# Question 776dd

##### 1 Answer
Dec 8, 2017

The ratio of expansion will be,

$\setminus \Delta {V}_{1} : \setminus \Delta {V}_{2} = 1 : 1$

#### Explanation:

Because the rods are made of same material, material properties such as specific heat capacity ($c$), mass density ($\setminus \rho$) and coefficient of volume expansion ($\setminus \gamma$) are the same.

If $Q$ is the heat entering a body of mass $m$, volume $V$, density $\setminus \rho$ and specific heat capacity $c$, then the change in temperature of the body $\setminus \Delta T$ is related to other quantities as -

Q = mc\Delta T = (\rho.V)c\Delta T; \qquad

Rearranging,

$V \setminus \Delta T = \left(\setminus \frac{Q}{\setminus \rho . c}\right)$ ...... (1)

Since $\setminus \rho$ and $c$ are material constants, the the product of volume and change in temperature ($V \setminus \Delta T$) will be a constant, if $Q$ is a constant.

For a temperature change of $\setminus \Delta T$ the volume expansion is -

$\setminus \Delta V = \setminus \gamma \left(V \setminus \Delta T\right)$ ...... (2)

Let ${V}_{1}$ and ${V}_{2}$ be the volumes of the two rods and $\setminus \Delta {T}_{1}$ and $\setminus \Delta {T}_{2}$ are the change in their temperatures when they absorb the same heat $Q$. Then the change in their volumes are -

\DeltaV_1 = \gamma(V_1\DeltaT_1); \qquad \DeltaV_2 = \gamma (V_2\DeltaT_2)

But we know from (1): $\setminus \quad {V}_{1} \setminus \Delta {T}_{1} = {V}_{2} \setminus \Delta {T}_{2}$

Therefore, \quad \DeltaV_1 = \DeltaV_2; \qquad \rightarrow \DeltaV_1 : \DeltaV_2 = 1:1#.