Before You Begin

Question #42026

Difficulty: 8/10 | 🟢 Free

The Question

A particle starts from rest at the origin $x=0$ at time $t=0$ . It is subjected to a time-varying force such that its acceleration $a(t)$ is directly proportional to the instantaneous activity $A(t)$ of a radioactive sample. The sample has an initial number of active nuclei $N_0$ and a decay constant $\lambda$ . Given the proportionality constant is $k$ (such that $a(t) = k A(t)$ ), what is the displacement of the particle at the exact moment the activity of the sample reduces to $\frac{1}{e}$ of its initial value?

Options

A.
$\frac{k N_0}{e \lambda}$
B.
$\frac{k N_0}{\lambda} \left( 1 - \frac{1}{e} \right)$
C.
$\frac{k N_0}{\lambda}$
D.
$\frac{k N_0}{e}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Velocity-Time Slope-Area Duality|Radioactive Decay Law Dynamics|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the duality between kinematics and radioactive decay laws, specifically focusing on time-varying acceleration.

The primary trap lies in the boundary conditions during integration.

Students often incorrectly integrate $a(t)$ without applying $v(0)=0$ , erroneously obtaining $v(t) \propto e^{-\lambda t}$ instead of $1 - e^{-\lambda t}$ .

Consequently, they miss the linear time term in the displacement function and arrive at Option B.

It forces students to reconcile stochastic decay statistics with rigorous kinematic calculus, punishing those who rely on constant-acceleration formulas or sloppy integration.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	Velocity-Time Slope-Area Duality
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free