Question

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Answer 1

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

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