JEE Mains · Physics · STD 12 - 3. current electricity
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 \(4 \frac{ VA }{\rho l}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4} \frac{ VA }{\rho l}\)
Step-by-step Solution
Detailed explanation
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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