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) \[\rho =\frac{{{\rho }_{1}}{{x}_{1}}+{{\rho }_{2}}{{x}_{2}}}{{{x}_{1}}+{{x}_{2}}}\]
Correct Answer: B
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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