Question on: JAMB Physics - 2020

The question provides the linear expansivity of brass and asks for the volume at a different temperature. We need to use the concept of *volume expansivity*, which is related to linear expansivity. Here's how to solve the problem: 1. **Relationship between linear and volume expansivity:** * Volume expansivity (\(\gamma\)) is approximately three times the linear expansivity (\(\alpha\)): \(\gamma = 3\alpha\) 2. **Calculate the volume expansivity:** * Given \(\alpha = 2 \times 10^{-5} \, ^\circ C^{-1}\) * \(\gamma = 3 \times (2 \times 10^{-5} \, ^\circ C^{-1}) = 6 \times 10^{-5} \, ^\circ C^{-1}\) 3. **Apply the volume expansion formula:** * \(\Delta V = V_0 \gamma \Delta T\) * Where: * \(\Delta V\) is the change in volume * \(V_0\) is the original volume (\(2 \times 10\) cm\(^3\) = 20 cm\(^3\)) * \(\gamma\) is the volume expansivity (\(6 \times 10^{-5} \, ^\circ C^{-1}\)) * \(\Delta T\) is the change in temperature (\(100 ^\circ C - 0 ^\circ C = 100 ^\circ C\)) 4. **Calculate the change in volume:** * \(\Delta V = (20 \, cm^3) \times (6 \times 10^{-5} \, ^\circ C^{-1}) \times (100 \, ^\circ C)\) * \(\Delta V = 0.12 \, cm^3\) 5. **Calculate the final volume:** * \(V = V_0 + \Delta V\) * \(V = 20 \, cm^3 + 0.12 \, cm^3 = 20.12 \, cm^3\) Therefore, the final volume of the brass at \(100 ^\circ C\) is \(20.12 \, cm^3\). None of the given options match this result.