Let the distance between the cars when they come to rest be \( d \). Each car has an initial speed of 20 m/s. The deceleration \( a \) is \(-2 \, \text{m/s}^2\).
Using the equation of motion:
\( v^2 = u^2 + 2ad, \)
where \( v = 0 \) (final speed), \( u = 20 \, \text{m/s} \) (initial speed), and \( a = -2 \, \text{m/s}^2 \), we get:
\( 0 = 20^2 + 2(-2)d \quad \Rightarrow \quad 0 = 400 - 4d \quad \Rightarrow \quad d = 100 \, \text{m}. \)
Since the two cars are moving towards each other, the total distance covered by both cars is \( 100 + 100 = 200 \, \text{m} \), so the distance between them when they come to rest is 100 m.
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)