Question

A copper wire of resistance 5 ohm is stretched to reduce its diameter to 1/2 of its initial value calculate the new resistance

Answer 1

Resistance = resistivity * Length / cross-sectional area 

Let L be the length, d be the diameter, then volume of the wire = pi*d^2*L/4 

Now when the wire is stretched to half of its diameter, d' =d/2, but volume will remain the same 

So new length L' = ?? 

pi*d^2*L/4 = pi* (d/2)^2 * L' /4 

=> L = L'/4 


Now R = ro * L/A => ro = RA/L 

So new R' = ro * L'/A' = RA/L *(L'/A') 

now L' = 4L, 
d' = d / 2 

So A' = pi*d'^2 / 4 = pi* (d/2)^2 /4 = (pi*d^2/4)/4 = A/4 
So R' = 5 * A/L * 4L /(A/4) = 5 ohms 

Thus, the resistance stays the same.

Answer 2

Answer:

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