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Question #42368

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass $M$ rests on a smooth horizontal surface. A uniform rod of mass $m$ and length $L$ is hinged at its upper end $A$ to the block. The lower end $B$ of the rod is connected to a fixed vertical wall to the right by a horizontal massless inextensible string. Initially, the rod is vertical and the string is taut. A constant horizontal force $F$ is applied to the block towards the right. Find the tension in the string immediately after the force is applied.

Options

A.
$\frac{m F}{2(3M + m)}$
B.
$\frac{2 m F}{6M + m}$
C.
$\frac{m F}{2M + m}$
D.
$\frac{m F}{3(2M + m)}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Torque about non-fixed axes|Constraint Relations via String Length|Physical Pendulum Inertia

Why This Question Is Tricky

Common Mistakes

This question tests the ability to handle torque in non-inertial frames and constraint relations.

The lethal trap is calculating torque about the moving hinge $A$ without including the pseudo-force torque ( $-ma$ acting at the CM).

Students who ignore the pseudo-force write $\tau_A = T L = I_A \alpha$ , leading to $T = ma/3$ and an incorrect final answer of $\frac{2mF}{6M+m}$ .

The rigorous approach requires either applying the pseudo-force correctly ( $\tau_A = ma(L/2) - TL = I_A \alpha$ ) or recognizing that the constrained bottom end $B$ ( $a_B=0$ ) serves as an instantaneous inertial pivot, drastically simplifying the torque equation.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	System of Particles and Rotational Motion - Class 11 Part 1
Concept	Torque about non-fixed axes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free