JEE Main 2022 · 24 Jun · Shift 2 · Physics Q42

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PhysicsSingle correct+4 / -1Question 42 of 90

An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use $g=10\ \mathrm{ms^{-2}}$ ].

1:1

\sqrt2:\sqrt3

\sqrt3:\sqrt2

2:3

Answer and worked solution

Official answer: B

Accelerations. The drag is $\frac{10}5=2$ m s $^{-2}$ per unit mass. Going up: $a_1=10+2=12$ m s $^{-2}$ . Coming down: $a_2=10-2=8$ m s $^{-2}$ .

Same height. $h=\frac12a_1t_1^2=\frac12a_2t_2^2$ , so $\dfrac{t_1}{t_2}=\sqrt{\dfrac{a_2}{a_1}}=\sqrt{\dfrac8{12}}=\dfrac{\sqrt2}{\sqrt3}$ .

(B)

Topics:Motion in a straight line Newton's laws

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JEE Main 2022 · 24 Jun · Shift 2, Physics question 42