Answer
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Hint: We know the expression of kinetic energy in terms of mass and velocity. We need to calculate the rate of change of kinetic energy where the mass per unit length is varying for the water that is being pumped out of the hose.
Detailed step by step solution:
The expression for the kinetic energy is given as
$K = \dfrac{1}{2}m{{\text{v}}^2}{\text{ }}...{\text{(i)}}$
We are given that an engine pumps out water continuously through a hose with a velocity v. The water has kinetic energy given by equation (i). The engine provides kinetic energy to the water that is being pumped out. We need to find the rate at which this kinetic energy is being imparted to the water. This rate can be given as $\dfrac{{dK}}{{dt}}$. Using the equation (i), we get the rate of change of kinetic energy to be
$\dfrac{{dK}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{1}{2}m{{\text{v}}^2}} \right){\text{ }}...{\text{(ii)}}$
The velocity that is being imparted to the water by the engine is constant. We are given that the mass per unit length of the water. If it is denoted by m then we can write
$
m = \dfrac{M}{l} \\
\Rightarrow M = ml \\
$
Using this expression in equation (ii), we get
$
\dfrac{{dK}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{1}{2}ml{{\text{v}}^2}} \right) \\
= \dfrac{1}{2}{\text{m}}{{\text{v}}^2}\dfrac{{dl}}{{dt}} \\
$
The rate of change of length with time is equal to the velocity of the water that is being pumped out. Therefore, we can write
$\dfrac{{dK}}{{dt}} = \dfrac{1}{2}{\text{m}}{{\text{v}}^3}$
Hence, the correct answer is option B.
Note: Energy can be of two types: Kinetic energy and potential energy. Kinetic energy is the energy possessed by a body by the virtue of its motion. Potential energy is the energy possessed by a body virtue of its position. A body at rest has zero kinetic energy but non-zero potential energy.
Detailed step by step solution:
The expression for the kinetic energy is given as
$K = \dfrac{1}{2}m{{\text{v}}^2}{\text{ }}...{\text{(i)}}$
We are given that an engine pumps out water continuously through a hose with a velocity v. The water has kinetic energy given by equation (i). The engine provides kinetic energy to the water that is being pumped out. We need to find the rate at which this kinetic energy is being imparted to the water. This rate can be given as $\dfrac{{dK}}{{dt}}$. Using the equation (i), we get the rate of change of kinetic energy to be
$\dfrac{{dK}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{1}{2}m{{\text{v}}^2}} \right){\text{ }}...{\text{(ii)}}$
The velocity that is being imparted to the water by the engine is constant. We are given that the mass per unit length of the water. If it is denoted by m then we can write
$
m = \dfrac{M}{l} \\
\Rightarrow M = ml \\
$
Using this expression in equation (ii), we get
$
\dfrac{{dK}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{1}{2}ml{{\text{v}}^2}} \right) \\
= \dfrac{1}{2}{\text{m}}{{\text{v}}^2}\dfrac{{dl}}{{dt}} \\
$
The rate of change of length with time is equal to the velocity of the water that is being pumped out. Therefore, we can write
$\dfrac{{dK}}{{dt}} = \dfrac{1}{2}{\text{m}}{{\text{v}}^3}$
Hence, the correct answer is option B.
Note: Energy can be of two types: Kinetic energy and potential energy. Kinetic energy is the energy possessed by a body by the virtue of its motion. Potential energy is the energy possessed by a body virtue of its position. A body at rest has zero kinetic energy but non-zero potential energy.
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