A wooden cylinder is floating on water in a beaker which is placed in a lift. When the lift is at rest, `1//3` of the volume of wood is exposed above the water . The lift now moves up with an ac celeration equal to `g//2`. What is the fraction of the volume exposed now?

Explanation

Let `upsilon` be the volume of wooden cylinder outside the water when lift is at rest and V be the total volume of he wooden cylinder. Fraction of the floating volume = `(upsilon)/(V) = 1//3`. Let `rho and rho_(w)` be the density of wood and water respectively. when the lift is at rest and wood is floating in water, then
`(V- upsilon) rho_(w) g = V rho g or (V-upsilon) rho_(w) = V(rho)` ..(i)
When lift is moving upwards with ac celeration `g//2`, then effective ac celeration due to gravity, `g' = g+ g//2 = 3g//2`. If `upsilon'` is the volume of the wood outside the water white floating, then
`(V-upsilon') rho_(w) 3g//2 = V rho 3g//2 or (V- upsilon') rho_(w) = Vrho` ..(ii)
From (i) and (ii) , we note that ` upsilon =upsilon'`
`:. (upsilon')/(V) = (upsilon)/(V) = 1/3`
So, the new fraction of the volume exposed = old fraction of the volume exposed = `1//3`.