Question

If a body floats with ${\left(\frac{p}{q}\right)}^{th}$ of its volume in above the surface of water, then the relative density of the body is

• A. $\frac{q+p}{q}$
• B. $1-\frac{p}{q}$
• C. $\frac{q-p}{q}$
• D. $\frac{p}{q}$

Answer 1

Answer: (B), (C)

The correct options are
B $1-\frac{p}{q}$
C $\frac{q-p}{q}$

Let the volume of body be $V$ and the density be $\rho$

Volume immersed in liquid is $V_1=V-\frac{Vp}{q}=V\frac{q-p}{q}$

Now buoyancy force must be equal to weight of body

$\rho Vg={\rho }_{water}{V}_{1}g$

$\rho V={\rho }_{water}V\frac{q-p}{q}$

Thus Relative density $=\frac{\rho }{{\rho }_{water}}=\frac{q-p}{q}$