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Answer» ONG>Answer: answer Explanation: Given: Let the INITIAL length of WIRE=l1 Final length of wire after stretching=l1+(10/100)11 =l1(1+10/100) =( 110/100 )11 = 11/10 11 Area of initial wire=A1 Area of final wire = A2 Volume of wire remains constant. 11A1=I2A2 A2=11A1/12 As we know that R=p l/A R1/R2 =( 11/A1 )/(12/A2) = 11A2/12A1 = 11 (11A1/12)/12A1 R1/R2 = 11^2/12^2=100/121 R2=R1x121/100 (AR/R)*100=(R2-R1/R)x100=21% The PERCENTAGE increase in resistance IS 21% I THINK so this is your answer tq byee |
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