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At sea level, a `N_(2)`, molecule in air has an average translational `KE=6.2xxx10^(-21)J`. Its mass is `4.7xx10^(-26)kg.` If the molecule shoots up straight without any resistance, it will risw to a height ofA. `135km`B. `13.5km`C. `1.35 km`D. `1350 km` |
Answer» Correct Answer - B Here,`K.E.=6.2xx10^(21)J, m=4.7xx10^(-26)kg` As `KE` of molecule will get converted into its `PE` as the water level rise. `:. PE=6.2 xx10^(21)J=mgh` or `h=(6.2xx10^(-21))/(mg)=(6.2xx10^(-21))/(4.7xx10^(-26)xx98)` `~~13.5xx10^(3)m =13.5km` |
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