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

The problem requires us to find the minimum heat current through the conductor. Let's start by breaking down the problem step-by-step.

The heat passing through the cross-section of a conductor is given by the equation: \(Q(t) = \alpha t - \beta t^2 + \gamma t^3\).

The heat current, which is the rate of change of heat with respect to time, can be defined as the derivative of \(Q(t)\) with respect to \(t\). Therefore, we differentiate \(Q(t)\) to find the heat current, \(I(t)\):

\(I(t) = \frac{dQ}{dt} = \frac{d}{dt}(\alpha t - \beta t^2 + \gamma t^3)\)

Let's differentiate each term:

• \(\frac{d}{dt}(\alpha t) = \alpha\)
• \(\frac{d}{dt}(-\beta t^2) = -2\beta t\)
• \(\frac{d}{dt}(\gamma t^3) = 3\gamma t^2\)

Combining these, we have:

\(I(t) = \alpha - 2\beta t + 3\gamma t^2\)

To find the minimum value of the heat current, we need to find the critical points by setting the derivative of \(I(t)\) with respect to \(t\) to zero:

\(\frac{dI}{dt} = \frac{d}{dt}(\alpha - 2\beta t + 3\gamma t^2) = -2\beta + 6\gamma t\)

Set the derivative to zero:

\( -2\beta + 6\gamma t = 0 \)

Simplifying gives:

\( t = \frac{\beta}{3\gamma} \)

Substitute this value of \(t\) back into the expression for \(I(t)\) to find the minimum heat current:

\(I(t) = \alpha - 2\beta \left(\frac{\beta}{3\gamma}\right) + 3\gamma \left(\frac{\beta}{3\gamma}\right)^2\)

Calculating the above expression:

\(I(t) = \alpha - \frac{2\beta^2}{3\gamma} + \frac{3\beta^2}{9\gamma}\)

Simplify further:

\(I(t) = \alpha - \frac{2\beta^2}{3\gamma} + \frac{\beta^2}{3\gamma}\)

This reduces to:

\(I(t) = \alpha - \frac{\beta^2}{3\gamma} \)

Thus, the minimum heat current through the conductor is \(\alpha - \frac{\beta^2}{3\gamma}\).

Therefore, the correct answer is \(\alpha-\frac{\beta^2}{3\gamma}\).