A hollow cylindrical container floats in water with half its length immersed. A liquid of specific gravity `sigma` is slowly poured into the container until the levels of the liquid inside the container and of water outside are the same. It is observed that the container is submerged to two-thirds its length. Then the value of `sigma` is

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `1/3`

Answer

C

Explanation

Since the container alone floats of half its length in water, its relative density `=0.5`. Since the levels of the liquid inside and water outside are the same, the volumes of the liquid and of the displaced water are the same. If this volume is `V` and the volume of the material of container is , then the total mass of floating body is `M=[Vsigma+v(0.5)rho]`, where `rho` is the density of water. By law of floatation, `M=(Vsigma+v(0.5)rho]=Vrho` giving `V(1-rho)=vxx0.5.`
Let the cross section of the container be `A`. Then the volume `V'` of water displaced `V` by the liqid is
`LA[2/3-1/2]=LA[1/6]`
but `LA=(3V)/2`
Hence `V'=(3V)/2xx1/6=V/4`
Hence `Vsigmarho=(V/4)rhoimplies sigma=1/4`