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EAMCET Medical EAMCET Medical Solved Paper-2008

  • question_answer
    A heater coil working on mains produces 100 cal of heat in a certain time. Now the heater coil is cut into three equal parts and one part is only used for heating. The quantity of heat produced (in calories) in the same time is

    A)  300                                       

    B)  200

    C)  100/3                                   

    D)  200/3

    Correct Answer: A

    Solution :

                     We know that, \[H=\frac{{{V}^{2}}}{R}.\] For the same potential difference, the heat produced per second \[H\propto \frac{1}{R}\] One part of the heater coil has a resistance R/3. Hence, \[\frac{{{H}_{2}}}{{{H}_{1}}}=\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{R}{R/3}\]                 \[\Rightarrow \]               \[\frac{{{H}_{2}}}{100}=\frac{3R}{R}\] \[\Rightarrow \]\[{{H}_{2}}=3\times 100=300\,cal.\]


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