A force applied by an engine of a train of mass 2.05 x 10⁶kg changes its velocity from 5 m/s to 25 m/s in 5 minutes. The power of the engine is

The correct option is 2.05 MW.

To find the power generated by the engine of the train, we can use the relation between power, work, and kinetic energy. Power is defined as the rate of doing work, or the rate of change of kinetic energy over time:

$P=\frac{W}{t}=\frac{\mathrm{\Delta K.E.}}{t}$

1. Identify the given values:
Mass of the train ($m$):
$m=2.05\times {10}^6\text{kg}$
Initial velocity ($u$):
$u=5\text{m/s}$
Final velocity ($v$):
$v=25\text{m/s}$
Time interval ($t$):
$t=5\text{minutes}=5\times 60\text{seconds}=300\text{s}$

2. Calculate the change in kinetic energy ($\mathrm{\Delta K.E.}$):
The work done by the engine is equal to the change in the kinetic energy of the train:
$\mathrm{\Delta K.E.}=\frac{1}{2}m(v^2-u^2)$

Substituting the given values into the equation:
$\mathrm{\Delta K.E.}=\frac{1}{2}\times (2.05\times {10}^6)\times ({25}^2-5^2)$

$\mathrm{\Delta K.E.}=\frac{1}{2}\times (2.05\times {10}^6)\times (625-25)$

$\mathrm{\Delta K.E.}=\frac{1}{2}\times (2.05\times {10}^6)\times 600$

$\mathrm{\Delta K.E.}=2.05\times {10}^6\times 300\text{J}$

3. Calculate the power ($P$):
Now, divide the change in kinetic energy by the time taken:
$P=\frac{2.05\times {10}^6\times 300}{300}$

$P=2.05\times {10}^6\text{W}$

Converting Watts to Megawatts (MW), where $1\text{MW}={10}^6\text{W}$:
$P=2.05\text{MW}$