
vA=dtdx=dtd[2lcos(θ)]=−2lsin(θ)dtdθ and
Vertical velocity of B, vB=dtdy=dtd[lsin(θ)]=lcos(θ)dtdθ
vBvA=−2tan(θ)⇒vB=−2vAcot(θ)
θ→increase
cotθ→decrease
Hence, ∣vB∣→decrease.
JEE Main 2017 — Physics Mechanics
The machine as shown has 2 rods of length 1m 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 roller that rolls along the floor in a slot. As the roller goes back and forth, a 2kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a :

Held on 9 Apr 2017 · Verified 6 Jul 2026.
Speed which is 43th of that of the roller when the weight is 0.4m above the ground
Constant speed
Decreasing speed
Increasing speed
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