A running man has the same kinetic energy as that of a boy of half his mass. The man speeds up by 2 m/s and the boy changes his speed by x m/s so that the kinetic energies of the boy and the man are again equal. Then x in m/s is
(a) $ - 2\sqrt 2 $
(b) $2\sqrt 2 $
(c) $\sqrt 2 $
(d) $2$

Hint: This question uses the concept of kinetic energy, so students should know the formula of kinetic energy. Firstly, students need to find out the relation between the speed of the man and boy, from the first case. Later by using the second case, you can easily evaluate the value of the increased speed of the boy by comparing the expressions.

Complete step by step answer:
The following are the data given in the question.
The speed of the man is two times the mass of the boy, and the kinetic energy of the man is equal to the kinetic energy of the boy.

Let the mass of the boy be m, so the mass of the man will be 2m.
We can calculate the kinetic energy of the man as,
$
{\left( {KE} \right)_m} = \dfrac{1}{2}\left( {2m} \right)v_m^2\\
\implies {\left( {KE} \right)_m} = mv_m^2
$

Similarly we can calculate the kinetic energy of the boy as,
${\left( {KE} \right)_b} = \dfrac{1}{2}mv_b^2$

As the kinetic energy of man and boy is equal. So on comparing both the expressions,
$
mv_m^2 = \dfrac{1}{2}mv_b^2\\
\implies {v_b} = \sqrt 2 {v_m}......\left( 1 \right)
$

So we can clearly see that the velocity of the boy is $\sqrt 2 $ time more than the velocity of the man.

Now we look at the second case of the question.
The velocity of the man is increased by 2 m/s, so we can calculate the kinetic energy as,
$
{\left( {KE} \right)_m} = \dfrac{1}{2} \times \left( {2m} \right) \times {\left( {{v_m} + 2} \right)^2}\\
\implies {\left( {KE} \right)_m} = m{\left( {{v_m} + 2} \right)^2}
$

Similarly, the velocity of the boy is increased by x m/s, so we can calculate the kinetic energy as,
${\left( {KE} \right)_b} = \dfrac{1}{2} \times m \times {\left( {{v_b} + x} \right)^2}$

Again the kinetic energy of both man and boy is equal, we can write it as,
$
m{\left( {{v_m} + 2} \right)^2} = \dfrac{1}{2} \times m \times {\left( {{v_b} + x} \right)^2}\\
\implies 2{\left( {{v_m} + 2} \right)^2} = {\left( {{v_b} + x} \right)^2}\\
\implies {v_b} + x = \sqrt 2 \left( {{v_m} + 2} \right)
$

Further solving the above expression,
$
{v_b} + x = \sqrt 2 \left( {{v_m} + 2} \right)\\
\implies {v_b} + x = \sqrt 2 {v_m} + 2\sqrt 2
$

Substituting equation (1) in above expression,
$
{v_b} + x = {v_b} + 2\sqrt 2 \\
\implies x = 2\sqrt 2
$

Therefore, the value of x in m/s is $2\sqrt 2 $.

Therefore, the option (B) is the correct answer.

Note:
Students can make a mistake while performing the calculation, as it is a little bit complex. As it is mentioned in the question that the kinetic energy of boy and man are equal in both the cases, so, you need to equate them, in order to get the answer.