A block of mass 2kg hangs from the rim of a wheel of radius 0.5m. On releasing from rest the block falls through 5m height in 2s. The moment of inertia of the wheel will be

A block of mass 2kg hangs from the rim of a wheel of radius 0.5m. On r

1.	A block of mass 2kg hangs from the rim of a wheel of radius 0.5m. On releasing from rest the block falls through 5m height in 2s. The moment of inertia of the wheel will be
Answer» `1 kg-m^(2)` `3.2 kg-m^(2)` `2.5kg-m^(2)` `1.5kg-m^(3)` Solution :On releasing from REST the block falls through 5m HEIGHT in 2 sec `5= 0 +(1)/(2) a(2)^(2)` [As `S= U t+ (1)/(2) a t^(2)`] `THEREFORE a= 2.5m//s^(2)` Substituting the value of a in the formula `a= (g)/(1 + (I)/(mR^(2)))` and by sloving we get `rArr 2.5= (10)/(1+ (I)/(2 xx (0.5)^(2))) rArr I= 1.5kg -m^(2)`