JEE Main 2017 · 9 Apr · Shift 1 · Physics Q4

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PhysicsSingle correct+4 / -1Question 4 of 90

The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a :

Two 1 m rods forming an inverted V with a 2 kg weight at the top pivot; the right rod's foot is a fixed pivot on the floor, the left rod's foot is a movable roller pushed right by F; x is the distance between the feet.

constant speed

decreasing speed

increasing speed

speed which is

\dfrac34

th of that of the roller when the weight is 0.4 m above the ground

Answer and worked solution

Official answer: B

With the feet $x$ apart, the height is $y=\sqrt{1-x^2/4}$ . Differentiating, $\dfrac{dy}{dt}=-\dfrac x{4y}\dfrac{dx}{dt}$ . As the roller moves right, $x$ falls and $y$ rises, so $\frac x{4y}$ falls: the weight rises with decreasing speed.

(At $y=0.4$ m, $\frac x{4y}=\frac{1.83}{1.6}\approx1.15$ , not $\frac34$ .)

Topics:Motion in a straight line

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 4