Question

The heat passing through the cross-section of a conductor, varies with time ‘t’ as \(Q(t) = αt – βt^2 + γt^3\). (\(α, β\) and \(γ\) are positive constants.) The minimum heat current through the conductor is

• A. \(\alpha-\frac{\beta^2}{2\gamma}\)
• B. \(\alpha-\frac{\beta^2}{3\gamma}\)
• C. \(\alpha-\frac{\beta^2}{\gamma}\)
• D. \(\alpha-\frac{3\beta^2}{\gamma}\)

Answer 1

The Correct Option is B

Given:

The heat passing through the cross-section of the rod is: \[ Q = \alpha t - \beta t^2 + \gamma t^3 \]

Step-by-Step Solution:

1. Heat Current:

The heat current (\( \frac{dQ}{dt} \)) is the rate of change of \( Q \) with respect to time: \[ \frac{dQ}{dt} = \alpha - 2\beta t + 3\gamma t^2 \]

2. Condition for Minimum Heat Current:

To find the time \( t \) at which the heat current is minimum, differentiate \( \frac{dQ}{dt} \) again: \[ \frac{d^2Q}{dt^2} = -2\beta + 6\gamma t \] Set \( \frac{d^2Q}{dt^2} = 0 \) to find the critical point: \[ -2\beta + 6\gamma t = 0 \] \[ t = \frac{2\beta}{6\gamma} = \frac{\beta}{3\gamma} \]

3. Minimum Heat Current:

Substitute \( t = \frac{\beta}{3\gamma} \) into the expression for \( \frac{dQ}{dt} \): \[ \frac{dQ}{dt}\big|_{\text{minimum}} = \alpha - 2\beta \cdot \frac{\beta}{3\gamma} + 3\gamma \cdot \left(\frac{\beta}{3\gamma}\right)^2 \] Simplify each term: \[ = \alpha - \frac{2\beta^2}{3\gamma} + \frac{\beta^2}{3\gamma} \] Combine terms: \[ \frac{dQ}{dt}\big|_{\text{minimum}} = \alpha - \frac{\beta^2}{3\gamma} \]

Final Answer:

The correct answer is (B): \( \alpha - \frac{\beta^2}{3\gamma} \).