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STD 11 - 9.1 , fluid mechanics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 9.1 , fluid mechanics4 Marks
MCQ
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

$\Delta L=\frac{F L}{A Y}$
Let density of liquid $=\rho$
Let density of object $=\sigma$
Mass of object $=M.$
$\Rightarrow$ Elongation $\propto$ force and force is due to weight
So elongation $\propto$ weight
$\Delta L_1 \propto$ weight $ \ldots (1)$
When not submerged in liquid 
$\Delta L_2 \propto$ apparant weight $\ldots .(2)$
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 $3: 1$

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