Heat and Thermodynamics 2 Question 7
10. A cube of coefficient of linear expansion $\alpha _s$ is floating in a bath containing a liquid of coefficient of volume expansion $\gamma _l$. When the temperature is raised by $\Delta T$, the depth upto which the cube is submerged in the liquid remains the same. Find the relation between $\alpha _s$ and $\gamma _l$ showing all the steps.
(2004, 2M)
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Answer:
Correct Answer: 10. $\beta _l=\beta\left(\frac{w _0-w _1}{w _0-w _2}\right)+\frac{\left(w _2-w _1\right)}{\left(w _0-w _2\right)\left(T _2-T _1\right)}$
Solution:
- When the temperature is increased, volume of the cube will increase while density of liquid will decrease. The depth upto which the cube is submerged in the liquid remains the same.
Upthrust $=$ Weight. Therefore, upthrust should not change
$$ \begin{array}{rlrl} F & =F^{\prime} \\ & \therefore \quad V _i \rho _L g & =V _i^{\prime} \rho^{\prime}{ } _L g \quad\left(V _i=\text { volume immersed }\right) \\ \therefore & \left(A h _i\right)\left(\rho _L\right)(g) & =A\left(1+2 \alpha _s \Delta T\right)\left(h _i\right)\left(\frac{\rho _L}{1+\gamma _l \Delta T}\right) g \end{array} $$
Solving this equation, we get $\gamma _l=2 \alpha _s$
