A cube with sides of length 2a and mass M is moving with an initial speed $v_{0}$ along a frictionless table. When the cube reaches the end of the table it is caught abruptly by a short lip and begins to rotate. The minimum speed $v_{0}$ such that the cube falls off the table is :-

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TestHub · Accepted Answer

Angular momentum about the lip is conserved

$M v_{0} a = \frac{8}{3} M a^{2} ω$

$\Rightarrow ω$

$= \frac{3 v _{0}}{8 a}$

It must just reach the position with the centre above the lip

$\frac{1}{2} I ω^{2} = M g (2 a - a)$

$v_{0} = \frac{16 a g ( 2 - 1 )}{3}$

Answer: A

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JEE Advanced 2027 (Class 12+)