Question 77 of 380

A conducting wire of length $l$, area of crosssection $A$ 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$\dfrac{1}{4}\dfrac{VA}{\rho l}$B$\dfrac{3}{4}\dfrac{VA}{\rho l}$C$\dfrac{1}{4}\dfrac{\rho l}{VA}$D$4\dfrac{VA}{\rho l}$

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Answer: (A) $\dfrac{1}{4}\dfrac{VA}{\rho l}$

As per the question Resistance $=\dfrac{\rho \left(2l\right)}{\left(A/2\right)}=\dfrac{4\rho l}{A}$$\Rightarrow$ Current $=\dfrac{V}{R}=\dfrac{VA}{4\rho l}$