A) \[{{11}^{o}}C\]
B) \[{{12.36}^{o}}C\]
C) \[{{13}^{o}}C\]
D) \[{{10.37}^{o}}C\]
Correct Answer: D
Solution :
At the bottom of the lake, volume of the bubble \[{{V}_{1}}=\frac{4}{3}\pi {{r}^{3}},=\frac{4}{3}\pi \,{{(0.18)}^{3}}\] Pressure on the bubble, \[{{p}_{1}}=\] atmospheric pressure + pressure due to a column of 250 cm of water. \[=76\times 13.6\times 980+250\times 1\times 980\] \[=(76\times 13.6+250)980\] dyne/\[c{{m}^{2}}\] At the surface of the lake, volume of the bubble \[{{V}_{2}}=\frac{4}{3}\pi r_{2}^{3}=\frac{4}{3}\pi {{(0.2)}^{3}}c{{m}^{3}}\] Pressure on the bubble, \[{{p}_{2}}=\] atmospheric pressure \[=(76\times 13.6\times 980)\] dyne/\[c{{m}^{2}}\] \[{{T}_{2}}=273+40=313\,K\] Now, \[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[\Rightarrow \] \[\frac{(76\times 13.6\times 250)\,980\times \left( \frac{4}{3} \right)\,\pi {{(0.18)}^{3}}}{{{T}_{1}}}\] \[=\frac{(76\times 13.6)980\left( \frac{4}{3} \right)\pi {{(0.2)}^{3}}}{313}\] \[\Rightarrow \] \[{{T}_{1}}=283.37\,K\] \[\therefore \] \[{{T}_{1}}=283.37-273={{10.37}^{o}}C\]
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