Question

If a sample of metal weighs $210g$ in air, $180g$ in water and $120g$ in a liquid

• A. RD of metal is $3$
• B. RD of metal is $7$
• C. RD of liquid is $7$
• D. RD of liquid is $\left(1/3\right)$

Answer 1

Answer: (B)

RD of metal is $7$

Let the relative density of metal be $r_m$ and that of the liquid be $r_l$

Density of water be $\rho$

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

Let the volume of the sample be $V$

Weight of the sample in air is $W_1=r_m\rho Vg=210\text{g}$

Weight of the sample in water is $W_2=(r_m-1)\rho Vg=180\text{g}$

Weight of the sample in the liquid is $W_3=(r_m-r_l)\rho Vg=120\text{g}$

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

$\frac{W_2}{W_1}=\frac{r_m-1}{r_m}=\frac{180}{210}\Rightarrow r_m=7$

Dividing $W_3$ by $W_1$, we get

$\frac{W_3}{W_1}=\frac{r_m-r_l}{r_m}=\frac{120}{210}\Rightarrow \frac{7-r_l}{7}=\frac{4}{7}\Rightarrow r_l=3$