# A liquid of temperature coefficient of volume expansion g = 4 10 5 / C is poured in a cylindrical...

## Question:

A liquid of temperature coefficient of volume expansion g = 4 10 5 / C is poured in a cylindrical container of metal. On increasing temperature height of liquid in container will remain same if temperature coefficient of linear expansion of container is approximately : (a) 2 10 5 / C (b) 3 10 5 / C (c) 8 10 5 / C (d) 6 10 5 / C

## Thermal Expansion:

Any object will experience a change in its linear dimensions upon heating. This phenomenon is called a thermal expansion. Most of the substances will increase their dimensions upon increasing temperature, but there are some notable exceptions. The relative change in the linear dimension/volume upon the increase of temperature by one degree is called a coefficient of linear/volume expansion: {eq}\alpha = \dfrac {1}{V_0} \dfrac {\partial V}{\partial T} {/eq}

## Answer and Explanation:

Let the initial volume be {eq}V_0 {/eq} and initial radius of the container to be {eq}R_0 {/eq}

Then the initial height is:

{eq}h_0 = \dfrac {V_0}{\pi R^2_0} {/eq}

After the temperature increase, the volume and radius are given by:

{eq}V = V_0 (1 + \alpha_V \Delta T) {/eq}

{eq}R = R_0 (1 + \alpha_R \Delta T) {/eq}

Here

- {eq}\alpha_V = 4\times 10^{-5} \ ^{\circ}C^{-1} {/eq} is the coefficient of volume thermal expansion of the fluid;

- {eq}\alpha_R {/eq} is the coefficient of linear thermal expansion of the container;

New height of the fluid is determined to be:

{eq}h = \dfrac {V}{\pi R^2} = \dfrac {V_0}{\pi R^2_0} \times \dfrac {1 + \alpha_V \Delta T}{(1 + \alpha_R \Delta T)^2} = h_0 \dfrac {1 + \alpha_V \Delta T}{(1 + \alpha_R \Delta T)^2} {/eq}

Neglecting the terms quadratic with respect to temperature change, since the coefficient of linear and volume expansion are small, we obtain:

{eq}h = \dfrac {V}{\pi R^2} \approx h_0 \dfrac {1 + \alpha_V \Delta T}{1 + 2\alpha_R \Delta T} {/eq}

The height remains the same if the following equality holds:

{eq}\dfrac {1 + \alpha_V \Delta T}{1 + 2\alpha_R \Delta T} = 1 {/eq}

This is correct if

{eq}\alpha_V = 2 \alpha_R {/eq}

Then

{eq}\alpha_R = \dfrac {\alpha_V}{2} {/eq}

Computation yields:

{eq}\alpha_R = \dfrac {4\times 10^{-5} \ ^{\circ}C^{-1}}{2} = \boxed {2\times 10^{-5} \ ^{\circ}C^{-1}} {/eq}

Hence, correct answer is **a)**

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Chapter 16 / Lesson 3