A cylindrical copper rod has length $L$ and resistance $R$. If it is melted and formed into another rod of length $2 L$, then the resistance will be ....... $R$
KVPY 2011, Medium
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(c)
As material of rod is not changed, resistivity of both rods is same.
Also, volume of material is same for both rods, so
$A_1 l_1=A_2 l_2$ $\text { or } A_1 L=A_2(2 L)$ $\Rightarrow A_2=\frac{A_1}{2}$
Now, using $R=\rho \frac{l}{A}$, we have
$R_2=\rho \frac{2 L}{\left(A_1 / 2\right)}=4\left(\frac{\rho L}{A_1}\right)$
or $R_2=4 R$
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