W}}{{\rho }_{W}})V={{H}_{Hg}}{{\rho }_{Hg}}\times 3V\] Þ \[{{H}_{Hg}}{{\rho }_{Hg}}+{{H}_{W}}\FRAC{{{\rho }_{Hg}}}{10}=3{{H}_{Hg}}{{\rho }_{Hg}}\] Þ\[{{H}_{W}}=2{{H}_{Hg}}\times 10\]\[=\frac{2\times 75\times 10}{100}=15M\]

"> W}}{{\rho }_{W}})V={{H}_{Hg}}{{\rho }_{Hg}}\times 3V\] Þ \[{{H}_{Hg}}{{\rho }_{Hg}}+{{H}_{W}}\FRAC{{{\rho }_{Hg}}}{10}=3{{H}_{Hg}}{{\rho }_{Hg}}\] Þ\[{{H}_{W}}=2{{H}_{Hg}}\times 10\]\[=\frac{2\times 75\times 10}{100}=15M\]

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The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake is                                 [AMU 1995]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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\[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]Þ \[({{H}_{Hg}}{{\rho }_{Hg}}+{{H}_{W}}{{\rho }_{W}})V={{H}_{Hg}}{{\rho }_{Hg}}\times 3V\] Þ \[{{H}_{Hg}}{{\rho }_{Hg}}+{{H}_{W}}\FRAC{{{\rho }_{Hg}}}{10}=3{{H}_{Hg}}{{\rho }_{Hg}}\] Þ\[{{H}_{W}}=2{{H}_{Hg}}\times 10\]\[=\frac{2\times 75\times 10}{100}=15M\]

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Posted on 24 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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