Question

Energy of the photon produced as the result of electron transition,
${{E}_{{{n}_{i}}\to {{n}_{f}}}}=Rhc\left( \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right)$
If the velocity of a body is doubled its kinetic energy
A. Gets doubled
B. Becomes half
C. Does not change
D. Becomes 4 times

Answer 1

Hint: Since we are supposed to find the trend shown by the kinetic energy when the velocity of the body is doubled, we could find the relation between the two as a first step. Then you could find the other relation by making necessary changes in the same. After that we could simply take the ratio of the two to get the answer.

Formula used:
Kinetic energy,
$K.E=\dfrac{1}{2}m{{v}^{2}}$

Complete step-by-step solution:
In the question, we are asked to find the trend shown by the kinetic energy of a body when the velocity of this body is doubled.
In order to find the trend shown by kinetic energy with change in the body’s velocity, we have to have an idea on the relation between the kinetic energy and velocity.
The kinetic energy is given in terms of velocity by the following relation,
$K.E=\dfrac{1}{2}m{{v}^{2}}$ ………………………………………………. (1)
We could take the above relation as the kinetic energy of the body when the velocity of the body is v.
Now, when the velocity is doubled, the kinetic energy of the body will become,
$K.E'=\dfrac{1}{2}m{{\left( 2v \right)}^{2}}=4\left( \dfrac{1}{2}m{{v}^{2}} \right)$……………………………… (2)
Now, we could take the ratio of the equations (1) and (2) to get,
$\dfrac{K.E}{K.E'}=\dfrac{1}{4}$
$\therefore K.E'=4K.E$
Therefore, we found that the kinetic energy of the body becomes 4times when the velocity of the body is doubled. Hence, option D is the correct answer.

Note: This is basically the steps followed when the trend shown by a certain parameter with another parameter is supposed to be found. We find the relation between the two and then find the change in the relation as mentioned and after that we take the ratio of the two to get the answer.