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A uniform rod of length L and mass M is pivoted freely at one end (at bottom level) and placed in vertical position. What is the tangential linear acceleration of the free end when the rod is horizontal?

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a
g/2
b
3g/2
c
g/3
d
2g/3

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detailed solution

Correct option is B

τ=

mg=L2sinθ=ML23α

α=3gsinθL
when rod is horizontal θ=π2
α=3gL x 1=3gL

ar=L2α=L23gL=3g2

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