A body of mass $$m = 10^{-2}$$ kg is moving in a medium and experiences a frictional force $$F = -kv^{2}$$. Its initial speed is $$v_{0} = 10$$ m s$$^{-1}$$. After 10 s its kinetic energy is $$\frac{1}{8}mv_{0}^{2}$$, then the value of $$k$$ will be:

We have a body of mass $$m = 10^{-2}\,{\rm kg}$$ that is moving through a medium where the resisting force is proportional to the square of the speed and is given by $$F = -k v^{2}$$. The negative sign only tells us that the force is opposite to the motion.

First we determine the speed of the body after 10 s from the information about its kinetic energy. The definition of kinetic energy is the well-known formula

$$K = \frac12 m v^{2}.$$

Initially the kinetic energy is

$$K_0 = \frac12 m v_0^{2}.$$

After a time of 10 s the kinetic energy is given to be

$$K = \frac18 m v_0^{2}.$$

Equating this to the general expression $$K=\frac12 m v^{2}$$ for the final speed $$v(t)$$, we get

$$\frac12 m v^{2} = \frac18 m v_0^{2}.$$

The mass cancels out on both sides, leaving

$$\frac12\,v^{2} = \frac18\,v_0^{2}$$

$$\Longrightarrow\; v^{2} = \frac14\,v_0^{2}$$

$$\Longrightarrow\; v = \frac{v_0}{2}.$$

Thus after 10 s the speed has dropped to one-half of its initial value:

$$v(10\,{\rm s}) = \frac{v_0}{2} = \frac{10}{2} = 5\ {\rm m\,s^{-1}}.$$

Now we turn to the dynamics. Newton’s second law gives the acceleration in terms of the force:

$$m\,\frac{dv}{dt} = -k v^{2}.$$

Dividing both sides by $$m$$ and rearranging the differentials we obtain

$$\frac{dv}{v^{2}} = -\frac{k}{m}\,dt.$$

We now integrate. At time $$t = 0$$ the speed is $$v = v_0$$, and at time $$t = T = 10\ {\rm s}$$ the speed is $$v = v(T) = v_0/2$$:

$$\int_{v_0}^{v_0/2}\!\frac{dv}{v^{2}} = -\frac{k}{m}\int_{0}^{T}\!dt.$$

On the left we use the integral $$\displaystyle\int\!v^{-2}\,dv = -\frac1v$$. Carrying out both integrations yields

$$\Bigl[-\frac{1}{v}\Bigr]_{v_0}^{v_0/2} = -\frac{k}{m}\,[\,t\,]_{0}^{T}$$

$$\left(-\frac1{v_0/2}\right) - \left(-\frac1{v_0}\right) \;=\; -\frac{k}{m}\,(T - 0).$$

Simplifying the bracket on the left:

$$-\frac{2}{v_0} + \frac{1}{v_0} = -\frac{k}{m}\,T$$

$$-\frac{1}{v_0} = -\frac{k}{m}\,T.$$

We can cancel the negative signs on both sides, giving the beautifully simple relation

$$\frac{1}{v_0} = \frac{k}{m}\,T.$$

Solving this for the unknown constant $$k$$, we find

$$k = \frac{m}{T\,v_0}.$$

All that remains is to substitute the numerical values:

$$m = 10^{-2}\ {\rm kg}, \qquad T = 10\ {\rm s}, \qquad v_0 = 10\ {\rm m\,s^{-1}}.$$

Substituting, we obtain

$$k = \frac{10^{-2}}{\,10 \times 10\,} = \frac{10^{-2}}{10^{2}} = 10^{-4}\ {\rm kg\,m^{-1}}.$$

Hence, the correct answer is Option D.