The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr^{3+ ion (Atomic no. : Cr = 24) is:
3.57 B.M.
Step 1: Understanding the magnetic moment. The magnetic moment depends on the number of unpaired electrons. For Cr^{3+} (with atomic number 24), the electron configuration is \( [Ar]3d^3 \), meaning there are 3 unpaired electrons.
Step 2: Calculation. The magnetic moment (\(\mu\)) is calculated using the formula: \[ \mu = \sqrt{n(n+2)} \, \text{B.M.} \] Where \( n \) is the number of unpaired electrons. For Cr^{3+} (3 unpaired electrons), we get: \[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{B.M.} \]
Step 3: Conclusion. Thus, the magnetic moment is 3.87 B.M., corresponding to option (B). \vspace{10pt}
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).