A piece of aluminium wire is streched to become it's diameter half of it's original value ,the new resistance will become ? Ans 16 times can u explain this?

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A piece of aluminium wire is streched to become it's diameter half of it's original value ,the new resistance will become ? Ans 16 times can u explain this?

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A piece of aluminium wire is streched to become it's diameter half of ...

R=(p*l)/A;

where

[R= resistance of the conductor

p= resistivity of the conductor

l= length of the conductor

A= Area of cross section of the conductor]

Now , R =(p*l*A)/(A*A);

[Multiplying the numerator and denominator with the area of the cross section of the conductor (A)]

Now, R=(p*V)/(A*A);

[as V=l*A;

where V is the volume of the conductor]

Now, when we are stretching the conductor the RESISISTIVITY (p), and the tolal VOLUME (V) remains constant.

Which means,

R is inversely proportional to A^2

Implies, R is inversely proportional to d^4

[assuming the cross section of the wire is circular

d is the diameter of the area of cross section]

So simply, if d reduces to d/2

d^4 reduces to (d^4)/16

R increases to 16R.

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A piece of aluminium wire is streched to become it's diameter half of it's original value ,the new resistance will become ? Ans 16 times can u explain this?

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