If an average jogs, he produces `14.5xx10^(3)` cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires `580xx10^(3)` cal for evaporation) isA. 0.25 KgB. 2.25 KgC. 0.05 kgD. 0.20 Kg

If an average jogs, he produces `14.5xx10^(3)` cal/min. This is remove

1.	If an average jogs, he produces `14.5xx10^(3)` cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires `580xx10^(3)` cal for evaporation) isA. 0.25 KgB. 2.25 KgC. 0.05 kgD. 0.20 Kg
Answer» Correct Answer - A Amount of sweat evaporated per minute `=("calories produed per mimute")/("no of calories required for evaporation per kg")` `=(14.5xx10^(4))/(580xx10^(3))` =0.25 Kg

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If an average jogs, he produces `14.5xx10^(3)` cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires `580xx10^(3)` cal for evaporation) isA. 0.25 KgB. 2.25 KgC. 0.05 kgD. 0.20 Kg

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