Two wires of equal diameters, of resistivities ρ1 and ρ2 and lengths l1 and l2, respectively, are joined in series. The equivalent resistivity of the combination is

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ρ1l1 + ρ2l2l1l2

ρ1l2 + ρ2l1l1−l2

ρ1l2 + ρ2l1l1+l2

ρ1l2 − ρ2l1l1−l2

answer is A.

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Detailed Solution

R1=ρ1l1 A and R2 ρ2l2A. In series Req =R1+R2ρeq.l1+l2A=ρ1l1 A+ ρ2l2A⇒ ρeq.= ρ1l1 +ρ2l2 l1+l2

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