Statement II: The number of unpaired electrons in a $Cr^{2+}$ ion (Z = 24) is the same as that of a $Nd^{3+}$ ion (Z = 60).
In the light of the above statements, choose the correct answer from the options given below:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec:
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :