Step 1: Use the equation of the parabola.
The equation of the parabola is: \[ x^2 = 4ay \] where \( a \) is the focal length. The point \( Q \) lies on the parabola, and the coordinates of \( Q \) are \( (x, y) \).
Step 2: Find the coordinates of point \( C \).
The point \( C \) divides the line segment \( OQ \) in the ratio 2:3. Using the section formula, we find the coordinates of \( C \). The section formula gives the point dividing the line in the ratio \( m:n \) as: \[ C = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]
Step 3: Apply the section formula.
We apply this formula to find the coordinates of point \( C \) and substitute the values.
Step 4: Find the equation of the chord.
Using the mid-point formula and simplifying, we obtain the equation of the chord of the parabola as: \[ 5x - 4y + 3 = 0 \]
In the figure, triangle ABC is equilateral. 
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]