mechanics JEE type

min friction means friction = 0 (when relative vel. becomes zero)

therefore only normal reaction and weight is acting on the mass

N=mgcosQ

mv²RsinQ = NsinQ =mgsin²QcosQ

v² = gR*sin²Q*cosQ .... (i)

now alpha =0.1
w(ang. vel.)v= 0.1*t
v_surface = w*RsinQ =0.1*t*R*sinQ .....(ii)

from eqns. i) and ii) find t (=100 sec)

hence k =2

answer given is 4

https://www.youtube.com/watch?v=Z47IRtuF0eI

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