Question

A vertical column 50 cm long at 50$^{\circ}$C balances another column of same liquid 60 cm long at 100$^{\circ}$C. The coefficient of absolute expansion of the liquid is

• A. $0.005 / ^{\circ} C$
• B. $0.0005/^{\circ}C$
• C. $0.002/ ^{\circ}C$
• D. $0.0002 / ^{\circ}C$

Answer 1

The Correct Option is A

$ \, \, \, \, \, \frac{h_1}{h_2}=\frac{\rho_2}{\rho_1} =\frac{(1+\gamma \theta_1)}{(1+\gamma\theta_2)} \, \, \, \, \, \, \, \, \, \, \bigg[ \because \, \rho =\frac{\rho_0}{(1+\gamma \theta)}\bigg]$ $\Rightarrow \, \, \, \, \frac{50}{60}=\frac{1+ \gamma \times 50}{1+\gamma \times 100} \Rightarrow \, 0.005 / ^{\circ}C$

Answer 2

Given:
The initial height of the liquid column: \(h_1 = 50 \, \text{cm}\) at \(\theta_1 = 50^\circ\text{C}\)
The final height of the liquid column: \(h_2 = 60 \, \text{cm}\) at \(\theta_2 = 100^\circ\text{C}\)
We know:
\(\frac{h_1}{h_2} = \frac{\rho_2}{\rho_1} = \frac{1 + \gamma \theta_1}{1 + \gamma \theta_2}\)

Where \(\gamma\) is the coefficient of absolute expansion.
Rearrange the formula:
\(\frac{50}{60} = \frac{1 + \gamma \cdot 50}{1 + \gamma \cdot 100}\)

\(\frac{5}{6} = \frac{1 + 50\gamma}{1 + 100\gamma}\)
Cross-multiply to solve for \(\gamma\):
\(5 (1 + 100\gamma) = 6 (1 + 50\gamma)\)
\(5 + 500\gamma = 6 + 300\gamma\)
\(500\gamma - 300\gamma = 6 - 5\)
\(200\gamma = 1\)
\(\gamma = \frac{1}{200} = 0.005\degree\text{C}^{-1}\)

So, the correct option is (A): \(0.005\degree\text{C}^{-1}\)