JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Electric Conduction Ohms and Resistance

  • question_answer
    If a wire of resistance R is melted and recasted to half of its length, then the new resistance of the wire will be [KCET (Med.) 2001]

    A)            R/4

    B)            R/2

    C)            R    

    D)            2R

    Correct Answer: A

    Solution :

                 \[R\propto {{l}^{2}}\Rightarrow \frac{{{R}_{1}}}{{{R}_{2}}}={{\left( \frac{{{l}_{1}}}{{{l}_{2}}} \right)}^{2}}\]\[\Rightarrow \,\frac{R}{{{R}_{2}}}={{\left( \frac{l}{l/2} \right)}^{2}}=4\Rightarrow {{R}_{2}}=\frac{R}{4}\].               


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