JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर JEE PYQ-Current Electricity Charging and Discharging of Capacitors

  • question_answer
    Time taken by a 836 W heater to heat one litre of water from\[10{}^\circ C\]to\[40{}^\circ C\]is     [AIEEE 2004]

    A) 50s

    B) 100s

    C) 150s

    D) 200s

    Correct Answer: C

    Solution :

    [c] Let time taken in boiling the water by the heater is t second. Then,
    \[Q=ms\Delta T\Rightarrow \frac{Pt}{J}ms\Delta T\]      \[(\because Q=W=Pt)\]
    \[\therefore \] \[\frac{836}{4.2}t=1\times 1000(40-10)\]      \[(1C=4.2J)\]
    \[\frac{836}{4.2}t=1000\times 30\]
    \[t=\frac{1000\times 30\times 4.2}{836}=150\,s\]


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