JEE Main 2017 · 2 Apr · Shift 1 · Physics Q3

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PhysicsSingle correct+4 / -1Question 3 of 90

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 $v_0=10$ ms $^{-1}$ . If, after 10 s, its energy is $\dfrac18mv_0^2$ , the value of k will be :

10^{-3}

kg m

^{-1}

10^{-3}

kg s

^{-1}

10^{-4}

kg m

^{-1}

10^{-1}

kg m

^{-1}

^{-1}

Answer and worked solution

Official answer: C

$m\dfrac{dv}{dt}=-kv^2$ gives $\dfrac1v-\dfrac1{v_0}=\dfrac{kt}m$ .

The energy is $\frac18$ of the initial, so $v=\dfrac{v_0}{2\sqrt2}$ and
$\frac{2\sqrt2-1}{10}=\frac{k\times10}{10^{-2}}\implies k=\frac{1.83\times10^{-2}}{100}\approx1.8\times10^{-4}\text{ kg m}^{-1}.$
The closest option is $10^{-4}$ kg m⁻¹.

Topics:Newton's laws Work-energy theorem

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JEE Main 2017 · 2 Apr · Shift 1, Physics question 3