A homogeneous solid cylinder of length L(LltH/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure `P_0`. Then density D of solid is given by
A homogeneous solid cylinder of length L(LltH/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure `P_0`. Then density D of solid is given by
Text Solution
Verified by Experts
a. For equilibrium of the cylinder,
Weight of the cylinder `=` weight of the liquid displaced
`implies L(A/5)Dg=(3L)/3(A/5)dg+L/4(A/5)2dg`
This gives `D=3/4d+d/2` or `D=5/4d`
b. Total pressure at the bottom of cylinder is
`P= ("weight of liquids" + "weight of cylinder")/A+P_(0)`
`=([Ad(H/2)g+A2d(H/2)g]+A/5(5/4d)Lg)/A+P_(0)`
`=((6H+L)/4)dg+P_(0)`
Here we are asked to find the total pressure at the bottom of the container and not only the hydrostatic pressue, hence, we need to consider the total weight of liquid and cylinder in the container.
Weight of the cylinder `=` weight of the liquid displaced
`implies L(A/5)Dg=(3L)/3(A/5)dg+L/4(A/5)2dg`
This gives `D=3/4d+d/2` or `D=5/4d`
b. Total pressure at the bottom of cylinder is
`P= ("weight of liquids" + "weight of cylinder")/A+P_(0)`
`=([Ad(H/2)g+A2d(H/2)g]+A/5(5/4d)Lg)/A+P_(0)`
`=((6H+L)/4)dg+P_(0)`
Here we are asked to find the total pressure at the bottom of the container and not only the hydrostatic pressue, hence, we need to consider the total weight of liquid and cylinder in the container.
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