A particle is moving in translator motion. If momentum of the particle decreases by $10\% $ , kinetic energy will decrease by:
A. $10\% $
B. $5\% $
C. $20\% $
D. $19\% $

Hint: To solve this question, first we will write the formula for momentum with respect to the Kinetic energy and then we will explain the relationship between the momentum and the Kinetic Energy. And, further calculations, we will get the change in kinetic energy and the percentage of it.

Complete step by step solution:
According to the question,
Momentum of the particle decreases by $10\% $ .
As we know, momentum of particle:
$p = \sqrt {2mK} $
here, $p$ is the momentum of the particle
and, $K$ is the Kinetic energy of the moving particle.
Now, as we know, momentum is directly proportional to the root of Kinetic energy:
$p\propto \sqrt K $
Similarly, the final momentum is directly proportional to the root of another kinetic energy:
${p_1}\propto \sqrt {{K_1}} $
And, the final momentum:
${p_1} = p - \dfrac{{10p}}{{100}} $
$\Rightarrow {p_1} = \dfrac{{90p}}{{100}} $
So,
$\dfrac{p}{{{p_1}}} = \sqrt {\dfrac{K}{{{K_1}}}} $
$\Rightarrow {(\dfrac{{100}}{{90}})^2} = \dfrac{K}{{{K_1}}} $
$\Rightarrow \dfrac{K}{{{K_1}}} = \dfrac{{10000}}{{8100}} = \dfrac{{100}}{{81}} $
$ \text{or } \dfrac{{{K_1}}}{K} = \dfrac{{81}}{{100}} $
Decreases both sides by 1:
$\Rightarrow 1 - \dfrac{{{K_1}}}{K} = 1 - \dfrac{{81}}{{100}} $
$\Rightarrow \dfrac{{K - {K_1}}}{K} = \dfrac{{100 - 81}}{{100}} $
$\Rightarrow \dfrac{{K - {K_1}}}{K} = \dfrac{{19}}{{100}} $
Now, the change in Kinetic Energy:
$(\dfrac{{K - {K_1}}}{K}) \times 100 = \dfrac{{19}}{{100}} \times 100 = 19\% $
Therefore, Kinetic Energy will decrease by $19\% $ .
Hence, the correct option is D. $19\% $.

Note:
Change in kinetic energy is the energy the body possesses by virtue of the change in motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.