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Step-by-Step Solution
Step 1: Understand the Problem
A body of mass $m = 10^{-2}$ kg moves under a frictional (drag) force $F = -k v^2$, starting at an initial speed $v_0 = 10\,\text{m/s}$. After $10\,\text{s}$, its kinetic energy becomes $\tfrac{1}{8}m v_0^2$. We need to find the constant $k$.
Step 2: Relate Final Kinetic Energy to Final Speed
The initial kinetic energy is $\tfrac{1}{2} m v_0^2$. The final kinetic energy (after 10 s) is given as $\tfrac{1}{8} m v_0^2$. Let $v_f$ be the final speed after 10 s.
Since kinetic energy $K = \tfrac{1}{2} m v^2$, we have:
$$
\tfrac{1}{2} m v_f^2 = \tfrac{1}{8} m v_0^2.
$$
Canceling $\tfrac{1}{2} m$ on both sides, we get
$$
v_f^2 = \tfrac{1}{4} v_0^2 \quad \Rightarrow \quad v_f = \tfrac{v_0}{2}.
$$
Substituting $v_0 = 10\,\text{m/s}$,
$$
v_f = \tfrac{10}{2} = 5\,\text{m/s}.
$$
Step 3: Write the Equation of Motion
The net force on the body is $F = m \dfrac{dv}{dt} = -k v^2.$ Hence,
$$
m \dfrac{dv}{dt} = -k v^2.
$$
Substitute $m = 10^{-2}\,\text{kg}$:
$$
10^{-2} \dfrac{dv}{dt} = -k v^2.
$$
Rearrange to separate variables:
$$
\dfrac{dv}{v^2} = -100\,k \, dt.
$$
Step 4: Perform the Integration
Integrate both sides between the appropriate limits: from $v = 10\,\text{m/s}$ to $v = 5\,\text{m/s}$ for velocity, and from $t = 0$ to $t = 10\,\text{s}$ for time:
$$
\int_{10}^{5} \dfrac{dv}{v^2} = -100\,k \int_{0}^{10} dt.
$$
The integral of $\dfrac{1}{v^2}$ with respect to $v$ is $-\dfrac{1}{v}$. Carefully evaluating each side:
$$
\left[-\dfrac{1}{v}\right]_{10}^{5} = -100\,k \times [\,t\,]_{0}^{10}.
$$
Thus,
$$
\Bigl(-\dfrac{1}{5} + \dfrac{1}{10}\Bigr) = -100\,k \times (10 - 0).
$$
Note that $-\tfrac{1}{5} + \tfrac{1}{10} = -\tfrac{2}{10} + \tfrac{1}{10} = -\tfrac{1}{10}.$ So we have
$$
-\tfrac{1}{10} = -100\,k \times 10.
$$
Step 5: Solve for $k$
Simplify the right-hand side:
$$
-\tfrac{1}{10} = -1000\,k.
$$
Divide both sides by $-1000$:
$$
k = \dfrac{1}{10} \times \dfrac{1}{1000} = 10^{-4}\,\text{kg m}^{-1}.
$$
Step 6: Conclusion
The value of $k$ is
$$
k = 10^{-4}\,\text{kg m}^{-1}.
$$
This matches the correct answer.
