A) \[\operatorname{t}=\sqrt{\frac{2h}{g}}\times \frac{{{\rho }_{1}}}{{{\rho }_{2}}}\]
B) \[\operatorname{t}=\sqrt{\frac{2h}{g}}\times \frac{{{\rho }_{2}}}{{{\rho }_{1}}}\]
C) \[\operatorname{t}=\sqrt{\frac{2h}{g}}\]
D) \[\operatorname{t}=\sqrt{\frac{2h}{g}}\times \frac{{{\rho }_{1}}}{{{\rho }_{2}}-{{\rho }_{1}}}\]
Correct Answer: D
Solution :
When the body falls from height h, the velocity by which it hit the surface of liquid \[=\sqrt{2gh}\] When the body immersed in liquid of density \[{{\rho }_{2}}\] \[\operatorname{Up} thrust = V{{\rho }_{1}}g\] \[\operatorname{Weight} of body = V{{\rho }_{1}}g\] \[\operatorname{Net} retardation=\frac{V{{\rho }_{2}}g-V{{\rho }_{1}}g}{V{{\rho }_{1}}}=\frac{\left( {{\rho }_{2}}-{{\rho }_{1}} \right)}{{{\rho }_{1}}}\operatorname{g}\] Time of immersion for body to acquire final velocity zero \[\operatorname{v}={{v}_{o}}+at\Rightarrow 0=\sqrt{2gh}-\left[ \frac{{{\rho }_{2}}-{{\rho }_{1}}}{{{\rho }_{1}}}\operatorname{g} \right]t\] \[\operatorname{t}=\frac{2gh}{h}\frac{{{\rho }_{1}}}{{{\rho }_{2}}-{{\rho }_{1}}}\Rightarrow \sqrt{\frac{2h}{g}}\frac{{{\rho }_{1}}}{{{\rho }_{2}}-{{\rho }_{1}}}\]
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