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Question #43119

Difficulty: 8/10 | 🟢 Free

The Question

A vessel has a variable cross-sectional area given by $A(y) = c y$ , where $y$ is the vertical height from its bottom and $c = 0.10 \text{ m}$ . The vessel contains water of refractive index $\mu = 4/3$ . A small hole of area $a = 4.0 \times 10^{-5} \text{ m}^2$ is opened at the bottom ( $y=0$ ), allowing water to leak out. A small coin is fixed at the bottom of the vessel. An observer in air looks at the coin from vertically above. When the height of the water is $h = 0.20 \text{ m}$ , what is the speed of the image of the coin relative to the stationary observer? (Assume the water surface remains horizontal and take $g = 10 \text{ m/s}^2$ )

Options

A.
$1.0 \text{ mm/s}$
B.
$3.0 \text{ mm/s}$
C.
$4.0 \text{ mm/s}$
D.
$7.0 \text{ mm/s}$

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

Apparent Depth/Shift|Torricelli’s Efflux Velocity|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly identify reference frames when combining fluid kinematics with ray optics.

A common trap is to differentiate the apparent depth formula $d' = h/\mu$ directly with respect to time, yielding $v_{image} = (1/\mu)(dh/dt) = 3.0 \text{ mm/s}$ .

However, this incorrectly calculates the velocity of the image relative to the moving water surface.

Because the surface itself is falling at $v_s = 4.0 \text{ mm/s}$ , the absolute velocity of the image must be found by defining its position coordinate from a stationary datum (the bottom of the tank).

By failing to account for the motion of the refracting boundary, students misapply the relative velocity vector addition.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Ray Optics and Optical Instruments - Class 12
Concept	Apparent Depth/Shift
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free