On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k$$\Omega$$. How much was the resistance on the left slot before interchanging the resistances?

We are dealing with a meter bridge (which is simply a Wheatstone bridge stretched out over a 100 cm long uniform wire). Whenever the bridge is balanced, the well-known Wheatstone-bridge relation is valid:

$$\frac{R}{S}\;=\;\frac{l}{100-l}$$

Here

$$R=\text{resistance placed in the left slot},$$

$$S=\text{resistance placed in the right slot},$$

$$l$$ = balance length from the left end in cm (first observation).

After interchanging the two resistances, the new balance point is observed 10 cm to the left of the old one. Hence the new balance length becomes

$$l-10\ \text{cm}.$$

Applying the same Wheatstone formula to the interchanged situation, but now with $$S$$ on the left and $$R$$ on the right, we obtain

$$\frac{S}{R}\;=\;\frac{l-10}{100-(l-10)} \;=\;\frac{l-10}{110-l}.$$

From the first balance condition we already have

$$\frac{R}{S}\;=\;\frac{l}{100-l}\;.$$

Taking the reciprocal of this gives

$$\frac{S}{R}\;=\;\frac{100-l}{l}\;.$$

Both expressions represent the same $$\dfrac{S}{R}$$, so we equate them directly:

$$\frac{100-l}{l} \;=\;\frac{l-10}{110-l}.$$

Cross-multiplying every term gives a purely algebraic equation:

$$\bigl(100-l\bigr)\bigl(110-l\bigr)\;=\;l\,(l-10).$$

We now expand every product, showing each multiplication step:

Left side:

$$\bigl(100-l\bigr)\bigl(110-l\bigr) = 100\times110 - 100\,l - 110\,l + l^2 = 11000 - 210\,l + l^2.$$

Right side:

$$l\,(l-10) = l^2 - 10\,l.$$

Equating left and right sides:

$$11000 - 210\,l + l^2 \;=\; l^2 - 10\,l.$$

The $$l^2$$ terms cancel out immediately, leaving a simple linear equation:

$$11000 - 210\,l \;=\; -10\,l.$$

We bring the $$l$$ terms to one side by adding $$210\,l$$ to both sides:

$$11000 \;=\; 200\,l.$$

Dividing both sides by 200 yields the first balance length unambiguously:

$$l \;=\;\frac{11000}{200} \;=\; 55\ \text{cm}.$$

So the first balance point is at 55 cm, and after interchanging it indeed shifts to

$$55 - 10 = 45\ \text{cm},$$

which is self-consistent.

Now we use $$l=55$$ cm in the original Wheatstone ratio to pin down the ratio $$R:S$$:

$$\frac{R}{S} \;=\;\frac{l}{100-l} = \frac{55}{100-55} = \frac{55}{45} = \frac{11}{9}.$$

Hence

$$R = \frac{11}{9}\,S.$$

We are furthermore told that the two resistances in series add up to 1 k$$\Omega$$, i.e.

$$R + S = 1\ \text{k}\Omega = 1000\ \Omega.$$

Substituting $$R = \dfrac{11}{9}S$$ into the series‐sum equation:

$$\frac{11}{9}S + S = 1000.$$

Combining the two $$S$$ terms requires a common denominator:

$$\left(\frac{11}{9} + 1\right)S = \left(\frac{11}{9} + \frac{9}{9}\right)S = \frac{20}{9}S = 1000.$$

Now isolate $$S$$:

$$S = 1000 \times \frac{9}{20} = 1000 \times 0.45 = 450\ \Omega.$$

Finally, substitute this $$S$$ back into $$R = \dfrac{11}{9}S$$ to obtain $$R$$ explicitly:

$$R = \frac{11}{9}\times 450 = 11 \times 50 = 550\ \Omega.$$

The question asks for the resistance that was originally in the left slot, which is exactly $$R$$. We have obtained

$$R = 550\ \Omega.$$

Hence, the correct answer is Option D.