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Given quantity of water is boiled by an electric heater in 5 min. If supply voltage of heater reduces to half then time taken to boil the same quantity of water will be ........... min. (Assume the resistance of the heater remaining constant) |
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Answer» Solution :20 H = `I^(2) RT` = = `(V^(2))/(R^(2)) XX Rt = (V^(2))/(R) t ` In `t = (HR)/(V^(2)) `, HR REMAINS same. `therefore tprop (1)/(V^(2))` `therefore (t_(2))/(t_(1)) = ((V_(1))/(V_(2)))^(2)` = `((V)/((V)/(2)))^(2)` = 4 `therefore t_(2) = t_(1)xx 4 = 5 xx 4 = 20 ` min |
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