• # question_answer An object with uniform density p 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) $\rho =\frac{{{\rho }_{1}}{{x}_{1}}-{{\rho }_{2}}{{x}_{2}}}{{{x}_{1}}-{{x}_{2}}}$ 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) $\rho =\frac{{{\rho }_{1}}{{x}_{1}}+{{\rho }_{2}}{{x}_{2}}}{{{x}_{1}}+{{x}_{2}}}$

Solution :

 for equilibrium of block hung from string, spring force+ Buoyant force =weight of block So, we have $k{{x}_{1}}+{{\rho }_{1}}Vg=\rho Vg...(i)$ $\operatorname{and}k{{x}_{2}}+{{\rho }_{2}}Vg=\rho Vg...(ii)$ Eliminating$k$, we get $\rho =\frac{{{\rho }_{1}}{{x}_{2}}-{{\rho }_{2}}{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

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