A boat with base area `8m ^(2)` floating on the surface of a still river is intended to move with a constant speed of 2 m/s by the application of a horizontal force. If the river bed is 2m deep find the force needed, (assuming a constant velocity gradient) Coefficient of viscosity of water is `0.90 xx 10 ^(-2)` poise.

Options

A. 729 dyne
B. 620 dyne
C. 520 dyne
D. 360 dyne

Explanation

To find the force needed to move the boat with a constant speed of 2 m/s in the river, we can use the concept of viscosity and the formula for viscous drag. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have a boat with a base area of \( A = 8 \, m^2 \) floating on a river. The boat is moving with a constant speed \( v = 2 \, m/s \). The coefficient of viscosity of water is given as \( \eta = 0.90 \times 10^{-2} \, \text{poise} = 0.90 \times 10^{-2} \times 0.1 \, \text{Pa.s} = 0.009 \, \text{Pa.s} \).

### Step 2: Calculate the velocity gradient
Since the riverbed is 2 m deep and the boat is moving at the surface, the velocity gradient \( \frac{du}{dy} \) can be approximated as:
\[ \frac{du}{dy} = \frac{v}{h} = \frac{2 \, m/s}{2 \, m} = 1 \, s^{-1} \]
where \( h \) is the depth of the river (2 m).

### Step 3: Calculate the viscous force
Using the formula for viscous force \( F = \eta A \frac{du}{dy} \), we can substitute the known values:
\[ F = \eta A \frac{du}{dy} = (0.009 \, \text{Pa.s}) \times (8 \, m^2) \times (1 \, s^{-1}) \]
Calculating this gives:
\[ F = 0.009 \times 8 = 0.072 \, \text{N} \]

### Step 4: Convert the force to the appropriate unit
Since the force is often expressed in dynes in some contexts, we can convert it:
\[ 1 \, N = 10^5 \, dynes \]
Thus,
\[ F = 0.072 \, N \times 10^5 \, \text{dynes/N} = 7200 \, \text{dynes} \]

### Step 5: Final answer
The force needed to move the boat at a constant speed of 2 m/s is \( 7200 \, \text{dynes} \).

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