A charged capacitor of capacitance C is discharged through a resistance R. A radio active sample decays with an average life T. Find R in terms of C and t in order that the ratio of the electrostatic energy stored in the capacitor to the activity of the radio active sample remains constant with time

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2TC

2T/C

C/2T

answer is A.

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Detailed Solution

$\begin{array}{l} U = \frac{1}{2} \frac{q^{2}}{c} where q = q_{0} e^{- t / RC} \\ U = \frac{1}{2} \frac{q_{0}^{2}}{c} e^{- 2 t / RC}; A = Nλ = \frac{N}{T} (∵ τ = T = \frac{1}{λ}) \\ \Rightarrow \frac{U}{A} = \frac{1}{2} \frac{q_{0}^{2}}{C} \frac{(e^{- 2 t / RC})}{\frac{N}{T}} = \frac{1}{2} \frac{q_{0}^{2}}{C} \frac{(e^{- 2 t / RC})}{N_{0} e^{- \frac{t}{T}}} T \end{array}$
Since U/A is constant with time
$\begin{array}{l} \Rightarrow \frac{d}{dt} (\frac{U}{A}) = 0 \Rightarrow \frac{q_{0}^{2} T}{2 N_{0} C} \frac{d}{dt} [(e^{- (2 t / RC) + (t / T)})] = 0 \\ \Rightarrow [(e^{- (2 t / RC) + (t / T)})] (\frac{- 2 t}{R C} + \frac{t}{T}) = 0 \\ \Rightarrow [\frac{- 2 t}{RC} + \frac{t}{T}] = 0 \Rightarrow R = \frac{2 T}{C} \end{array}$