91Ó°ÊÓ

In liquid A, B, and C the block floats above the water surface at the following heights

• $h/2$
  
• $2h/3$
  
• $h/4$
  

When the block is stationary, the net force acting on it is zero. Using this fact, we derive the relation between the total height of the block and the height of the block above the water surface. Using this data, we find the relative densities.

Formula:

$F_b={\mathrm{ÒÏ}}_f{V}_{f}g$

As the block is stationary

$F_b$- $F_g$=0

${\mathrm{ÒÏ}}_f(I×b×h)g$-$\mathrm{ÒÏ}(L×B×H)g$= 0

${\mathrm{ÒÏ}}_f(I×b×h)g$= $\mathrm{ÒÏ}(L×B×H)g$

Here,

$F_b$is buoyant force,$F_g$is weight of the object.

I is length, b is the breadth, h is the height of displaced fluid. L is the length, B is the breadth, H is the height of the object.

Rearranging the equation,

$h=\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_f}H$

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_f}=\frac{m}{H}$

The block is fully submerged in water, hence$h=H$ (1)

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_w}=1$

Therefore, the density of the object relative to the density of water is 1 .

Applying the above equation to liquid A

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_a}=\frac{H/2}{H}$

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_a}=\frac{1}{2}$ (2)

Dividing equation (1) and (2)

$\frac{{\mathrm{ÒÏ}}_a}{{\mathrm{ÒÏ}}_w}=2$

Therefore, the density of the fluid A relative to the water is 2 .

Applying the above equation to liquid B

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_b}=\frac{H/3}{H}$

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_b}=\frac{1}{3}$ (3)

Dividing equation (1) and (3)

$\frac{{\mathrm{ÒÏ}}_b}{{\mathrm{ÒÏ}}_w}=3$

Therefore, the density of fluid B relative to the water is 3 .

Applying the above equation to liquid C

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_c}$=$\frac{3H/4}{H}$

$\frac{\mathrm{ÒÏ}}{{\mathrm{ÒÏ}}_c}$=$\frac{3}{4}$ (4)

Dividing equation (1) and (4)

Therefore, the density of the fluid C relative to the water is 4 /3.