A rocket rises vertically up from the surface of earth so that it is distance veries from the earth's surface as `l=bt^(2)` where 'b' is a constant. After `10 sec` the rocket has travelled `2 Km`. Determine it's speed at that moment.

Options

A. `100 m//s`
B. `200 m//s`
C. `400 m//s`
D. `600 m//s`

Explanation

To solve the problem step by step, we will follow the given information and derive the necessary values to find the speed of the rocket after 10 seconds.

### Step 1: Understand the given equation
The distance \( l \) of the rocket from the Earth's surface is given by the equation:
\[ l = bt^2 \]
where \( b \) is a constant and \( t \) is the time in seconds.

### Step 2: Substitute the known values
We know that after \( t = 10 \) seconds, the rocket has traveled \( 2 \) km, which is equal to \( 2000 \) meters. Therefore, we can set up the equation:
\[ l = b(10^2) = 2000 \]
This simplifies to:
\[ l = b(100) \]

### Step 3: Solve for the constant \( b \)
From the equation \( 100b = 2000 \), we can solve for \( b \):
\[ b = \frac{2000}{100} = 20 \]

### Step 4: Find the velocity of the rocket
The velocity \( v \) of the rocket can be found by taking the derivative of the distance \( l \) with respect to time \( t \):
\[ v = \frac{dl}{dt} = \frac{d}{dt}(bt^2) \]
Using the power rule of differentiation:
\[ v = 2bt \]

### Step 5: Substitute \( b \) and \( t \) to find the velocity at \( t = 10 \) seconds
Now we substitute \( b = 20 \) and \( t = 10 \) into the velocity equation:
\[ v = 2(20)(10) \]
\[ v = 400 \text{ m/s} \]

### Final Answer
The speed of the rocket after 10 seconds is:
\[ \boxed{400 \text{ m/s}} \]

---