A solid cylinder of mass M and radius R released from rest rolls down a rough inclined plane inclined at an angle of  45∘ with the horizontal. The coefficient of friction between the cylinder and the inclined plane if the heat produced is maximum is 1−n. Find n.

Acceleration of cylinder  =g(sin45o−μcos45o)=g2(1−μ) Angular acceleration of cylinder  =τfrictionI=Rμmgcos45o(mR22) =2μgRGain in KE of cylinder travelling a distance L=12mv2+12Iω2=12m(2g2(1−μ)L)+12(mR22)22μgR2μL1-μR =mgL2[(1−μ)+2μ21−μ] Loss in PE of cylinder = mgh=mgL2Heat generated = Loss in PE - Gain in KE = mgL2-mgL2[(1−μ)+2μ21−μ]Since, this heat loss generated due to kinetic friction depends on μ,Differentiating this equation wrt μ we get,  μ=1−23