A potential V₀ is applied across a uniform wire of resistance R. The power dissipation is P₁. The wire is then cut into two equal halves and a potential of V₀ is applied across the length of each half. The total power dissipation across two wires is P₂. The ratio P₂: P₁ is √x: 1. The value of x is ___________.

Solution

Physics sir · 3 steps

Same voltage, half the resistance: use P = V²/R, so each half burns twice the original power.

1. Original wire. The voltage is fixed, so use P = V²/R. For the whole wire, P₁ = V₀²/R.
2. Each half. Resistance is proportional to length, so each half has resistance R/2. With the full V₀ across it, each half dissipates V₀²/(R/2) = 2V₀²/R.

   P₂ = 2 × 2V₀²/R = 4V₀²/R = 4P₁

3. So. P₂ : P₁ = 4 : 1 = √16 : 1, so x = 16.

The common slip: 4. Stopping at the ratio P₂ : P₁ = 4 gives 4, but the question writes the ratio as √x, so x = 4² = 16.

The idea to keep. When the voltage across a resistor is held fixed, reach for P = V²/R: halving the resistance doubles the power. Always read whether the question asks for the ratio or something inside a root.

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