Heat And Thermodynamics Ques 45
- The apparatus shown in figure consists of four glass columns connected by horizontal sections. The height of two central columns $B$ and $C$ are $49 cm$ each. The two outer columns $A$ and $D$ are open to the atmosphere.
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Solution:
Formula:
Calorimetry, Thermal Expansion, Types Of Thermometers :(/important-formula/physics/heat_and_thermodynamics)
- Density of a liquid varies with temperature as
$$ \rho _{t^{\circ} C}=\left(\frac{\rho _{0{ }^{\circ} C}}{1+\gamma t}\right) $$
Here, $\gamma$ is the coefficient of volume expansion with respect to temperature.
In the figure shown
$$ h _1=52.8 cm, h _2=51 cm \text { and } h=49 cm $$
Now, pressure at $B=$ pressure at $C$
$$ \begin{aligned} p _0+h _1 \rho _{95^{\circ}} g-h \rho _{5^{\circ}} g & =p _0+h _2 \rho _{5^{\circ}} g-h \rho _{95^{\circ}} g \\ \Rightarrow \quad \rho _{95^{\circ}}\left(h _1+h\right) & =\rho _{5^{\circ}}\left(h _2+h\right) \\ \Rightarrow \quad \frac{\rho _{95^{\circ}}}{\rho _{5^{\circ}}}=\frac{h _2+h}{h _1+h} & \Rightarrow \quad \frac{\frac{\rho _{0^{\circ}}}{1+95 \gamma}}{\frac{\rho _{0^{\circ}}}{1+5 \gamma}}=\frac{h _2+h}{h _1+h} \end{aligned} $$
$\Rightarrow \quad \frac{1+5 \gamma}{1+95 \gamma}=\frac{51+49}{52.8+49}=\frac{100}{101.8}$
Solving this equation, we obtain
$$ \gamma=2 \times 10^{-4} , ^{\circ}C^{-1} $$
$\therefore$ Coefficient of linear expansion of material,
$$ \alpha=\frac{\gamma}{3}=6.7 \times 10^{-5} /{ }^{\circ} C $$
BETA
