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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

  • question_answer
        A wire when connected to 220V mains supply has power dissipation\[{{P}_{1}}\]. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is\[{{P}_{2}}\]. Then\[{{P}_{2}}:{{P}_{1}}\]is

    A)  1                                            

    B)  4

    C)  2                                            

    D)  3

    Correct Answer: B

    Solution :

                    In 1st case: Using the formula \[P=\frac{{{V}^{2}}}{R}\] where\[R\]is resistance of wire, V is voltage across wire and P is power dissipation in wire and                 \[R=\frac{\rho l}{A}\] From eqs. (1) and (2)                 \[=\frac{{{V}^{2}}}{\frac{\rho l}{A}}=\frac{{{V}^{2}}}{\rho l}.A\]                 \[{{P}_{1}}=\frac{{{V}^{2}}}{\rho l}.A\]             ...(1) In 2nd case: Let \[{{R}_{2}}\]is net resistance                 \[{{R}_{2}}=\frac{R.R}{R+R}=\frac{R}{2}\]                 \[{{R}_{2}}=\frac{\rho .\left( \frac{l}{2} \right)}{A.2}\]                 \[=\frac{\rho l}{4A}\] \[\therefore \]  \[{{P}_{2}}=\frac{{{V}^{2}}}{\rho l}.4A\]           ...(2) Hence, from eqs. (3) and (4)                 \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{1}{4}\] \[\Rightarrow \]               \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{4}{1}\]


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