A wooden block is floating on water kept in a beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to `g//2`. The block will then

Options

A. sink
B. float with 10% above the water surface
C. float with 40% above the water surface
D. float with 70% above the water surface

Answer

B

Explanation

According to Archimedes' principal, weight of the woodn block=weight of the liquid displaced
`Rightarrow mg=((60)/(100))Vxxp_(1)xxg=(3)/(5)Vxxp_(l)xxg...(i)`
[Here, V=total volume of block `p_(1)`, =density of liquid]
When beaker is kept in a lift,
Net weight of the block =mg+mg//2 =`(3mg)/(2) [therefore R=mg=(mg)/(2)]`

Again for float condition
Net weight of the block=Buoyant force
`Rightarrow (3mg)/(2)=V_(i)xxp_(i)xxg....(ii)`
Hence, `V_(i)` =Volume of block inside liquid from Eqs. (i) and (ii), we get
`((3)/(2))xx(3)/(5)(Vxxp_(i)xxg)=V_(i)xxp_(i)xxg`
`Rightarrow V_(i)=(9)/(10)V`
So, volume of the block above liquid. `(1)/(10)V=10%"of total volume"`