| An object floats in water such that \[\frac{1}{4}th\] of its volume is above the surface of water. |
| When the same object is made to float in sample of impure alcohol, the volume that remains outside the liquid is \[\frac{1}{10}\] of total volume. The relative density of the sample of alcohol will be |
A) 84
B) 8.4
C) 0.84
D) 0.084
Correct Answer: C
Solution :
According to Archimedes's principle, buoyant force, \[{{\text{F}}_{B}}=\]Volume of the object immersed in liquid \[\times \] density of liquid \[\times \]acceleration due to gravity Let V be the volume of object. For water: \[{{F}_{B}}=\frac{3V}{4}\times {{\rho }_{w}}\times g\] Or \[{{F}_{B}}=\frac{3V}{4}\times {{\rho }_{w}}\times 10=\frac{15}{2}V{{\rho }_{w}}\] - (i) For alcohol: \[{{F}_{B}}=\frac{9V}{10}\times {{\rho }_{a}}\times g=\frac{9V}{10}\times {{\rho }_{a}}\times 10=9V{{\rho }_{a}}\] - (ii) From eqn. (i) and (ii), \[\frac{15}{2}V{{\rho }_{w}}=9V{{\rho }_{a}}\therefore \frac{{{\rho }_{a}}}{{{\rho }_{w}}}=\frac{15}{2\times 9}=\frac{15}{18}=0.84\]
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