# A cylindrical vessel contains liquid of volume v_0 and area of cross-section a_0 at temperature...

## Question:

A cylindrical vessel contains liquid of volume {eq}v_0 {/eq} and area of cross-section {eq}a_0 {/eq} at temperature {eq}t_1 {/eq}. Find the expression for the rise in the level of the liquid at {eq}t_2 {/eq}.

## Cubical Expansion of Liquids

Matter existing in the liquid phase has comparatively weak inter molecular molecular forces. This gives them the fluidity. So liquids when they receive heat energy their molecular vibrations will be more vigorous and they expand more than the matter existing solid phase. Normally all liquids expand considerably with temperature than matter existing in the solid phase. Cubical expansion or volume expansion of a liquid with increase in temperature is given by the equation {eq}V_1 = V \times [ 1 + \gamma \Delta T ] {/eq}.

• V - reference volume.
• {eq}\gamma {/eq} - cubical expansion coefficient
• {eq}\Delta T {/eq} - increase in temperature.

Given data

• Volume of the liquid at some reference temperature {eq}V_0 {/eq}
• Area of cross section of the cylinder {eq}a_0 {/eq}
• Reference temperature {eq}t_1 {/eq}
• Required final temperature {eq}t_2 {/eq}

Let {eq}\gamma, \ \ h_0 {/eq} be the cubical expansion coefficient of the liquid and initial height of the liquid in the cylinder.

Let {eq}2 \alpha {/eq} be the area expansion coefficient of the cylinder.

Initial volume of the liquid in the cylinder at reference temperature {eq}V_0 = a_0 h_0 {/eq}

Let {eq}a_1, \ \ h_1 {/eq} be the new area of cross section and height of liquid in the cylinder at the given final temperature.

Then the new area of cross section {eq}a_1 = a_0 \times [1 + 2 \alpha \times ( t_2- t_1 ) ] {/eq}

Volume of the liquid after expansion {eq}V_1 = a_1 h_1 {/eq}

The same cubical expansion can be expressed as {eq}V_1 = V_0 \times [ 1 + \gamma \times ( t_2 - t_1 ) ] \\ \implies a_1 h_1 = a_0 h_0 \times [ 1 + \gamma \times ( t_2- t_1 ) ] {/eq}

Therefore the rise in liquid level in the cylinder {eq}h_1 = \dfrac { a_0 h_0 \times [ 1 + \gamma \times ( t_2- t_1 0) ] } { a_1 } \\ h_1 = \dfrac { a_0 h_0 \times [ 1 + \gamma \times ( t_2 - t_1 ) ] } { a_0 \times [ 1 + 2 \alpha \times ( t_2 - t_1 ) ] } \\ h_1 = h_0 \times \left[ \dfrac { 1 + \gamma \times ( t_2- t_1 ) } { 1 + 2 \alpha \times ( t_2 - t_1 ) }\right] {/eq}