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Question #78716

Difficulty: 5/10 | 🟢 Free

The Question

A particle of mass $m$ is constrained to move along the $+x$ -axis. It experiences a conservative force such that its acceleration is given by $a(x) = \alpha x^2 - \beta x^3$ , where $\alpha$ and $\beta$ are positive constants with appropriate SI units. The particle is placed at the origin ( $x=0$ ) and given an infinitesimal positive displacement. Which of the following options correctly identifies the dimensional formula for $\beta$ , the position of stable equilibrium $x_0$ , and the maximum velocity $v_{max}$ acquired by the particle during its subsequent motion?

Options

A.
$[\text{L}^{-2}\text{T}^{-2}], \; \frac{\alpha}{\beta}, \; \frac{\alpha^2}{\sqrt{6\beta^3}}$
B.
$[\text{L}^{-2}\text{T}^{-2}], \; \frac{2\alpha}{3\beta}, \; \frac{\alpha^2}{\sqrt{2\beta^3}}$
C.
$[\text{M L}^{-2}\text{T}^{-2}], \; \frac{\alpha}{\beta}, \; \frac{\alpha^2}{\sqrt{6\beta^3}}$
D.
$[\text{L}^{-2}\text{T}^{-2}], \; \frac{\alpha}{\beta}, \; \frac{\alpha^2}{\sqrt{12\beta^3}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Equilibrium Stability|Conservative Field Path Independence|Non-Uniform Acceleration Constraint|Integration

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the student's ability to map a non-linear acceleration constraint to potential energy and stability, while avoiding common algebraic and dimensional traps.

Students often incorrectly assume the potential energy $U(x)$ is directly proportional to the acceleration function, leading them to find the minimum of $a(x)$ (at $x = 2\alpha/3\beta$ ) instead of the zero of $a(x)$ for equilibrium.

Furthermore, they frequently forget that $a(x)$ is independent of mass, trapping them into assigning force dimensions $[\text{M L}^{-2}\text{T}^{-2}]$ to $\beta$ .

Finally, calculating $v_{max}$ requires correctly applying the Work-Energy theorem and recognizing that maximum velocity occurs exactly at the stable equilibrium point where $a(x)=0$ , not at the maximum of the $a(x)$ curve.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Work, Energy, and Power - Class 11 Part 1
Concept	Equilibrium Stability
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free