A) \[10.5m\]
B) \[8.7m\]
C) \[11.2m\]
D) \[9.5m\]
Correct Answer: D
Solution :
Pressure at bottom \[({{P}_{1}})={{P}_{atm}}+\rho gh+\frac{4T}{{{R}_{1}}}..........(1)\] Pressure at top \[({{P}_{2}})={{P}_{atm}}+\frac{4T}{{{R}_{2}}}..........(2)\] Given\[{{R}_{1}}=r\] \[{{R}_{2}}=\frac{5r}{4}\] So \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[({{P}_{1}})\frac{4}{3}\pi {{r}^{3}}=({{P}_{2}})\frac{4}{3}\frac{125{{r}^{3}}}{64}\] Dividing (1) and (2) \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{P}_{atm}}+\rho gh+\frac{4T}{r}}{{{P}_{atm}}+\frac{4T\times 4}{5r}}=\frac{125}{64}\] \[\frac{\rho g(10)+\rho gh}{\rho g(10)}=\frac{125}{64}\] \[640+64h=1250\] On solving we get \[h=9.5m\]
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