MCQ
An object with uniform density $\rho$ is attached to a spring that is known to stretch linearly with applied force as shown below.When the spring object system is immersed in a liquid of density $\rho_1$ as shown in the above figure, the spring stretches by an amount $x_1\left(\rho > \rho_1\right)$. When the experiment is repeated in a liquid of density $\left(\rho_2 < \rho_1\right)$, the spring stretches by an amount $x_2$. Neglecting any buoyant force on the spring, the density of the object is
  • A
    $\rho=\frac{\rho_1 x_1-\rho_2 x_2}{x_1-x_2}$
  • $\rho=\frac{\rho_1 x_2-\rho_2 x_1}{x_2-x_1}$
  • C
    $\rho=\frac{\rho_1 x_2+\rho_2 x_1}{x_1+x_2}$
  • D
    $\rho=\frac{\rho_1 x_1+\rho_2 x_2}{x_1+x_2}$

Answer

Correct option: B.
$\rho=\frac{\rho_1 x_2-\rho_2 x_1}{x_2-x_1}$
b
(b)

For equilibrium of block hung from string, Spring force + Buoyant force $=$ Weight of block

So, we have

$k x_1+\rho_1 V g=\rho V g\quad \dots(i)$

$\text { and }$ $k x_2+\rho_2 V g=\rho V g \quad \dots(ii)$

Eliminating $k$, we get

$\rho=\frac{\rho_1 x_2-\rho_2 x_1}{x_2-x_1}$

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