JEE Main 2016 · 9 Apr · Shift 1 · Physics Q2

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PhysicsSingle correct+4 / -1Question 2 of 90

Which of the following option correctly describes the variation of the speed $v$ and acceleration ' $a$ ' of a point mass falling vertically in a viscous medium that applies a force F = −kv, where 'k' is a constant, on the body ? (Graphs are schematic and not drawn to scale)

v rising ever more steeply, a falling

Against t: v starts at zero and rises ever more steeply; a starts high and falls.

v rising to a constant value, a falling to zero

Against t: v rises from zero and levels off at a constant value; a starts high and decays towards zero.

a rising to a constant value, v falling

Against t: a rises from zero and levels off; v starts high and decays.

v rising to a constant value, a falling linearly to zero

Against t: v rises from zero and levels off; a falls along a straight line to zero.

Answer and worked solution

Official answer: B

$m\frac{dv}{dt}=mg-kv$ : v rises from zero towards the terminal value $\frac{mg}k$ , and $a=g-\frac km v$ decays exponentially to zero. That is graph (2).

Topics:Viscosity and Stokes' law Motion graphs

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JEE Main 2016 · 9 Apr · Shift 1, Physics question 2