Water filled in an empty tank of area 10 A through a tap of cross sectional area A. The speed of water flowing out of tap is given by v(m/s)`=10(1-sin((pi)/(30)t)` where `t` is in second. The height of water level from the bottom of the tank at `t=15` second will be:

Options

A. `10m`
B. `[15+(30)/(pi)]m`
C. `(5)/(4)m`
D. `[15-(30)/(pi)]`

Answer

To find the height of the water level in the tank at \( t = 15 \) seconds, we can follow these steps:

### Step 1: Understand the flow rate
The speed of water flowing out of the tap is given by:
\[ v(t) = 10(1 - \sin\left(\frac{\pi}{30} t\right)) \]
The flow rate \( Q \) (volume per unit time) through the tap of cross-sectional area \( A \) is given by:
\[ Q = A \cdot v(t) \]

### Step 2: Express the volume of water filled in the tank
The volume of water \( dV \) that flows into the tank during a small time interval \( dt \) is:
\[ dV = Q \cdot dt = A \cdot v(t) \cdot dt \]

### Step 3: Integrate to find the total volume
To find the total volume \( V \) of water in the tank from \( t = 0 \) to \( t = 15 \) seconds, we integrate:
\[ V = \int_0^{15} A \cdot v(t) \, dt = A \int_0^{15} 10(1 - \sin\left(\frac{\pi}{30} t\right)) \, dt \]

### Step 4: Calculate the integral
We can break this integral into two parts:
\[ V = A \left( 10 \int_0^{15} dt - 10 \int_0^{15} \sin\left(\frac{\pi}{30} t\right) \, dt \right) \]
Calculating the first integral:
\[ \int_0^{15} dt = 15 \]
Calculating the second integral:
\[ \int \sin\left(\frac{\pi}{30} t\right) dt = -\frac{30}{\pi} \cos\left(\frac{\pi}{30} t\right) \]
Thus,
\[ \int_0^{15} \sin\left(\frac{\pi}{30} t\right) \, dt = -\frac{30}{\pi} \left[ \cos\left(\frac{\pi}{30} \cdot 15\right) - \cos(0) \right] \]
Calculating \( \cos\left(\frac{\pi}{30} \cdot 15\right) = \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ \int_0^{15} \sin\left(\frac{\pi}{30} t\right) \, dt = -\frac{30}{\pi} (0 - 1) = \frac{30}{\pi} \]

### Step 5: Substitute back into the volume equation
Now substituting back:
\[ V = A \left( 10 \cdot 15 - 10 \cdot \frac{30}{\pi} \right) = A \left( 150 - \frac{300}{\pi} \right) \]

### Step 6: Find the height of water in the tank
The area of the tank is \( 10A \), so the height \( h \) of the water is given by:
\[ h = \frac{V}{\text{Area of tank}} = \frac{A \left( 150 - \frac{300}{\pi} \right)}{10A} = \frac{150 - \frac{300}{\pi}}{10} \]
This simplifies to:
\[ h = 15 - \frac{30}{\pi} \]

### Final Answer
Thus, the height of the water level from the bottom of the tank at \( t = 15 \) seconds is:
\[ h = 15 - \frac{30}{\pi} \]