1. Given relation

The speed of the particle depends on position: $v=b\sqrt{x}$

Also, $v=\frac{dx}{dt}$

So, $\frac{dx}{dt}=b\sqrt{x}$

2. Separate the variables

$\frac{dx}{\sqrt{x}}=b\,dt$

3. Integrate both sides

Using the initial condition: at $t=0$, the particle is at origin, so $x=0$.

$\int_0^x \frac{dx}{\sqrt{x}} = \int_0^t b\,dt$

$2\sqrt{x}=bt$

Thus, $\sqrt{x}=\frac{bt}{2}$

and hence $x=\frac{b^2 t^2}{4}$

4. Find speed as a function of time

Since $v=b\sqrt{x}$ substitute $\sqrt{x}=\frac{bt}{2}$:

$v=b\left(\frac{bt}{2}\right)=\frac{b^2 t}{2}$

5. At $t=\tau$

$v(\tau)=\frac{b^2\tau}{2}$

6. Match with options

This corresponds to:

Option C: $\boxed{\frac{b^2\tau}{2}}$