Question

A cube floats in water with ${\left(\frac{1}{4}\right)}^{\text{th}}$ part outside the surface of water. If it floats in another liquid of relative density $\frac{4}{3}$ with $x^{th}$ part outside the liquid, then the value of $x$ is

• A. $\frac{7}{16}$
• B. $\frac{1}{2}$
• C. $\frac{8}{15}$
• D. $\frac{9}{16}$

Answer 1

The correct option is A $\frac{7}{16}$

Volume of a cuboid shape block is:
$V=(A\times H)$, where $A=$base area, $H=$height
In water
$F_B=\frac{3H}{4}\times A\times {\rho }_w\times g$
For floating of object,
$F_B=mg$
$\frac{3H}{4}\times A\times {\rho }_w\times g=mg...\left(i\right)$
When $x^{th}$ part is outside liquid of relative density $D$,
$F_B^{\prime }=(H-Hx)\times A\times \left(D{\rho }_w\right)\times g...\left(ii\right)$
For floating of object,
$F_B^{\prime }=mg$
$(1-x)H\times A\times \left(D{\rho }_w\right)\times g=mg...\left(iii\right)$

Dividing Eq $\left(i\right)\&\left(iii\right)$:
$\frac{3}{4(1-x)\times D}=1$
where relative density, $D=\frac{4}{3}$
$\frac{16}{3}(1-x)=3$
$\Rightarrow 1-x=\frac{9}{16}$
$\therefore x=1-\frac{9}{16}=\frac{7}{16}$