Two metal rods P and Q of the same original length and at the same initial temperature are heated through the same temperature change. If the linear expansivity of P is twice that of Q, what is the ra

The change in length ($\Delta L$) of a material due to a temperature change ($\Delta T$) is given by:

where $L_0$ is the original length and $\alpha$ is the coefficient of linear expansivity.

• Linear expansivity of P is twice that of Q: $\alpha_P = 2\alpha_Q$
• Same temperature change: $\Delta T_P = \Delta T_Q = \Delta T$
• Same increase in length: $\Delta L_P = \Delta L_Q$
• Let the original lengths be $L_{0P}$ and $L_{0Q}$.

For metal P:

For metal Q:

Since $\Delta L_P = \Delta L_Q$:

Substitute $\alpha_P = 2\alpha_Q$:

Divide both sides by $\alpha_Q \Delta T$ (assuming $\alpha_Q \neq 0$ and $\Delta T \neq 0$):

Rearrange to find the ratio $\frac{L_{0P}}{L_{0Q}}$:

The ratio of the original length of P to that of Q is 1:2.

The final answer is $\text{B. 1:2}$.