If a wire of resistance 'R' is stretched twice its original length keeping volume constant then what will be its new resistance? (let new resistance as R') .A 2R B. R/2 C. R/3 D. 4R

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If a wire of resistance &#039;R&#039; is stretched twice its original length keeping volume constant then what will be its new resistance? (let new resistance as R&#039;) .A 2R B. R/2 C. R/3 D. 4R

eyrurin3wj · Accepted Answer

The volume of wire   =  length  ×  cross sectional area. Volume being a constant Accordingly, let length of wire  = L let area of wire = A So resistance   R  =      ρ        L    A  , Now according to given condition length of stretched wire  = 2L Hence area of stretched wire = A/2 Thus, the stretched wire&#039;s resistance   R      ′    =      ρ              2      L              A              /            2       or,  R      ′    =      ρ              2      ×      2      L        A    R      ′    =      ρ              4      L        A    R      ′    =  4  (                    ρ            L        A    ) or,       R    ′    =  4  (  R  ) or, the new resistance (R&#039;) of the stretched wire is 4 times the value of resistance of the unstretched wire (R).

If a wire of resistance 'R' is stretched twice its original length keeping volume constant then what will be its new resistance? (let new resistance as R') .A 2R B. R/2 C. R/3 D. 4R

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