JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 8)
The charge flowing in a conductor changes with time as $$\mathrm{Q}(\mathrm{t})=\alpha \mathrm{t}-\beta \mathrm{t}^{2}+\gamma \mathrm{t}^{3}$$. Where $$\alpha, \beta$$ and $$\gamma$$ are constants. Minimum value of current is :
$$\beta-\frac{\alpha^{2}}{3 \gamma}$$
$$\alpha-\frac{3 \beta^{2}}{\gamma}$$
$$\alpha-\frac{\beta^{2}}{3 \gamma}$$
$$\alpha-\frac{\gamma^{2}}{3 \beta}$$
Explanation
$$Q(t) = \alpha t - \beta {t^2} + \gamma {t^3}$$
$$i(t) = \alpha - 2\beta t + 3\gamma {t^2}$$
$${{di} \over {dt}} = - 2\beta + 6\gamma t = 0$$ (for max/min of i)
at $$t = {\beta \over {3r}}$$ (i is minimum as i is an upward parabola)
$$i\left( {{\beta \over {3\gamma }}} \right) = \alpha - 2\beta \left( {{\beta \over {3\gamma }}} \right) + {{3\gamma {\beta ^2}} \over {9{\gamma ^2}}}$$
$$ = \alpha {{ - {\beta ^2}} \over {3\gamma }}$$
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