A) \[2\rho h\]
B) \[4\rho h\]
C) \[8\rho h\]
D) \[7\rho h\]
Correct Answer: D
Solution :
[d] From Boyle's law (T= constant) \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[\therefore \left( H{{d}_{waer}}+h{{d}_{mercury}} \right)g\left( \frac{4}{3}\pi {{r}^{3}} \right)\] \[=h{{d}_{mercury}}g\left( \frac{4}{3}\pi {{\left( 2r \right)}^{3}} \right)\] \[\Rightarrow H{{d}_{water}}=8h{{d}_{mercury}}-h{{d}_{mercury}}\] \[\Rightarrow H=7h\frac{{{d}_{mercury}}}{{{d}_{water}}}\] \[\therefore H=7hp\]
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