The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x × 10⁻³ Ω cm. The value of x is ____________. [Take π = 3.14]

Solution

Physics sir · 4 steps

The slope of a V–I graph is the resistance, and a 45° line means R is exactly 1 ohm.

1. Read the graph. Potential is on the vertical axis and current on the horizontal axis, so the slope is V/I = R. The line makes 45° with the current axis, so R = tan 45° = 1 Ω.
2. Area of cross-section. The diameter is d = 2.4 cm, so A = (π d²)/4 = (3.14 × 5.76)/4 = 4.5216 cm².

   ρ = RA/L = (1 × 4.5216)/31.4 Ω cm

3. Put in numbers. Since 31.4 = 10π, the π cancels: ρ = 5.76/40 = 0.144 Ω cm. Keeping lengths in cm gives ρ directly in Ω cm.
4. So. ρ = 144 × 10⁻³ Ω cm, so x = 144.

The common slip: 576. Using 2.4 cm as the radius instead of the diameter makes the area four times too large and gives 576.

The idea to keep. On a V–I graph the slope is R; on an I–V graph it is 1/R. Always check which quantity sits on which axis before reading a slope as resistance.

Answer from the NTA final answer key. Solution by PrepWiser.