Question

A metre rod of silver at $0^\circ {\rm{C}}$ is heated to $100^\circ {\rm{C}}$. It's length is increased by $0.19{\rm{ cm}}$. Coefficient of volume expansion of the silver rod is
(1) $5.7 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}}$
(2) $0.63 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}}$
(3) $1.9 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}}$
(4) $16.1 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}}$

Answer 1

The increase in the length of the silver rod is given by the product of its length, increase in temperature, and coefficient of linear expansion of its material when the rod's temperature is increased from some initial value to final value. We will be using this concept to find the coefficient of volume expansion of the silver rod.

Complete step by step answer:
Given:
The actual length of the rod is $l = 1{\rm{ m}}$.
The initial temperature of the rod is ${T_1} = 0^\circ {\rm{C}}$.
The final temperature of the rod is ${T_2} = 100^\circ {\rm{C}}$.
The increased length of the rod is $l = 0.19{\rm{ cm}} = 0.19{\rm{ cm}} \times \left( {\dfrac{{\rm{m}}}{{100{\rm{ cm}}}}} \right) = 0.0019{\rm{ m}}$.

Let us write the expression for the increase in the length of the rod when its temperature increases and denoted by a temperature difference $\Delta T^\circ {\rm{C}}$
$\Delta l = \alpha l\Delta T$
Here $\alpha$ is the coefficient of linear expansion of rod material silver.

We know that the temperature of the rod is increased from ${T_1}$ to ${T_2}$. Therefore we can substitute ${T_2} - {T_1}$ for $\Delta T$ in the above expression.

\Delta l = \alpha l\left( {{T_2} - {T_1}} \right)\\\ \alpha = \dfrac{{\Delta l}}{{l\left( {{T_2} - {T_1}} \right)}} \end{array}$$ On substituting $$1{\rm{ m}}$$ for l, $$0.0019{\rm{ m}}$$ for $$\Delta l$$, $$0^\circ {\rm{C}}$$ for $${T_1}$$ and $$100^\circ {\rm{C}}$$ for $${T_2}$$ in the above expression, we get: $$\begin{array}{c} \alpha = \dfrac{{0.0019{\rm{ m}}}}{{\left( {1{\rm{ m}}} \right)\left( {100^\circ {\rm{C}} - 0^\circ {\rm{C}}} \right)}}\\\ = 1.9 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}} \end{array}$$ We know that the coefficient of volume expansion of the rod's material is three times of its linear expansion. Therefore, we can write: $$\beta = 3\alpha $$ Here $$\beta $$ is the coefficient of volume expansion of the silver rod. On substituting $$1.9 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}}$$ for $$\alpha $$ in the above expression, we get: $$\begin{array}{c} \beta = 3\left( {1.9 \times {{10}^{ - 5}}{\rm{/}}^\circ {\rm{C}}} \right)\\\ = 5.7 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}} \end{array}$$ Therefore, we can say that the silver rod's coefficient of volume expansion is $$5.7 \times {10^{ - 5}}{\rm{/}}^\circ {\rm{C}}$$ **So, the correct answer is “Option 1”.** **Note:** We can remember that the coefficient of volume expansion of the rod material is twice its coefficient of linear expansion. Do not forget to convert all the substituted values into the same system of units for uniformity of units so that the final expression of the coefficient of volume expansion is obtained in terms of per degree Celsius.