Let the resistances be $R$ and $S$.

Initially, suppose $R$ is in the left gap and $S$ is in the right gap of the meter bridge.

Balance condition of meter bridge

If the balance point is at $l$ cm from the left end, then $\frac{R}{S}=\frac{l}{100-l}$

After interchanging the resistances

Now $S$ is on the left and $R$ is on the right. Let the new balance point be $l'$. Then $\frac{S}{R}=\frac{l'}{100-l'}$

Since $\frac{S}{R}=\frac{100-l}{l}$ we get $\frac{l'}{100-l'}=\frac{100-l}{l}$

This gives $l'=100-l$

Use the given shift

On interchanging, the balance point shifts to the left by $10$ cm. So the new point is $10$ cm less than the old point: $l' = l-10$

But also, $l' = 100-l$

Therefore, $100-l = l-10$ $110 = 2l$ $l = 55 \text{ cm}$

Hence initially the balance point was at $55$ cm.

Find the ratio of resistances

$\frac{R}{S}=\frac{55}{45}=\frac{11}{9}$

Use the series combination

Given: $R+S=1000\,\Omega$

Let $R=11x,\quad S=9x$ Then $20x=1000$ $x=50$

So, $R=11\times 50=550\,\Omega$ $S=9\times 50=450\,\Omega$

Identify the resistance on the left slot before interchanging

Before interchanging, the left slot had $R$. Hence the required resistance is $550\,\Omega$

Check with options

This matches Option D.