Question

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 $x_1(\rho >{\rho }_1)$. 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

• A. $\rho =\frac{{\rho }_1{x}_1-{\rho }_2{x}_2}{x_1-x_2}$
• B. $\rho =\frac{{\rho }_1{x}_2-{\rho }_2{x}_1}{x_2-x_1}$
• 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 1

The correct option is B $\rho =\frac{{\rho }_1{x}_2-{\rho }_2{x}_1}{x_2-x_1}$

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 $x_1(\rho >{\rho }_1)$. If $v$ be the volume.

$k{x}_1+{\rho }_{1}vg=\rho vg$

In case 2,

$k{x}_2+{\rho }_{2}vg=\rho vg$

Subtracting the two equations

$k=\frac{({\rho }_2-{\rho }_1)vg}{(x_1-x_2)}$

Putting the value of k in any of the equation we get.

$\rho =\frac{({\rho }_2{x}_1-{\rho }_1{x}_2)}{x_1-x_2}=\frac{({\rho }_1{x}_2-{\rho }_2{x}_1)}{x_2-x_1}$