The relation 3t = √(3x) + 6 describes the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero, is

The correct option is Zero.

We are given the relation between the displacement $x$ (in metres) and the time $t$ (in seconds):
$3t=\sqrt{3x}+6$
We need to determine the displacement of the particle when its velocity is zero.

First, we isolate the square root term containing the displacement $x$:
$\sqrt{3x}=3t-6$
Next, squaring both sides of the equation gives:
$3x={\left(3t-6\right)}^2$
Expanding the right side, we get:
$3x=9t^2-36t+36$
Dividing the entire equation by 3, we obtain the expression for displacement as a function of time:
$x=3t^2-12t+12$

Velocity $v$ is defined as the rate of change of displacement with respect to time, which is the derivative of $x$ with respect to $t$:
$v=\frac{dx}{dt}=\frac{d}{dt}\left(3t^2-12t+12\right)$
Differentiating term by term:
$v=6t-12$

To find the time $t$ when the velocity is zero, we set $v=0$:
$6t-12=0$
$6t=12$
$t=2\text{seconds}$

Now, we substitute $t=2$ back into our displacement equation to find the displacement at this instant:
$x=3{\left(2\right)}^2-12\left(2\right)+12$
$x=3\left(4\right)-24+12$
$x=12-24+12$
$x=0\text{metres}$

Therefore, the displacement of the particle when the velocity is zero is Zero.