d
\(\text { Given, } Q=a t-b t^{2}\)

\(\therefore \quad I=\frac{d Q}{d t}=a-2 b t\)

At \(t=0, Q=0 \Rightarrow I=0\)

Also, \(I=0\) at \(t=a / 2 b\)

\(\therefore \quad\) Total heat produced in resistance \(R\)

\(H = \int\limits_0^{a/2b} {{I^2}Rdt = R\int\limits_0^{a/2b} {{{(a - 2bt)}^2}dt} } \)

\({ = R\int\limits_0^{a/2b} {\left( {{a^2} + 4{b^2}{t^2} - 4abt} \right)dt} }\)

\(=R\left[a^{2} t+4 b^{2} \frac{t^{3}}{3}-4 a b \frac{t^{2}}{2}\right]_{0}^{a / 2 b} \)

\(=R\left[a^{2} \times \frac{a}{2 b}+\frac{4 b^{2}}{3} \times \frac{a^{3}}{8 b^{3}}-\frac{4 a b}{2} \times \frac{a^{2}}{4 b^{2}}\right]\)

\(=\frac{a^{3} R}{b}\left[\frac{1}{2}+\frac{1}{6}-\frac{1}{2}\right]=\frac{a^{3} R}{6}\)