Question

A metallic bob weight $50$g in air. If it is immersed in a liquid at a temperature of $25C$, it weighs $45g.$When the temperature of the liquid is raised to $100C$, it weighs $45.1g$. Calculate the coefficient of cubical expansion of the liquid. Given that the coefficient of cubical expansion of the meta is $12 \times {10^6}{C^1}$.

Answer 1

Hint:In order to solve this problem, we are going to apply the concept of expansion i.e,change in the volume of a liquid with respect to temperature is called the coefficient of vertical expansion of liquid.

Formula used:
$\dfrac{{\Delta {W^1}}}{{\Delta W}} = \dfrac{{{F^1}}}{F} = \left( {\dfrac{{1 + {Y_s}\Delta \theta }}{{1 + {Y_l}\Delta \theta }}} \right)$

Complete step by step answer:
Given information in the question is
Weight of the metallic lab is $50g$
Weight of the metallic lab when weight is increased in the liquid $\Delta 25^\circ C$ is $45g$.
Weight of the metallic loob when is increased in the temperature of liquid is $100^\circ C$, is $45.1g$ efficient of vertical expansion of the metal is $12 \times {10^{ - 6}}{C^{ - 1}}$.
Now, we know that change in weight is equal to upthrust on $100\% $ volume which is given by
$\dfrac{{\Delta {W^1}}}{{\Delta W}} = \dfrac{{{F^1}}}{F}$ . . (1)
When,
$\Delta {W^1}$ be the final change in weight $\Delta W$ be the initial change in weight
${F^1}$ be the final upthrust
$F$ be the initial final upthrust
We also have,
$\dfrac{{{F^1}}}{F} = \dfrac{{1 + {Y_s}\Delta \theta }}{{1 + {Y_l}\Delta \theta }}$ . . . (2)
Where,
${Y_s}$ is the coefficient and vertical expansion and solid.
${Y_l}$ is the coefficient of vertical expansion of liquid.
$\Delta \theta $ be the change in temperature.
From equation (1) and (2), we get
$\dfrac{{\Delta {W^1}}}{{\Delta W}} = \dfrac{{1 + {Y_s}\Delta \theta }}{{1 + {Y_l}\Delta \theta }}$
By substituting the given value in the above equation we get.
$\dfrac{{50 - 45.1}}{{50 - 45}} = \dfrac{{1 + 12 \times {{10}^{ - 6}}(100 - 25)}}{{1 + {Y_e}(100 - 25)}}$
Can simplifying, we get
$ \Rightarrow \dfrac{{4.9}}{5} = \dfrac{{1 + 12 \times {{10}^{ - 6}} \times 75}}{{1 + {Y_e} \times 75}}$
By cross multiplying we get
$4.9(1 + {Y_e} \times 75) = 5(1 + 12 \times {10^{ - 6}} \times 75)$
Open the brackets.
$ \Rightarrow 4.9 + 4.9 \times 75 \times {Y_e}=5 + 5 \times 75 \times 12 \times {10^{ - 6}}$
By taking $4.9$ to the other side, we get
$49 \times 75 \times {Y_e} = (5 - 4.9) + 5 \times 75 \times 12 \times {10^{ - 6}}$
$ \Rightarrow {Y_e} = \dfrac{{0.1 + 4500 \times {{10}^{ - 6}}}}{{4.9 \times 75}}$
$ \Rightarrow {Y_e} = \dfrac{{0.1 + 45 \times {{10}^{ - 4}}}}{{367.5}}$
By simplifying, we get
$ = \dfrac{{0.1 + \dfrac{{45}}{{{{10}^4}}}}}{{367.5}}$
$ = \dfrac{{1000 + 45}}{{367.5}} \times {10^{ - 4}}$
$ = \dfrac{{1045}}{{367.5}} \times {10^{ - 4}}$
By finding, we get
$ \Rightarrow {Y_e} = 2.84 \times {10^{ - 4}}^\circ {C^{ - 1}}$
Therefore, the coefficient of vertical expansion of the liquid is a
${Y_e} = 2.84 \times {10^{ - 4}}^\circ {C^{ - 1}}$.

Note:Basically upthrust means that how much amount of force is exerted by an object when we place it on the surface of liquid and according to the Archimedes principle,it is equal to the weight of the displaced liquid.