Q 11-04-200JEE MainJEE Main 2017 (2 Apr)Medium
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 $v_0 = 10\ \text{m s}^{-1}$. After $10$ s its kinetic energy is $\frac18 mv_0^2$. Then the value of $k$ will be:
Answer: (D) $10^{-4}\ \text{kg m}^{-1}$
After $10$ s, $\frac12mv^2 = \frac18 mv_0^2$ gives $v^2 = \frac{v_0^2}{4}$, so $v = \dfrac{v_0}{2} = 5$ m/s.
Equation of motion:
$$m\frac{dv}{dt} = -kv^2 \;\Rightarrow\; \frac{1}{v} - \frac{1}{v_0} = \frac{k}{m}t$$
$$\frac15 - \frac1{10} = \frac{k}{10^{-2}}\times10 \;\Rightarrow\; 0.1 = 1000k \;\Rightarrow\; k = 10^{-4}\ \text{kg m}^{-1}$$
(The unit follows from $kv^2$ being a force.)
Solution by Sreeraj P, M.Sc Physics