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A wire of resistance 4 \Omega is stretched to twice its original length. The resistance of stretched wire would be:

  • Option 1)

    16 \Omega

  • Option 2)

    2 \Omega

  • Option 3)

    4 \Omega

  • Option 4)

    8 \Omega

 

Answers (1)

best_answer

i=2l

Volume of wire remains constant

lA=l{}'A{}'=A{}'=\frac{l}{l{}'}A=\frac{l}{l{}'}A=\frac{l}{2l}A=\frac{A}{2}

New resistance 

R{}'=\frac{\int l{}'}{A{}'}=\frac{\int 2l}{\left( \frac{A}{2} \right )}=4\frac{\int l}{A}

R{}'=4(4\pi )=16\pi


Option 1)

16 \Omega

This option is correct 

Option 2)

2 \Omega

This option is incorrect 

Option 3)

4 \Omega

This option is incorrect 

Option 4)

8 \Omega

This option is incorrect 

Posted by

Plabita

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