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Given, vA = 10m/sec, α = 60°. Let length of the rod = L. From figure, x^2 + y^2 = L. Differentiation with respect to time 't', ⇒ 2x dx/dt + 2y dy/dt = 0 ⇒ x dx/dt + y dy/dt = 0, where dx/dt = vA and dy/dt = vB. So, x vA + y vB = 0. vB = −x/y vA, where x/y = cot α. So, vB = −vA cot α = −10 cot 60° = −10/√3 = −5.773 = 5.8m/sec.

1. (a) Given, vA = 10m/sec, α = 60°
2. (b) Let length of the rod = L
3. (c) From figure, x^2 + y^2 = L
4. (d) Differentiation with respect to time 't', ⇒ 2x dx/dt + 2y dy/dt = 0

Correct answer: (d) Differentiation with respect to time 't', ⇒ 2x dx/dt + 2y dy/dt = 0

Solution

Option (d) is correct because it represents the key step in deriving the relationship between the velocities of the two ends of the rod by differentiating the constraint equation with respect to time.

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