We apply a force of 200 Newtons on a wooden Box and push it on the floor at constant velocity. The marginal friction force will be :

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Correct — B, 200 Newtons. The decisive words in the stem are 'constant velocity', and everything else in the question is decoration. Constant velocity means the velocity is not changing, so the acceleration is zero; and by Newton's second law, F(net) = ma, zero acceleration forces the net external force on the box to be exactly zero. Resolve the forces on the box: vertically, its weight acts downwards and the floor's normal reaction acts upwards, and these two cancel because the box is not sinking or lifting off; horizontally, only two forces exist — your 200 N push forwards and the friction of the floor backwards. For the horizontal sum to vanish, friction must be 200 N, equal in magnitude and opposite in direction to the push. That is option (b), and no other option can be reconciled with steady motion. Notice how little the question actually gives you: no mass, no coefficient of friction, no material of the floor beyond 'wooden Box'. That absence is deliberate and it is the clue — the examiner has stripped out every quantity you would need to compute friction independently from f = (mu)N, because the answer is meant to come from the equilibrium condition alone. On terminology, one honest clarification, because it is what makes the question look harder than it is. The paper's phrase 'marginal friction force' is the Hindi seemant gharshan bal of the parallel Hindi text, the standard Indian-textbook term normally rendered in English as limiting friction. Strictly, limiting friction is the maximum value that static friction can reach at the instant sliding is about to begin, whereas the force acting on a body that is already sliding is kinetic or sliding friction, which for the same pair of surfaces is a little smaller. The box in the stem is already moving, so what balances your hand is kinetic friction — and because the box neither speeds up nor slows down, that kinetic friction is exactly 200 N. Whichever of the two names you attach to it, 200 N is the only value on the option list consistent with the motion described. A concrete feel for the number: if the wood-on-floor kinetic coefficient is a typical 0.2, then the normal reaction is N = f/(mu) = 200/0.2 = 1,000 N, which is a box of roughly 102 kg — a heavy crate, pushed steadily by one determined person.

Why the other options are wrong — (a) The intuitive answer, and the one this item is built to punish. It assumes half your push is 'used up' by friction and the other half is 'left over' to keep the box going. But an unopposed 100 N forward would be a net force, and a net force means acceleration: a 100 kg box would gain 1 m/s of speed every second and be moving at 10 m/s after ten seconds. That directly contradicts 'constant velocity'. This is Aristotle's physics — motion needs a surviving force — not Newton's, in which only a CHANGE of motion needs one.  (c) If friction were 300 N against a 200 N push, the net force would be 100 N backwards, the box would decelerate and would come to rest within seconds. A body cannot slide at unchanging speed while a net backward force acts on it. Candidates arrive here by treating friction as an active opposing agent that always 'wins' over the applied force, instead of a passive contact force whose value during steady sliding is pinned by the balance condition to exactly match the push.  (d) Twice the applied force — usually chosen by half-remembering f = (mu)N and treating some remembered number as a multiplier of the push, which the friction law never is: the normal reaction N, not the applied force, sets the friction. It also fails a second, simpler test. If this floor could really supply 400 N of resistance, then the limiting static friction would exceed your 200 N push and the box would never have started moving at all — yet the stem says it is moving.

Concept: Friction is the contact force that opposes relative sliding between two surfaces, and it comes in three regimes that examiners deliberately blur. Static friction is self-adjusting: push a stationary crate with 40 N and static friction returns exactly 40 N, push with 90 N and it returns 90 N, up to a ceiling called limiting friction, equal to (mu-s)N. Push harder than that ceiling and the crate breaks away; from then on the resistance is kinetic (sliding) friction, f = (mu-k)N, which stays roughly constant while sliding and is smaller than the limiting value — which is precisely why a heavy almirah is hard to start and noticeably easier to keep moving. The empirical rules are the Amontons-Coulomb laws, set out by Guillaume Amontons in 1699 and put on a quantitative footing by Charles-Augustin de Coulomb in 1785: friction is proportional to the normal reaction, is independent of the apparent area of contact, and is nearly independent of the speed of sliding. The apparent-area rule looks absurd until you know the microscopic reason — two surfaces touch only at the tips of microscopic asperities, so the REAL contact area is a tiny fraction of the visible one and grows in proportion to the load, not to how the block is laid out. Sitting underneath all of this is Newton's first law, from the Principia of 1687: rest and uniform motion in a straight line are the same dynamical state, and a body stays in it unless an unbalanced external force acts. A box sliding at steady speed is therefore in equilibrium exactly as a box standing still is — a condition physicists call dynamic equilibrium.

State PCS 2025 Q121 — Science & Technology PYQ with Answer

State PCS 2025 Q121 — Science & Technology PYQ with Answer