A 10 Ω resistor is connected across a 5 V battery. The current flowing is MCQ with Answer and Explanation

Explanation:
η = 1 - T₂/T₁ = 1 - (300/400) = 1 - 0.75 = 0.25 = 25%. Wait, T₁ high temp 400 K, T₂ low 300 K, so η = 1 - 300/400 = 0.25 = 25%. Option A is 25%. Let's correct answer A. I'll set option A: 25%.

Explanation:
Resistance R = rhoL/A. If the wire is stretched so length becomes L' = 2L, the volume (V = AL) must remain constant. Therefore, the new area A' must be A/2 to maintain V. The new resistance R' = rho(2L)/(A/2) = 4 * (rhoL/A) = 4R. Stretching a wire to double its length increases its resistance strictly by a factor of 4. This is a classic trap question compared to merely 'changing the wire'.

Explanation:
ΔPE = mgΔh = 4×10×(8-2) = 4×10×6 = 240 J.