Question

A spherical shell of copper is completely filled with a liquid at a temperature ${t^ \circ }C$. The bulk modulus of the liquid is $K$ and the coefficient of volume expansion is$\gamma $. If the temperatures of the liquid and the shell is increased by $\Delta T$, then the outward the bulk modulus of the liquid is pressure $\Delta V$ on the shell that results from the temperature increase is given by ($\alpha $ is the coefficient of linear expansion of the material of the shell):
(A). $\dfrac{K}{2}\left( {\gamma - 3\alpha } \right)\Delta T$
(B). $K\left( {\gamma - 3\alpha } \right)\Delta T$
(C). $K\left( {k - \gamma } \right)\Delta T$
(D). $\gamma \left( {3\alpha - K} \right)\Delta T$

Answer 1

Hint: Circular shell implies it is empty inside. On the off chance that shell and circle are made of metal, at that point there is no difference. On the off chance that they are made of non-metal and charge homogeneous circulated, at that point additionally they can be dealt with similarly however if non-homogeneous than not equivalent.

A Circular shell implies it is empty inside. On the off chance that shell and circle are made of metal, at that point there is no difference. On the off chance that they are made of non-metal and charge homogeneous conveyed, at that point likewise they can be dealt with similarly however on the off chance that non-homogeneous than not equivalent.

Formula used:
According to hooke's law,
${F_s} = - K $$x$
${F_s} = $ Spring force
$K $$ = $ Spring constant
$x$$ = $ Spring stretch or compression
Bulk modulus of the liquids:
 $K = \dfrac{{dp}}{{\dfrac{{\Delta V}}{V}}}$
Complete step by step answer:
Let $V, {V_1}, {V_2}$ are volumes
Let us consider the terms we get,
Outward pressure $ = $ $\Delta P$
Bulk modulus of the liquids $ = $ $K$
The temperatures of the liquid and the shell is increased by $\Delta T$,
Coefficient of volume expansion is $\gamma $.
$\alpha $ is the coefficient of linear expansion
Then, $\Delta {V_2} = V \times \gamma \Delta t$ $\Delta {V_1} = V \times 3\alpha \Delta t$
Also, $\Delta {V_2} = V \times \gamma \Delta t$
The developed stress increases the volume of the vessel and decreases the volume of liquid.
Let the volume change in due to stress $ = $ $\Delta V$
The final volume of both vessel and liquid is the same
$\therefore $ The final volume of vessel $ \Rightarrow $ The final volume of the vessel $ \Rightarrow V + \Delta {V_1} + \Delta V$ ${V_1} = V + \Delta {V_1}$
And the final volume of liquid $ \Rightarrow $${V_2} = V + \Delta {V_2} - \Delta V$
But we know ${V_1} = {V_2}$
$V + \Delta {V_1} + \Delta V$$ = $$V + \Delta {V_2} - \Delta V$
$\therefore 2\Delta V = \Delta {V_2} - \Delta {V_1}$ $\therefore 2\Delta V = \Delta {V_2} - \Delta {V_1}$
Cancel out the $V$, then
$\therefore $ From Hooke’s law,
$K = \dfrac{{dp}}{{\dfrac{{\Delta V}}{V}}}$
$dp = K \dfrac{{V\Delta V}}{V}$
$ \Rightarrow K \dfrac{{\dfrac{{V\Delta T}}{2}\left( {\gamma - 3\alpha } \right)}}{V}$ ($\therefore $ coefficient of volume expansion is $\gamma $)($\therefore $$\alpha $ is the coefficient of linear expansion)
$dp = \dfrac{K}{2}\left( {\gamma - 3\alpha } \right)\Delta T$

$\therefore $ The correct option is option (A).

Note: It is equivalent to the CSA of the internal circle deducted from the CSA of the external circle. All out surface territory of empty circle: The surface zone of an empty circle is equivalent to the CSA of an empty circle as an empty circle has just one surface that comprises it. The surface territory of a circle is characterized as the area covered by its external surface in three-dimensional space.