Apparent Weight In A Lift

As explained in NCERT Class 11, Chapter 5 (Laws of Motion), the reading on a weighing scale is the normal reaction N it exerts on the person, which is the apparent weight. For a lift accelerating upward, applying Newton's Second Law in the vertical direction gives N − mg = ma, so N = m(g + a). Substituting, N = 70 × (10 + 2) = 70 × 12 = 840 N. The person feels heavier because the floor must both support the weight and provide the extra upward force needed for acceleration. Option 700 N is the true weight mg, valid only when the lift moves at constant velocity or is at rest. Option 560 N uses (g − a), which applies to downward acceleration, not upward. Option 770 N corresponds to an acceleration of only 1 m/s², not the given 2 m/s². It is worth noting that the actual gravitational force on the person, mg = 700 N, is unchanged throughout; only the support force from the floor changes, which is why apparent weight differs from true weight inside accelerating systems. In a freely falling lift, a would equal g and the scale would read zero, the familiar state of weightlessness. Plausibility check: upward acceleration must increase the apparent weight above 700 N, and 840 N exceeds mg by exactly ma = 140 N, confirming the result.