Two rods A and B of identical dimensions are at temperature $$30°C$$. If A is heated upto $$180°C$$ and B upto $$T°C$$, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is

Let us denote the original length of each rod by $$L$$. Both rods initially are at the same temperature $$30^\circ\text{C}$$, so their initial lengths are equal.

When a solid rod is heated through a temperature difference $$\Delta T$$, its new length is given by the linear expansion formula

$$L_\text{new}=L\,(1+\alpha\,\Delta T),$$

where $$\alpha$$ is the coefficient of linear expansion for that material. We now apply this formula separately to rods A and B.

For rod A we have:

$$\Delta T_A = 180^\circ\text{C}-30^\circ\text{C}=150^\circ\text{C},$$

so its new length becomes

$$L_A = L\,(1+\alpha_A\,\Delta T_A).$$

For rod B, its final temperature is $$T^\circ\text{C}$$, hence

$$\Delta T_B = T^\circ\text{C}-30^\circ\text{C},$$

and its new length is

$$L_B = L\,(1+\alpha_B\,\Delta T_B).$$

According to the problem, these new lengths are the same, so

$$L_A = L_B.$$

Substituting the two expressions, we get

$$L\,(1+\alpha_A\,\Delta T_A) = L\,(1+\alpha_B\,\Delta T_B).$$

Because the original length $$L$$ is common and non-zero, we can cancel it, giving

$$1+\alpha_A\,\Delta T_A = 1+\alpha_B\,\Delta T_B.$$

Now subtract 1 from both sides:

$$\alpha_A\,\Delta T_A = \alpha_B\,\Delta T_B.$$

The ratio of the coefficients of linear expansion is given to be

$$\frac{\alpha_A}{\alpha_B} = \frac{4}{3}.$$

Substituting this ratio into the equality of expansions, we find

$$\frac{4}{3}\,\Delta T_A = \Delta T_B.$$

We already know $$\Delta T_A = 150^\circ\text{C}$$, so

$$\Delta T_B = \frac{4}{3}\times 150^\circ\text{C} = 200^\circ\text{C}.$$

Finally, the actual final temperature of rod B is obtained by adding its temperature rise to the initial temperature:

$$T = 30^\circ\text{C} + 200^\circ\text{C} = 230^\circ\text{C}.$$

Hence, the correct answer is Option A.