NEET-XII-Physics
32: Electric Current in Conductors
- #7A uniform wire of resistance 100 Ω is melted and recast as a wire whose length is double that of the original. What would be the resistance of the wire?Ans : Let
Resistivity of the wire = ρ
Original length of the wire = l
New length of the wire = l'
Original area of the wire = A
New area of the wire = A'
Original resistance of the wire = R = 100 Ω
New resistance of the wire = R'
Given:
l' = 2l
The volume of the wire remains constant. So,
`` Al=A\text{'}l\text{'}``
`` \Rightarrow A\text{'}=\frac{A}{2}``
We know:
`` R=\frac{\rho l}{A}``
`` \Rightarrow R\text{'}=\frac{\rho l\text{'}}{A\text{'}}``
`` \Rightarrow \frac{R\text{'}}{R}=\frac{l\text{'}A}{lA\text{'}}``
`` \because l\text{'}=2l,A\text{'}=\frac{A}{2}``
`` \Rightarrow \frac{R\text{'}}{R}=\frac{\left(2l\right)A}{l\left({\displaystyle \frac{A}{2}}\right)}=4``
`` \Rightarrow R\text{'}=4R=400\,\mathrm{\,\Omega \,}``
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