Given: - Original resistance \( R_1 = 5 \, \Omega \) - The new length \( L_2 = 2L_1 \) (where \( L_1 \) is the original length) - The new cross-sectional area \( A_2 = \frac{1}{2} A_1 \) (where \( A_1 \) is the original area) Step 1: Recall the formula for resistance The resistance \( R \) of a wire is given by: \(R = \rho \frac{L}{A}\) where: - \( \rho \) is the resistivity, - \( L \) is the length, - \( A \) is the cross-sectional area. Step 2: Calculate the new resistance The new resistance \( R_2 \) can be calculated using the new length and new area: \(R_2 = \rho \frac{L_2}{A_2}\) Substituting for \( L_2 \) and \( A_2 \): \(R_2 = \rho \frac{2L_1}{\frac{1}{2}A_1} = \rho \frac{2L_1 \cdot 2}{A_1} = \rho \frac{4L_1}{A_1}\) Step 3: Relate \( R_1 \) and \( R_2 \) From the original resistance \( R_1 \): \(R_1 = \rho \frac{L_1}{A_1}\) Now, we can express \( R_2 \) in terms of \( R_1 \): \(R_2 = 4 \left(\rho \frac{L_1}{A_1}\right) = 4R_1\) Step 4: Substitute the value of \( R_1 \) Now substitute \( R_1 = 5 \, \Omega \): \(R_2 = 4 \times 5 \, \Omega = 20 \, \Omega\) Therefore, the new resistance is: \(\boxed{20 \, \Omega}\)