Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A radioactive sample decays with a mean life of $20 \mathrm{~ms}$ . A capacitor of capacitance $100 \mu \mathrm{F}$ is charged to some potential and then the plates are connected through a resistance $R$ , The resistance $\mathrm{R}$ for which the ratio of the charge on the capacitance to the activity of the radioactive sample remains constant in time is found $\mathrm{y} x \, 100 \Omega$ . Value of $\mathrm{y}$ is
Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

Answers (1)

best_answer

The activity of radioactive sample of decay constant $\lambda$ at time $t$ is given by

$\mathrm{A}=\mathrm{A}_0 \mathrm{e}^{-\lambda t}$ $........(1)$

$\text { Where } A_0 \text { is initial activity }$

The charge on capacitor in series
RC-circuit after time t is

$\mathrm{Q}=\mathrm{Q}_0 \mathrm{e}^{-\frac{\mathrm{t}}{\mathrm{RC}}}$ $........(2)$

Where $Q_0$ is initial charge

Dividing (2) by (1) , we get

$\frac{\mathrm{Q}}{\mathrm{A}}=\frac{\mathrm{Q}_0}{\mathrm{~A}_0} \frac{\mathrm{e}^{\mathrm{t} / \mathrm{RC}}}{\mathrm{e}^{-\lambda t}}$

$\Rightarrow \frac{Q}{A}=\frac{Q_0}{A} e^{\left(\lambda-\frac{1}{R C}\right) t}$

Clearly this ratio will be independent of t if

$\lambda-\frac{1}{\mathrm{RC}}=0 \Rightarrow \lambda=\frac{1}{\mathrm{RC}} \Rightarrow \mathrm{R}=\frac{1}{\mathrm{C} \lambda}$

$\text { or as } \lambda=\frac{1}{\tau}$

$\begin{aligned} & \Rightarrow R=\frac{\tau}{C}=\frac{20 \times 10^{-3} \mathrm{~s}}{100 \times 10^{-6} \mathrm{~F}}=200 \Omega \\ & \Rightarrow y=2 \end{aligned}$

Posted by

Rishi

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Karnataka: NEET candidate dies by suicide; no note recovered

anam-khan

CBSE Re-evaluation Row: Dharmendra Pradhan seeks detailed report on technical failures

anam-khan

Before NEET, CMC Vellore’s unique MBBS admissions tested aptitude along with merit; paper-leak restarts debate

Latest Question