Question

a wire of resistance of 16 ohm. it is metal and drawn into a wire of half its length .calculate the resistance of the new wire. what is the percentage change in the resistance?

Answer 1

R1 = 1 Ohm
R2 = ?
L1 = L
L2 = 2L1 = 2L
Now, when the wire is melted and recast into a wire of twice length, its volume will remain unchanged
So, L1A1 = L2A2
      L1A1 = 2L1A2
          A2 = A1/2 ------------------------ (1)
Now, resistance R = RHO*L/A
R1 = RHO * L1/A1
R2 = RHO * L2/A2
R1/R2 = RHO(L1/A1)/ RHO(L2A2) = LIA2/L2A1
From equation 1 R1/R2 = L1A1/2*2L1A1 = 1/4
R2 = 4R1 = 16*4 = 64 Ohm
Therefore, change in resistance = R2-R1 = 64-16 = 48 Ohm
PERCENTAGE CHANGE IN RESISTANCE = 48/16 = 3*100 = 300%