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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 figure, the spring stretches by an amount $x_{1}\left(\rho>\rho_{1}\right)$. When the experiment is repeated in a liquid of density $\rho_{2} < \rho_{1}$, the spring stretches by an amount $x_{2}$. Neglecting any buoyant force on the spring, the density of the object is
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When the spring-object system is immersed in a liquid of density $\rho_{1}$ as shown in the figure, the spring stretches by an amount $x_{1}\left(\rho>\rho_{1}\right)$. When the experiment is repeated in a liquid of density $\rho_{2} < \rho_{1}$, the spring stretches by an amount $x_{2}$. Neglecting any buoyant force on the spring, the density of the object is
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Verified Answer
The correct answer is:
$\rho=\frac{\rho_{1} x_{2}-\rho_{2} x_{1}}{x_{2}-x_{1}}$
F.B.D
$\begin{array}{l}
\mathrm{kx}_{1}+\rho_{1} \mathrm{Vg}=\rho \mathrm{Vg} ...(1) \\
\mathrm{kx}_{2}+\rho_{2} \mathrm{Vg}=\rho \mathrm{Vg} ...(2)
\end{array}$
from (1) and (2)
$\rho=\frac{\rho_{2} \mathrm{x}_{1}-\rho_{1} \mathrm{x}_{2}}{\mathrm{x}_{1}-\mathrm{x}_{2}}=\frac{\rho_{1} \mathrm{x}_{2}-\rho_{2} \mathrm{x}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}$
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