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If the new length ( L' = 3L ) (three times original length), then to keep volume constant, the new area ( A' ) must satisfy: ncert solutions
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Original resistance: [ R = \rho \frac{L}{A} = 20 \ \Omega ] Original resistance: [ R = \rho \frac{L}{A} =
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When the wire is stretched, its volume remains constant. Volume ( V = A \times L = \text{constant} )
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