A conducting wire of length $ l$ area of cross-section $A$ and electric resistivity $\rho$ is connected between the terminals of a battery. $A$ potential difference $V$ is developed between its ends, causing an electric current.If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be
A$\frac{1}{4} \frac{ VA }{\rho l}$
B$\frac{3}{4} \frac{ VA }{\rho l}$
C$\frac{1}{4} \frac{\rho l}{ VA }$
D$4 \frac{ VA }{\rho l}$
JEE MAIN 2021, Medium
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A$\frac{1}{4} \frac{ VA }{\rho l}$
a As per the question
Resistance $=\frac{\rho(2 l)}{( A / 2)}=\frac{4 \rho l}{ A }$
$\Rightarrow$ Current $=\frac{ V }{ R }=\frac{ VA }{4 \rho l}$
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