\( M_{OA} = \dfrac{2}{5} \)
\( M_{OB} = \dfrac{a}{2} \)
\( = \tan \dfrac{\pi}{4} = \left| \dfrac{\dfrac{2}{5}}{1 + \dfrac{a}{2}} \right| \)
\( \Rightarrow 1 = \left| \dfrac{4 - 5a}{10 + 2a} \right| \Rightarrow a = \dfrac{-6}{7}, \dfrac{14}{3} \)
Product \( \Rightarrow -4, \text{Abs. value} = 4 \)

Consider the points \( A(5, 2) \) and \( B(2, a) \) joined by line segments from the origin \( O \). The given condition states that these segments subtend an angle \( \frac{\pi}{4} \) at the origin. Let the slopes of the lines \( OA \) and \( OB \) be \( M_{OA} \) and \( M_{OB} \) respectively.
The slope of line \( OA \) is: \[ M_{OA} = \frac{2}{5}. \]
The slope of line \( OB \) is: \[ M_{OB} = \frac{a}{2}. \]
Since the angle between the lines is \( \frac{\pi}{4} \),
we use the formula: \[ \tan \left( \frac{\pi}{4} \right) = \left| \frac{M_{OB} - M_{OA}}{1 + M_{OA} \cdot M_{OB}} \right|. \]
Given \( \tan \left( \frac{\pi}{4} \right) = 1 \),
we have: \[ 1 = \left| \frac{\frac{a}{2} - \frac{2}{5}}{1 + \frac{2}{5} \cdot \frac{a}{2}} \right|. \]
Simplifying the expression: \[ \left| \frac{5a - 4}{10 + 2a} \right| = 1. \]
Clearing the fractions: \[ \left| \frac{5a - 4}{10 + 2a} \right| = 1. \]
This gives two cases: \[ \frac{5a - 4}{10 + 2a} = 1 \quad \text{or} \quad \frac{5a - 4}{10 + 2a} = -1. \]
Case 1: \[ \frac{5a - 4}{10 + 2a} = 1. \]
Cross-multiplying: \[ 5a - 4 = 10 + 2a. \]
Rearranging terms: \[ 3a = 14 \implies a = \frac{14}{3}. \]
Case 2: \[ \frac{5a - 4}{10 + 2a} = -1. \]
Cross-multiplying: \[ 5a - 4 = -10 - 2a. \]
Rearranging terms: \[ 7a = -6 \implies a = -\frac{6}{7}. \]
The product of all possible values of \( a \) is: \[ a_1 \times a_2 = \left( \frac{14}{3} \right) \times \left( -\frac{6}{7} \right) = -4. \]
The absolute value of the product is: \[ |a_1 \times a_2| = 4. \]
Therefore: \[ 4. \]
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
