Given:
The potential energy \( U \) of a magnetic dipole in a uniform magnetic field is given by the formula:
\[ U = -MB \cos \theta \]
Since the magnetic moment is initially aligned with the magnetic field, \( \theta = 0^\circ \), and thus:
\[ U = -MB \cos(0^\circ) = -MB \]
Substitute the values:
\[ U = -5 \times 0.4 = -2 \, \text{J} \]
Thus, the potential energy of the bar magnet is -2 J.
When the magnet is turned by 180°, the angle between the magnetic moment and the magnetic field becomes \( \theta = 180^\circ \). The new potential energy is:
\[ U' = -MB \cos(180^\circ) = +MB \]
Substitute the values:
\[ U' = +5 \times 0.4 = +2 \, \text{J} \]
The work done in turning the magnet is the change in potential energy:
\[ W = U' - U = 2 - (-2) = 4 \, \text{J} \]
Thus, the work done in turning the magnet by 180° is 4 J.
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 \(P,O,Q\).