1. Use the relation between extension and tension

For a metallic wire obeying Hooke’s law,

$\text{extension} \propto \text{tension}$

If the original length is $l_0$, then for a given tension $T$,

$l = l_0 + \Delta l$

and since $\Delta l \propto T$, we can write

$l = l_0 + kT$

where $k$ is a constant for the wire.

2. Write equations for the two given cases

When tension is $T_1$, length is $l_1$:

$l_1 = l_0 + kT_1$

When tension is $T_2$, length is $l_2$:

$l_2 = l_0 + kT_2$

3. Eliminate $k$ to find $l_0$

From the two equations,

$l_1 - l_2 = k(T_1 - T_2)$

So,

$k = \frac{l_1 - l_2}{T_1 - T_2}$

Now substitute into

$l_0 = l_1 - kT_1$

Hence,

$l_0 = l_1 - T_1\cdot \frac{l_1 - l_2}{T_1 - T_2}$

Take LCM:

$l_0 = \frac{l_1(T_1 - T_2) - T_1(l_1 - l_2)}{T_1 - T_2}$

Simplify numerator:

$l_0 = \frac{l_1T_1 - l_1T_2 - T_1l_1 + T_1l_2}{T_1 - T_2}$

$l_0 = \frac{T_1l_2 - T_2l_1}{T_1 - T_2}$

Multiplying numerator and denominator by $-1$,

$l_0 = \frac{T_2l_1 - T_1l_2}{T_2 - T_1}$

4. Match with the options

This matches:

$\boxed{\frac{T_2l_1 - T_1l_2}{T_2 - T_1}}$

So the correct option is D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.