Question

An object suspended by a wire stretches it by $10\, mm$. When object is immersed in a liquid the elongation in wire reduces by $\frac{10}{3} mm$. The ratio of relative densities of the object and liquid is

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

Answer 1

Answer: (A)

$\Delta L=\frac{F L}{A Y}$
$\Rightarrow $ Elongation $\propto$ force and force is due to weight
So elongation $\propto$ weight
$\Delta L_{1} \propto \text { weight }$ ...(1) {When not submerged in liquid}
$\Delta L_{2} \propto \text { apparant weight}$ ...(2) {When submerged in liquid}
Dividing (1) by (2)
$\frac{10}{10-\frac{10}{3}}=\frac{M g}{M g-\frac{M g \rho}{\sigma}}$
$\Rightarrow \frac{1}{1-\frac{1}{3}}=\frac{1}{1-\frac{\rho}{\sigma}}$
Solving this we get
$\frac{\rho}{\sigma}=\frac{1}{3}$
So relative densities of object $(\sigma)$ and liquid $(\rho)$ is $3: 1$
Let density of liquid $=\rho$
Let density of object $=\sigma$
Mass of object $=M$