A) \[\text{11}{}^\circ \text{C}\]
B) \[\text{12}.\text{36}{}^\circ \text{C}\]
C) \[\text{13}{}^\circ \text{C13}{}^\circ \text{C}\]
D) \[\text{1}0.\text{37}{}^\circ \text{C}\]
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. \[=76\times 13.6\times 980+250\times 1\times 980\] \[\text{= (76 }\!\!\times\!\!\text{ 13}\text{.6+250) 980 dyne/c}{{\text{m}}^{\text{2}}}\] 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, \[{{\text{p}}_{\text{2}}}=\]atmospheric pressure \[=\left( \text{76}\times \text{13}.\text{6}\times \text{98}0 \right)\text{ dyne}/\text{c}{{\text{m}}^{\text{2}}}\] \[{{T}_{2}}=273+40=313K\] Now, \[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[\Rightarrow \] \[\frac{(76\times 13.6\times 250)980\times \left( \frac{4}{3} \right)\pi {{(0.18)}^{3}}}{{{T}_{1}}}\] \[=\frac{(76\times 13.6)\times 980\left( \frac{4}{3} \right)\pi {{(0.2)}^{3}}}{313}\] \[\Rightarrow \] \[{{T}_{1}}=283.37K\] \[\therefore \] \[{{T}_{1}}=283.37-273={{10.37}^{\circ }}C\]
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