A block of wood is floating in water such that `1//2` of it is submerged in water when the same block is floated in alcohol, `1//3^(rd)` of it's volume is submerged Now a mixture of water and alcohol is made taking equal volume of both and block is floated in it. What is the % of it's volume that is now submerged?

Options

A. `50`
B. `40`
C. `25`
D. `60`

Explanation

If `rho` and `V_(0)` be density and volume of wooden block,
In water `rho_(w)(V_(0))/(2)=rhoV_(0)rArrrho=(rho_(w))/(2)`
In alcohol `rho_(a)(V_(0))/(3)=rhoV_(0)rArrrho_(a)=(3rho_(w))/(2)`
Now, effective density of "alcohol + water " mixture
`rho_(m)=(rho_(a)+rho_(w))/(2)=(5)/(4)rho_(w)`
Hence, Required `%=(rho)/(rho_(m))xx100=((rho_(w//2))/(5(rho_(w))/(4)))xx100=40%`