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JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Critical Thinking

  • question_answer
    Water of volume 2 litre in a container is heated with a coil of 1 kW at 27 oC. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27oC to 77oC [Given specific heat of water is 4.2 kJ/kg]                                                       [IIT-JEE (Screening) 2005]

    A)            8 min 20 s                              

    B)            6 min 2 s

    C)            7 min                                       

    D)            14 min

    Correct Answer: A

    Solution :

                       Heat gained by water = Heat supplied by container - heat lost Þ mSDq = 1000t ? 160t Þ \[t=\frac{2\times 4.2\times 1000\times 50}{840}\] = 8 min 20 sec


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