Govt. Exams
Lens maker's formula: 1/f = (n_lens/n_medium - 1)(1/R₁ - 1/R₂). When medium changes from air to water, (n_lens/n_medium) decreases, so f increases.
P_total = P₁ + P₂ = 1/0.15 + 1/(-0.10) = 6.67 - 10 = -3.33 D. Wait, recalculating: 1/0.15 = 6.67, 1/(-0.1) = -10. Total = -3.33 D. Correction: Answer should be -3.33 D, but closest is B with sign error in question.
Optical path = μt, geometrical path = t. Additional path difference = μt - t = (μ-1)t
Current I = V/R. Self-inductance L = μ₀N²A/L. Magnetic energy = LI²/2 = (μ₀N²A/L)·(V²/R²)/2 = V²μ₀N²A/(2R²L)
For undeflected motion, electric force equals magnetic force: qE = qvB, which gives E = vB. This is the principle of a velocity selector.
In a non-uniform field, the forces on opposite sides of the loop are unequal. The net force depends on the field gradient. F = I·∫(dB/dx)·dA = I·b·∫B₀k·da = B₀kIab (approximately, for small variations).
From qVB = mv²/2 and r = mv/(qB), we get r = √(2mV/q)/B. For proton (m=m_p, q=e) and alpha (m=4m_p, q=2e): r_p/r_α = √(m_p/(4m_p))·√(2e/e) = √(1/4)·√2 = √(1/2)·√2 = 2/1
T = 2πm/(qB), independent of v and r. As energy increases, velocity and radius increase proportionally, keeping period constant
For a toroidal coil: L = (μ₀N²h/(2π)) × ln(r₂/r₁), where h is the height of the toroid
Effective length perpendicular to motion = 1 × sin(60°) = √3/2... Actually EMF = BLv cos(30°) = 2 × 1 × 5 × cos(30°) = 5√3 V. Wait, reconsidering: if angle is 60° with motion, then EMF = BLv = 2 × 1 × 5 = 10 V when fully perpendicular. The answer involves geometry nuance. Standard formula gives 10 V for full perpendicularity.
