Indian Navy INET Physics — practice by topic

Escape velocity from Earth's surface: $v_e = \sqrt{2gR}$ (using the surface relation $GM/R^2 = g$). $v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = \sqrt{1.25 \times 10^8} \approx 1.12 \times 10^4$ m/s $= 11.2$ km/s. Option A ($7.9$ km/s) is the *orbital* velocity at Earth's surface, smaller by a factor of $\sqrt{2}$. That trap is identical to the one in the orbital-velocity question.

**For parallel:** voltage is same across each, currents add. 1/R_eq = Σ(1/Rᵢ). Equivalent is always LESS than any individual.

**In series:** E_eq = E₁ + E₂ + E₃ = 6V. We have r_eq = r₁ + r₂ + r₃. So r_eq = 1.5 Ω. (Series adds both emf and internal resistance.)