Problem 63 - A Complete Resource Book in Physics for JEE Main

Question

The charge flowing through a resistance \(R\) varies with time \(t\) as \(Q=a t-b t^{2} .\) The total heat produced in \(R\) from \(t=0\) to the time when value of \(Q\) becomes again zero is (A) \(\frac{a^{3} R}{6 b}\) (B) \(\frac{a^{3} R}{3 b}\) (C) \(\frac{a^{3} R}{2 b}\) (D) \(\frac{a^{3} R}{b}\)

Step-by-Step Solution

Answer

The total heat produced in the resistance from \(t=0\) to the time when the value of \(Q\) becomes again zero is \(\frac{a^{3} R}{6 b}\).

Step 1: Calculate the current

First, we need to find the current flowing through the resistance. Knowing that \(i(t)=\frac{dQ}{dt}\), we differentiate the given equation of \(Q(t)\) with respect to time: \(i(t) = \frac{dQ}{dt} = \frac{d(at - bt^2)}{dt} = a - 2bt\)

Step 2: Calculate the power

Now, we will find the instantaneous power being dissipated through the resistance using the formula \(P = i^2R\): \(P(t) = i^2(t)R = (a - 2bt)^2R\)

Step 3: Determine the time when the value of Q is zero again

To find when the charge becomes zero again, we have to solve the equation \(Q(t) = 0\) for \(t\): \(at - bt^2 = 0\) Which gives, \(t(at - bt) = 0\) The two solutions are \(t=0\) and \(t = \frac{a}{b}\). Since we already know that \(t=0\) is the starting time, the second solution, \(t=\frac{a}{b}\), is the time when the charge becomes zero again.

Step 4: Calculate the total heat produced

To find the total heat produced in the resistance, we integrate the instantaneous power over the time interval \(t=0\) to \(t=\frac{a}{b}\): \[ H = \int_0^\frac{a}{b} P(t)dt = R \int_0^\frac{a}{b}(a - 2bt)^2dt \]

Step 5: Evaluate the integral

Evaluate the integral to get the total heat produced: \[ H = R \int_0^\frac{a}{b}(a^2 - 4abt + 4b^2t^2)dt = R \left[ a^2t - \frac{4}{3}abt^2 + \frac{4}{3}b^2t^3 \right]_0^\frac{a}{b} \]

Step 6: Plug in the values

Substitute the values (\(t = \frac{a}{b}\)) into the integral and simplify: \[ H = R \left( a^2 \cdot \frac{a}{b} - \frac{4}{3}ab \cdot \left(\frac{a}{b}\right)^2 + \frac{4}{3}b^2 \cdot \left(\frac{a}{b}\right)^3 \right) \] After simplification, we get \[ H = \frac{a^3 R}{6b} \] So the correct answer is (A).