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NEET AIPMT SOLVED PAPER SCREENING 2004

  • question_answer
    A 6 V battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of \[100\,\,\Omega \]. The difference of potential between two points on the wire separated by a distance or 50 cm will be:

    A)                                                                                                                                                                 2 V                 

    B) 3 V                 

    C) 1 V       

    D) 1.5 V

    Correct Answer: C

    Solution :

    Total current drawn from the battery \[i=\frac{E}{R+r}=\frac{6}{100+0}=0.06\,\text{A}\] Resistance of 50 cm wire is \[R'=\frac{\rho l'}{A}=\left( \frac{\rho }{A} \right)l'\] \[=\left( \frac{R}{l} \right)l'\]                              \[\left( \because R=\frac{\rho l}{A} \right)\] \[=\frac{100}{300}\times 50\] So,  \[R'=\frac{50}{3}\Omega \]         Hence, the potential difference between two points on the wire separated by a distance \[l'\]is \[V=iR'=0.06\times \frac{50}{3}=1\,\text{V}\]


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