Question

A wire 1 m long has resistance of 1 ohm. It is uniformly stretched, so that its length increases by 25 percent, then its resistance increased by

Answer 1

R = pL/A

L= 1 m

1*25/100 = 0.25

New l= 1.25 m

Volume = L*A (constant here)

So if length becomes 1.25 m, area should decrease by 1.25.

R'= p 1.25(L)/A/1.25

= p (1.25)^2 L/A

= (1.25)^2 [p L/A]

= 1.5625 R (R = pL/A)

Therefore, increase in resistance = 56.25%

Answer 2

Given :

Length of the wire (L) = 1 m

Resistance (R) = 1 Ω

To Find :

Increase in Resistance

Solution :

Let 'L' be the initial length and 'A' be the initial area of the wire

∴ Initial Resistance of the wire (R) = ρ × $\frac{L}{A}$ = 1 Ω

Final length of the wire (L') = L + $\frac{25}{100}$×L

                                            = L + 0.25L

                                            = 1.25L m

∴ Initial volume of the wire   =    Final Volume of the wire

⇒ Initial Area (A) × L              =    Final Area (A') × L'

⇒                     A × 1               =          A' × 1.25

∴                           A'             =   $\frac{A}{1.25}$

Final resistance of the wire (R') = ρ× $\frac{L'}{A'}$

                                                   = ρ×$\frac{1.25L}{A/1.25}$

                                                   =  ρ $\frac{L}{A}$ × (1.25)²

                                                   =  1 × 1.5625

                                                  = 1.5625 Ω

Increase in resistance (ΔR) = 1.56 - 1 = 0.56 Ω

Therefore the resistance is increased by 0.5625 Ω

Percentage increase in resistance = $\frac{0.5625}{1}$×100%

                                                          = 56.25%

Hence, the resistance is increased by 56.25%.