In the figure shown, both the blocks are released from rest. The time (in $s$ ) to cross each other is

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Solution

Step1: Find the acceleration of the blocks.
For the mass of $4 k g$ ,
$4 g - T = 4 a \dots (i)$

For the mass of $1 k g$
$T - g = a \dots (i i)$
From the equation $(i)$ and $(i i)$
$3 g = 5 a$
$\Rightarrow a = \frac{3 g}{5}$
Relative acceleration $= 2 a = \frac{6 g}{5}$

Step2: Find the time taken by the blocks to cross each other.
Formula Used: $s = u t + \frac{1}{2} a t^{2}$

From $2^{n d}$ equation of motion
$s = u t + \frac{1}{2} a t^{2} \dots (i i i)$
( $∵$ Blocks starts from rest, therefore $u = 0$ )

Substituting the values in equation $(i i i)$ , we get
$6 = \frac{1}{2} \times \frac{6 g}{5} \times t^{2}$

$\Rightarrow t^{2} = 1$

$\Rightarrow t = 1 s$

Final Answer: $1$

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