Question

Two cubical blocks identical in dimensions float in water in such a way that the first block floats with half part immersed in water and the second block floats with $3/4\text{th}$ of its volume immersed inside the water. The ratio of density of first block to that of second block is

• A. $2:3$
• B. $3:4$
• C. $1:3$
• D. $1:4$

Answer 1

The correct option is A $2:3$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> For floating of block,
$F_b=mg$
$V_f{\rho }_{f}g=V_b{\rho }_{1}g...\left(i\right)$
$V_f\to$ volume of fluid displaced
$V_b\to$volume of both blocks
Both blocks will have same volume due to same dimensions.
${\rho }_1\to$ density of first block
According to question, for first block:
$V_f=\frac{V_b}{2}$
From Eq.$\left(i\right)$,
$\frac{1}{2}{V}_b{\rho }_{f}g=V_b{\rho }_{1}g$
$\therefore {\rho }_f=2{\rho }_1...\left(ii\right)$
And for second block:
$V_f=\frac{3}{4}{V}_b$
From Eq.$\left(i\right)$,
$\frac{3}{4}{V}_b{\rho }_{f}g=V_b{\rho }_{2}g$
$\therefore {\rho }_f=\frac{4}{3}{\rho }_2...\left(iii\right)$

From Eq.$\left(ii\right)\&\left(iii\right)$,
$2{\rho }_1=\frac{4}{3}{\rho }_2$
$\therefore \frac{{\rho }_1}{{\rho }_2}=\frac{2}{3}$