NEET AIPMT SOLVED PAPER 2003

  • question_answer                 An electric kettle has two heating coils. When one of the coils is connected to an AC source, the water in the kettle boils in 10 min. When the other coil is used the water boils in 40 min. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be:                                     

    A)                 25 min                  

    B)                 15 min  

    C)                 8 min

    D)                 4 min

    Correct Answer: C

    Solution :

                           Let \[{{R}_{1}}\] and \[{{R}_{2}}\] be the resistances of the coils, V the supply voltage, H the heat required to boil the water.                 For first coil,       \[H=\frac{{{V}^{2}}{{t}_{1}}}{{{R}_{1}}}....(i)\]                 For second coil, \[H=\frac{{{V}^{2}}{{t}_{2}}}{{{R}_{2}}}.....(ii)\]                 Equating Eqs. (i) and (ii), we get                 \[\frac{{{t}_{1}}}{{{R}_{1}}}=\frac{{{t}_{2}}}{{{R}_{2}}}\]                 \[i.e.,\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{40}{10}=4\]                 \[\Rightarrow {{R}_{2}}=4{{R}_{1}}\]                                                       ....(iii)                 When the two heating coils are in parallel, \[R=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{{{R}_{1}}+4{{R}_{1}}}{{{R}_{1}}+4{{R}_{1}}}=\frac{4{{R}_{1}}}{5}\]                 and        \[H=\frac{{{V}^{2}}t}{R}....(iv)\]                 Comparing Eqs. (i) and (iv), we get                 \[\frac{{{V}^{2}}{{t}_{1}}}{{{R}_{1}}}=\frac{{{V}^{2}}t}{R}\]                 \[\Rightarrow t=\frac{R}{{{R}_{1}}}\times {{t}_{1}}=\frac{4}{5}\times 10=8\,\min \]

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