BPSC 71st CCE Prelims General Studies · Question 122 of 150

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.

• (a)11000 Newtons — 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)12000 Newtons — 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)13000 Newtons — 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.

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.

Work this in four moves. First, extract the data and notice what is absent: u = 0 from 'starts from rest', s = 400 m, t = 20 s, and no final velocity anywhere — so use s = ut + (1/2)at squared. Second, solve for a: with u = 0 the equation collapses to a = 2s/t squared = 800/400 = 2 m/s squared. Third, convert the tonnage, 7 t = 7,000 kg. Fourth, apply F = ma to get 14,000 N. The single discriminating fact is the one the Commission had to state explicitly: the '7 tons' in the stem is a MASS, not a weight, because the tonne is a mass unit and because the Hindi text of the same question prints dravyaman, the word for mass. Everything else in the question is routine; that one reading is what separates 14,000 N from a number no option carries. There is also a useful reasoning habit here that generalises beyond this paper. When a numerical looks dimensionally strange, test the suspect reading against the option list. Divide the bare 7,000 by g to 'recover' a mass and the answer comes out near 1,429 N with nothing on the paper resembling it — the option list has just told you that route is wrong, while both coherent readings, 7 tonnes of mass and 7 tonne-force of weight, give 7,000 kg and 14,000 N. The other trap is quieter: 'the force acting on it' means the net force, the resultant of gravity along the slope minus friction and rolling resistance. Candidates who go looking for the slope angle, or who compute the full weight of 68,600 N, are answering a question that was not asked. And a sanity check on the scenario: after 20 s the truck is doing v = at = 40 m/s, which is 144 km/h — a genuinely runaway vehicle, and the reason ghat roads carry gravel arrester beds.

• Equations of uniformly accelerated motion: v = u + at, s = ut + (1/2)at squared, v squared = u squared + 2as. Starting from rest (u = 0) with s = 400 m and t = 20 s, the second equation gives a = 2s/t squared = 800/400 = 2 m/s squared.
• Newton's second law gives F = ma for constant mass. The newton is defined as the force that accelerates one kilogram at one metre per second squared, a unit name adopted by the 9th General Conference on Weights and Measures in 1948. Here F = 7,000 x 2 = 14,000 N = 14 kN.
• Disposing of candidate objections on 31 October 2025, the Bihar Public Service Commission ruled that the '7 tons' in this stem is stated in a unit of mass and not of weight, noted that the Hindi version uses dravyaman, meaning mass, and directed that the truck be treated as 7,000 kg.
• One metric tonne is exactly 1,000 kg and is the unit used in Indian statutory and commercial practice; the US short ton is 907.18 kg and the UK long ton is 1,016.05 kg. Choosing the short ton here would give 6,350 kg and 12,700 N, which is why 13000 Newtons appears on the option list.
• Mass is a scalar in kilograms and is the same everywhere; weight is the force mg, a vector in newtons, and changes with location. A 7,000 kg truck weighs about 68,600 N on Earth at g = 9.8 m/s squared and about 11,400 N on the Moon, where g is roughly one-sixth of the Earth value.
• The 14,000 N is the net force along the slope, not the truck's weight. On a frictionless incline it would correspond to sin(theta) = a/g = 2/9.8, a slope of about 12 degrees; and after 20 s the truck's speed is v = at = 40 m/s, or 144 km/h.

• Dividing the bare number 7,000 by g to 'recover' a mass — that gives about 714 kg and a force near 1,429 N, which matches no option; read it as 7 tonnes of mass, which is the Commission's ruling, or even as 7 tonne-force of weight, and the answer is 14,000 N either way
• Forgetting the factor of one-half in s = ut + (1/2)at squared, which doubles the acceleration and the force
• Answering with the truck's weight, about 68,600 N, instead of the net force that accelerates it — the stem asks for 'the force acting on it', meaning the resultant along the slope

BPSC still sets the plug-in numerical: one kinematics equation, then F = ma, four round options three thousand newtons apart, and a unit conversion hidden in translated English — here the English said 'weight' where the Hindi said mass, and the Commission had to publish a remark to settle it. UPSC set exactly this shape in the 1990s and early 2000s, asking in 1998 for the velocity of a dropped ball after three seconds and in 2001 whether a 100 kg body keeps its mass on the Moon, but has since moved almost entirely to statement-based and assertion-reason formats, so a UPSC aspirant now meets F = ma as a concept to apply rather than a formula to substitute into.

• practice — not a real PYQ

  A body of mass 5 kg, starting from rest, attains a velocity of 20 m/s in 4 seconds under a constant force. The magnitude of the force is

  • (a)5 Newtons
  • (b)15 Newtons
  • (c)25 Newtons
  • (d)100 Newtons

  Answer(c) 25 Newtons — the acceleration is a = (v - u)/t = (20 - 0)/4 = 5 m/s squared, so F = ma = 5 x 5 = 25 N. The 100 N option is what you get by multiplying mass by velocity instead of by acceleration.

• practice — not a real PYQ

  A body is carried from the Earth to the Moon, where the acceleration due to gravity is about one-sixth of its value on the Earth. Which one of the following is correct ?

  • (a)Both its mass and its weight remain unchanged
  • (b)Its mass remains unchanged but its weight becomes about one-sixth
  • (c)Its mass becomes about one-sixth but its weight remains unchanged
  • (d)Both its mass and its weight become about one-sixth

  Answer(b) Its mass remains unchanged but its weight becomes about one-sixth — mass measures the quantity of matter and the inertia of a body and does not depend on location, whereas weight is the gravitational force mg and falls in proportion to g.