Question

A certain block weighs $22N$ in air. It weighs $17N$ when immersed in water. When immersed in another liquid it weighs $18N$. Calculate the relative density of the other liquid.

• A. $0.8$
• B. $0.67$
• C. $0.2$
• D. $1$

Answer 1

The correct option is A $0.8$

Let the relative density of the block be $r_b$ and that of the liquid be $r_l$

Density of water be $\rho$

$\Rightarrow$ Density of metal is $r_b\rho$ and density of liquid is $r_l\rho$

Let the volume of the block be $V$

Weight of the block in air is $W_1=r_b\rho Vg=22N$

Weight of the block in water is $W_2=(r_b-1)\rho Vg=17N$

Weight of the block in the liquid is $W_3=(r_b-r_l)\rho Vg=18N$

Thus, dividing $W_2$ by $W_1$, we get

$\frac{W_2}{W_1}=\frac{r_b-1}{r_b}=\frac{17}{22}\Rightarrow r_b=\frac{22}{5}=4.4$

Dividing $W_3$ by $W_1$, we get

$\frac{W_3}{W_1}=\frac{r_b-r_l}{r_b}=\frac{18}{22}\Rightarrow \frac{4.4-r_l}{4.4}=\frac{9}{11}\Rightarrow r_l=0.8$