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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 or water will be ...... min. (Assume the resistance of the beater remaining constant) |
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Answer» 40 H = `I^(2) RT= (V^(2))/(R^(2)) xx Rt= (V^(2))/(R) t ` In t =`(HR)/(V^(2))`, HR REMAINS same. `therefore t PROP (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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