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Question #42438

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass m=5 kgm = 5 \text{ kg} and variable internal density (configured such that its center of mass is at its base to prevent toppling) rests on a fixed rough wedge of angle θ=37∘\theta = 37^\circ. The block is connected to a hanging mass M=4 kgM = 4 \text{ kg} via a light inextensible string passing over a frictionless pulley at the top of the wedge, with the string parallel to the incline. The coefficient of static friction between the block and the wedge is μ=0.5\mu = 0.5. Initially, the system is in static equilibrium. Water of density ρw=1000 kg/m3\rho_w = 1000 \text{ kg/m}^3 slowly fills the container holding the wedge. Find the submerged volume of the block (in liters) at the exact moment it begins to slip up the incline. (Take g=10 m/s2g = 10 \text{ m/s}^2, sin⁡37∘=0.6\sin 37^\circ = 0.6, cos⁡37∘=0.8\cos 37^\circ = 0.8)

Options

  • A.

    1.0 L1.0 \text{ L}

  • B.

    1.67 L1.67 \text{ L}

  • C.

    2.5 L2.5 \text{ L}

  • D.

    1.36 L1.36 \text{ L}

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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

Static Friction Threshold|Archimedes' Buoyancy Dynamics|Wedge Constraint Logic

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of Archimedes' principle with constrained frictional motion.

The primary trap lies in the vector nature of the buoyant force.

Students often incorrectly assume that buoyancy only reduces the normal reaction NN (yielding 2.5 L2.5 \text{ L}) or only reduces the driving force down the incline (yielding 1.67 L1.67 \text{ L}).

In reality, the vertical buoyant force perfectly opposes gravity, meaning the effective weight vector magnitude decreases to (mg−Fb)(mg - F_b), which proportionally reduces both the normal force and the gravitational pull along the incline.

The variable density detail serves as a conceptual distractor to ensure the center of mass is low, preventing toppling and isolating the translational slip condition.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterLaws of Motion - Class 11 Part 1
ConceptStatic Friction Threshold
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree