Question

A piece of metal weighs \(45\,g\) in air and \(25\,g\) in a liquid of density \(1.5 \times 10^3\,kg\,m^{-3}\) kept at \(30^\circ C\). When the temperature of the liquid is raised to \(40^\circ C\), the metal piece weighs \(27\,g\) in the density of liquid at \(40^\circ C\) is \(1.25 \times 10^3\,kg\,m^{-3}\). The coefficient of linear expansion of metal is

• A. \(1.3 \times 10^{-3}/^\circ C\)
• B. \(5.2 \times 10^{-3}/^\circ C\)
• C. \(2.6 \times 10^{-3}/^\circ C\)
• D. \(0.26 \times 10^{-3}/^\circ C\)

Hint:

Buoyant force depends on \(\rho V\). If density decreases, apparent weight loss changes, helping find expansion of volume.

Answer 1

The Correct Option is C

Step 1: Use apparent weight loss due to buoyancy.
Apparent loss in weight = buoyant force = weight of displaced liquid.
At \(30^\circ C\):
\[ \Delta W_1 = 45g - 25g = 20g \]
At \(40^\circ C\):
\[ \Delta W_2 = 45g - 27g = 18g \]
Step 2: Relate buoyant force to density and volume.
\[ \Delta W \propto \rho V \]
So:
\[ \frac{\Delta W_1}{\Delta W_2} = \frac{\rho_1 V_1}{\rho_2 V_2} \]
Step 3: Substitute values.
\[ \frac{20}{18} = \frac{(1.5\times 10^3)V_1}{(1.25\times 10^3)V_2} \]
\[ \frac{20}{18} = \frac{1.5}{1.25}\cdot \frac{V_1}{V_2} \]
\[ \frac{V_2}{V_1} = \frac{1.5}{1.25}\cdot \frac{18}{20} \]
\[ \frac{V_2}{V_1} = 1.2 \cdot 0.9 = 1.08 \]
So volume increases by \(8%\).
Step 4: Relate volume expansion with linear expansion.
\[ \frac{\Delta V}{V} = 3\alpha \Delta T \]
Here:
\[ \frac{\Delta V}{V} = 0.08 \quad , \quad \Delta T = 10^\circ C \]
So:
\[ 0.08 = 3\alpha(10) \Rightarrow \alpha = \frac{0.08}{30} = 2.67\times 10^{-3}/^\circ C \approx 2.6\times 10^{-3}/^\circ C \]
Final Answer: \[ \boxed{2.6 \times 10^{-3}/^\circ C} \]