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(A) Given initial resistance $R = 4 \, \Omega$.When a wire is doubled on itself,its length $L$ becomes half $(L_n = L/2)$ and its area of cross-sectio

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$1 \, \Omega$

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(A) Given initial resistance $R = 4 \, \Omega$.When a wire is doubled on itself,its length $L$ becomes half $(L_n = L/2)$ and its area of cross-section $A$ becomes double $(A_n = 2A)$.The resistance of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity of the material.The new resistance $R_n$ is given by:$R_n = \rho \frac{L_n}{A_n} = \rho \frac{L/2}{2A} = \frac{1}{4} \left( \rho \frac{L}{A} \right) = \frac{R}{4}$.Substituting the value of $R = 4 \, \Omega$:$R_n = \frac{4 \, \Omega}{4} = 1 \, \Omega$.Thus,the new resistance of the wire is $1 \, \Omega$.

$A$ $4 \, \Omega$ resistance wire is doubled on itself. Calculate the new resistance of the wire.

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