A truck starts from rest down a hill with a constant acceleration. It achieves 400 meters in 20 seconds. If the weight of the truck is 7 tons then what will be the force acting on it?

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Correct — D, 14000 Newtons. The question is two clean steps with one deliberate landmine between them. Step one is kinematics. 'Starts from rest' fixes the initial velocity u = 0, and constant acceleration lets you use s = ut + (1/2)at squared. Substituting s = 400 m, u = 0 and t = 20 s gives 400 = (1/2) x a x 400, so a = 2s/t squared = 800/400 = 2 m/s squared. Cross-check it a second way, because it costs three seconds: the average speed is 400/20 = 20 m/s, and for a body uniformly accelerated from rest the final speed is exactly twice the average, so v = 40 m/s, and a = v/t = 40/20 = 2 m/s squared. Same answer, so the kinematics is safe. Step two is Newton's second law, F = ma, with the truck's 7 tons read as a mass of 7,000 kg — one metric tonne is 1,000 kg exactly. That gives F = 7,000 x 2 = 14,000 N, or 14 kilonewtons, which is option (d). Now the landmine, and it is the reason this question drew objections. The English stem says 'the weight of the truck is 7 tons', but weight is a force, measured in newtons, while the tonne is a unit of MASS. The Commission settled the point itself when it disposed of candidate objections on 31 October 2025: its published remark states that 7 tons is expressed in the unit of mass and not in the unit of weight, that the Hindi version of this very question uses the word dravyaman, which means mass, and that the mass of the truck is therefore to be taken as 7 tons, that is 7,000 kg. Read the sentence that way and F = ma gives 14,000 N at once. There is in fact a second, stricter reading that lands on the same figure, and knowing it removes any residual doubt about the key: if you insist that 'weight' must denote a force and read 7 tons as 7 tonne-force, that is 7,000 kgf or about 68,650 N, then the mass recovered from m = W/g is 68,650/9.81 = 7,000 kg all over again, and F = ma is unchanged at 14,000 N. Both defensible readings converge. The only route that fails is the incoherent middle one — dividing the bare number 7,000 by 9.8 to 'recover' a mass of about 714 kg and a force near 1,429 N, which matches no option on the paper, and the option list is itself telling you that this was not the intended reading. One last point of physics that is worth more than the mark: the 14,000 N is the NET or resultant force along the slope, not the truck's weight. A 7,000 kg truck weighs about 68,600 N on Earth; only the component of that weight along the hill, reduced by whatever friction and braking resist it, is left over as the 14,000 N that actually accelerates the vehicle.

Why the other options are wrong — (a) At the correctly computed acceleration of 2 m/s squared, 11,000 N would require a mass of 5,500 kg, that is 5.5 tonnes rather than 7 — so this option corresponds to misreading the tonnage, not to any standard slip in the kinematics. It sits on the option list mainly as a decoy: the four choices are clustered within 3,000 N of each other precisely so that a candidate who estimates rather than computes cannot arrive at the right one.  (b) This is the value for a 6,000 kg truck at the same 2 m/s squared, or equivalently for 7,000 kg at about 1.7 m/s squared. Neither figure comes out of the data given. It is the classic 'round it down and move on' choice, and it exists to punish candidates who compute the acceleration correctly and then rush the multiplication under time pressure in a 150-question, 120-minute paper.  (c) The most defensible wrong answer, and the one with a real reason behind it. If 'ton' is read as the US short ton of 907.18 kg instead of the metric tonne, the mass becomes 7 x 907.18 = 6,350 kg and the force 12,700 N, which rounds to this option. India, however, uses the metric tonne of exactly 1,000 kg in all statutory and commercial usage, so 7 tons here is 7,000 kg and the force is 14,000 N.

Concept: Two standard blocks of school mechanics meet in this question. The first is uniformly accelerated motion, described by three equations that follow from the definition of acceleration: v = u + at, s = ut + (1/2)at squared, and v squared = u squared + 2as, where u is the initial velocity, v the final velocity, a the constant acceleration, t the time and s the displacement. Choosing among them is purely a matter of which quantity is missing — here the final velocity is not given and is not asked for, so the second equation is the one to reach for. The second block is Newton's second law: the net external force on a body equals the rate of change of its momentum, which for constant mass reduces to F = ma. The SI unit of force is the newton, defined as the force that gives a mass of one kilogram an acceleration of one metre per second squared, a name formally adopted by the 9th General Conference on Weights and Measures in 1948. Underneath both blocks sits the distinction the question turns on. Mass is a scalar measure of the quantity of matter and of a body's inertia; its SI unit is the kilogram and it does not change with location. Weight is the gravitational force on that mass, W = mg; it is a vector, its unit is the newton, and it varies with where you are — the same truck that weighs about 68,600 N on Earth would weigh roughly 11,400 N on the Moon, where g is about one-sixth. Everyday Indian speech uses 'weight' for what physics calls mass, which is exactly the ambiguity that generated objections here.

State PCS 2025 Q122 — Science & Technology PYQ with Answer

State PCS 2025 Q122 — Science & Technology PYQ with Answer