1. Speed of efflux from the hole

Since the hole is at the bottom of the jar and the water level is at height $H$ above it, by Torricelli’s theorem the speed of water as it comes out is

$v_0 = \sqrt{2gH}.$

The hole has radius $r$, so the area of the stream at the hole is

$A_0 = \pi r^2.$

Hence the volume flow rate at the hole is

$Q = A_0 v_0 = \pi r^2 \sqrt{2gH}.$

2. Speed of the water stream when it reaches the ground

After coming out, the water falls through a vertical distance $h$ before hitting the ground.

Using energy conservation / kinematics:

$v^2 = v_0^2 + 2gh.$

Substitute $v_0^2 = 2gH$:

$v^2 = 2gH + 2gh = 2g(H+h).$

So the speed just before hitting the ground is

$v = \sqrt{2g(H+h)}.$

3. Use continuity of flow

For the falling stream, the volume flow rate remains constant:

$A_0 v_0 = A v,$

where $A = \pi x^2$ is the cross-sectional area of the stream at the ground.

Thus,

$\pi r^2 \sqrt{2gH} = \pi x^2 \sqrt{2g(H+h)}.$

Cancel $\pi$ and $\sqrt{2g}$:

$r^2 \sqrt{H} = x^2 \sqrt{H+h}.$

So,

$x^2 = r^2 \sqrt{\frac{H}{H+h}}.$

Taking square root,

$x = r\left(\frac{H}{H+h}\right)^{1/4}.$

4. Match with options

This corresponds to:

$\boxed{x = r\left(\frac{H}{H+h}\right)^{1/4}}$

So the correct option is C.

5. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.