Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating place is pulled horizontally with a constant velocity \mu _0. Which of the following statements is (are) true?

The resistive force of liquid on the plate is inversely proportional to h, The tangential (shear) stress on the floor of the tank increases with \mu _0, The tangential (shear) stress on the plate varies linearly with the viscosity \eta of the liquid

The resistive force of liquid on the plate is inversely proportional to h

The resistive force of liquid on the plate is independent of the area of the plate

The tangential (shear) stress on the floor of the tank increases with \mu _0

The tangential (shear) stress on the plate varies linearly with the viscosity \eta of the liquid

Properties of Matter PYQ — JEE Advanced 2018 · Shift 2 · Q38 with Solution

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Correct answer: A, D, C

Model the situation

A thin square plate floats on the surface of a viscous liquid and is pulled horizontally with constant speed $u_0$ .

Since the tank is large and the liquid depth $h$ is small compared to tank width, we can model the flow beneath the plate as simple laminar shear flow (like Couette flow):

Velocity at the bottom of the tank: $u=0$
Velocity at the plate: $u=u_0$
Liquid depth: $h$

Hence the velocity gradient is approximately

\frac{du}{dy} = \frac{u_0}{h}

Use Newton's law of viscosity

The shear stress in a viscous liquid is

\tau = \eta \frac{du}{dy}

So here,

\tau = \eta \frac{u_0}{h}

This shear stress acts throughout the liquid layer, including:

on the underside of the plate,
on the floor of the tank.

Force on the plate

If the area of the plate is $A$ , then resistive force by the liquid on the plate is

F = \tau A = \eta \frac{u_0}{h} A

Now check each statement.

Option A: The resistive force of liquid on the plate is inversely proportional to $h$

From

F = \eta \frac{u_0}{h} A

we see that

F \propto \frac{1}{h}

So A is true.

Option B: The resistive force of liquid on the plate is independent of the area of the plate

But

F = \eta \frac{u_0}{h} A

Thus force is directly proportional to area $A$ .

So B is false.

Option C: The tangential (shear) stress on the floor of the tank increases with $u_0$

Shear stress on the floor is

\tau_{\text{floor}} = \eta \frac{u_0}{h}

Clearly this increases linearly with $u_0$ .

So C is true.

Option D: The tangential (shear) stress on the plate varies linearly with the viscosity $\eta$ of the liquid

Shear stress on the plate is

\tau_{\text{plate}} = \eta \frac{u_0}{h}

Thus

\tau_{\text{plate}} \propto \eta

So D is true.

Final derived answer

Correct options are:

\boxed{A,\ C,\ D}

Comparison with stored correct answer

Stored correct answer: $A, D, C$

This is the same set as $A, C, D$ .

So the derived answer agrees with the stored answer.

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View written solutionFree

Correct answer: A, D, C

Model the situation

A thin square plate floats on the surface of a viscous liquid and is pulled horizontally with constant speed $u_0$ .

Since the tank is large and the liquid depth $h$ is small compared to tank width, we can model the flow beneath the plate as simple laminar shear flow (like Couette flow):

Velocity at the bottom of the tank: $u=0$
Velocity at the plate: $u=u_0$
Liquid depth: $h$

Hence the velocity gradient is approximately

\frac{du}{dy} = \frac{u_0}{h}

Use Newton's law of viscosity

The shear stress in a viscous liquid is

\tau = \eta \frac{du}{dy}

So here,

\tau = \eta \frac{u_0}{h}

This shear stress acts throughout the liquid layer, including:

on the underside of the plate,
on the floor of the tank.

Force on the plate

If the area of the plate is $A$ , then resistive force by the liquid on the plate is

F = \tau A = \eta \frac{u_0}{h} A

Now check each statement.

Option A: The resistive force of liquid on the plate is inversely proportional to $h$

From

F = \eta \frac{u_0}{h} A

we see that

F \propto \frac{1}{h}

So A is true.

Option B: The resistive force of liquid on the plate is independent of the area of the plate

But

F = \eta \frac{u_0}{h} A

Thus force is directly proportional to area $A$ .

So B is false.

Option C: The tangential (shear) stress on the floor of the tank increases with $u_0$

Shear stress on the floor is

\tau_{\text{floor}} = \eta \frac{u_0}{h}

Clearly this increases linearly with $u_0$ .

So C is true.

Option D: The tangential (shear) stress on the plate varies linearly with the viscosity $\eta$ of the liquid

Shear stress on the plate is

\tau_{\text{plate}} = \eta \frac{u_0}{h}

Thus

\tau_{\text{plate}} \propto \eta

So D is true.

Final derived answer

Correct options are:

\boxed{A,\ C,\ D}

Comparison with stored correct answer

Stored correct answer: $A, D, C$

This is the same set as $A, C, D$ .

So the derived answer agrees with the stored answer.

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