BPSC 71st CCE Prelims General Studies · Question 126 of 150

Correct — C, 2100 Joule. Work in physics is the force multiplied by the displacement produced along the line of that force: W = F × s, or more generally W = F s cos θ. The oxen pull the plough forward along the furrow, so the force and the displacement point the same way, θ = 0° and cos θ = 1. Substituting the only two numbers the stem supplies — F = 140 N and s = 15 m — gives W = 140 × 15 = 2100 N m = 2100 J. Nothing else in the question is usable, and that is the clue: there is no mass, so Newton's second law is not wanted; no time, so power is not wanted; no height, no velocity and no coefficient of friction. The 140 N is already stated as the force exerted on the plough, not the weight of the animals, so no free-body analysis is needed either. Two details repay attention. First, the newton-metre and the joule are the same unit — one joule is defined as the work done when a force of one newton displaces a body through one metre along the force's own line of action — so the answer converts itself and every option is dimensionally admissible; only the arithmetic separates them. Second, this item is not original. It is lifted almost word for word from NCERT Class IX Science, Chapter 11 'Work and Energy', where in-text question 4 after Section 11.1 reads: 'A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?' BPSC changed 'bullocks' to 'Oxen' and bolted four options onto it; the situation, the figures and the answer are the textbook's. That is the most useful thing to carry away from Q126 — the science block of the 71st CCE is a Class IX–X NCERT block, so the fastest preparation is the NCERT worked examples themselves rather than a general-science compendium. With +1 for a correct answer and, under the booklet's own instruction 9, −1/3 for a wrong one, 140 × 15 done carefully is a free mark.

• (a)1900 Joule — No legitimate operation on 140 and 15 produces 1900. It would require a force of about 126.7 N over the stated 15 m, or the stated 140 N over about 13.6 m, and neither figure appears anywhere in the question. Its only function is to be the bottom rung of a 1900–2000–2100–2200 ladder spaced 100 J apart, so that a candidate who has estimated 'roughly two thousand' rather than multiplied still finds somewhere plausible to land.
• (b)2000 Joule — The strongest wrong answer, and the reason is psychological rather than physical: 2000 is the roundest number on the ladder, and rounding feels like physics when 150 questions have to be cleared in 120 minutes. Arithmetically it corresponds to 140 N acting over 14.29 m, or to 133.3 N acting over 15 m — neither is stated. A candidate who reasons correctly all the way to 'about two thousand joules' and then picks the round figure loses the mark and a third of another.
• (d)2200 Joule — The over-shoot. It is where you land if the partial products are added wrongly — taking 140 × 15 as 1400 + 800 instead of 1400 + 700 — or if the 15 m furrow is quietly rounded up to 16 m, since 140 × 16 = 2240, which rounds towards 2200. The safe habit is to split the multiplication deliberately: 140 × 10 = 1400, 140 × 5 = 700, and 1400 + 700 = 2100.

'Work' in physics is a far narrower idea than work in ordinary speech, and the whole school chapter turns on that narrowing. Two conditions must both be satisfied before any work is done: a force must act on the object, AND the object must be displaced. A man who pushes a wall until he sweats does zero work on the wall, because the wall does not move; a porter who walks a level platform with a trunk on his head does zero work against gravity, because the force he applies is vertical while the displacement is horizontal, and cos 90° = 0. When force and displacement lie along the same line, W = F × s; in the general case W = F s cos θ, which is a dot product, so work is a scalar — it carries magnitude and sign but no direction. The sign is not decoration. Work is positive when the force has a component along the displacement, which is the case for the oxen on the plough; negative when the force opposes the displacement, as with friction, air drag, or NCERT's goalkeeper whose hands move back 15 cm while she pushes a ball forward with 200 N; and zero when the two are perpendicular. The SI unit is the joule, J = N m = kg m² s⁻², and energy carries the same unit because work is precisely energy transferred from one body to another. The rate of doing work is a separate quantity — power, measured in watts, 1 W = 1 J s⁻¹ — and it is the quantity a stem is asking for whenever it hands you a time.

The reasoning route here is short, and worth naming because BPSC sets three or four items of exactly this shape in every paper. Step one: inventory what the stem actually gives you. A force in newtons and a length in metres, and nothing else. No mass rules out F = ma; no time rules out power, P = W/t; no velocity rules out kinetic energy; no height rules out potential energy. A single force-and-distance pair acting along one line can only be a work question, because W = F s is the one formula that consumes exactly those two numbers and returns joules — and the options are in joules. That elimination, reading the units of what you are given against the units of the options until the formula picks itself, is the single discriminating move, and it is far more reliable than trying to recall which chapter the question came from. Step two: multiply in parts rather than as one lump — 140 × 10 = 1400, 140 × 5 = 700, total 2100. The four options are spaced exactly 100 J apart, which is deliberate design: they punish an arithmetic slip, not a conceptual one. The trap this item sets is therefore not physics at all. It is deciding, correctly, that the answer is 'about two thousand' and then choosing 2000 because it looks tidy. The second trap is over-reading the stem — worrying about the friction of the soil, the mass of the plough, or whether the oxen also do work lifting earth. The stem has already handed you the net force exerted on the plough; there is nothing left to model.

• W = F s cos θ. Here the pull and the displacement are collinear, so θ = 0°, cos θ = 1, and W = 140 N × 15 m = 2100 J.
• One joule is defined as the work done when a force of one newton displaces a body through one metre along the force's own line of action: 1 J = 1 N m = 1 kg m² s⁻², and 1 kJ = 1000 J.
• Work is a scalar — the dot product of force and displacement — so it has sign but no direction: positive when the two agree, negative when they oppose (NCERT's worked case, a goalkeeper's hand moving back 0.15 m against a 200 N push, gives 200 × (−0.15) = −30 J), and zero when they are perpendicular.
• The unit honours James Prescott Joule (1818–1889), who established the mechanical equivalent of heat through his paddle-wheel experiments; the rate of doing work is power, 1 watt = 1 joule per second, named after James Watt (1736–1819).
• The stem is NCERT Class IX Science, Chapter 11 'Work and Energy', in-text question 4 after Section 11.1 — 'A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long' — reproduced with only 'bullocks' changed to 'Oxen'. In NCERT's new Class IX 'Exploration' (First Edition, April 2026) the same material sits in Chapter 7, 'Work, Energy, and Simple Machines'.
• The same 71st CCE paper tested this unit twice more: Q123 required 250 domestic 'units' of electricity to be converted into joules — 250 kWh × 3.6 × 10⁶ = 9 × 10⁸ J, the working the Commission itself printed in its remarks — and Q125 asked whether a falling body's shifting potential and kinetic energy violates any law of physics, which it does not.

• Choosing 2000 Joule because it is the round number: the options sit exactly 100 J apart so that an arithmetic slip becomes a −1/3 penalty rather than a near miss
• Treating 140 newton as the weight of the oxen, or trying to subtract soil friction — the stem already states the force exerted while ploughing, so there is nothing to deduct
• Confusing work with power: power needs a time value and this stem gives none, so any stem that supplies seconds is asking for watts, not joules

BPSC's science block is a school-textbook block. It takes an NCERT Class IX–X worked example, keeps the original numbers, spaces four options a fixed step apart and asks for a one-line substitution — Q121 to Q131 of this very paper are all of that shape, and the Commission's published remarks on Q122, Q123 and Q131 are literally the arithmetic. UPSC has not set a bare plug-in numerical in Prelims for roughly two decades. Its work-and-energy items are conceptual and usually comparative — three vehicles with equal kinetic energies and equal braking forces, or what a simple machine does and does not save you. So for UPSC you must own the definition; for BPSC you must own the definition and still be able to multiply calmly against the clock.

• practice — not a real PYQ

  A porter carries a suitcase weighing 300 N on his head and walks 50 m at a steady speed along a level railway platform. The work he does against gravity on the suitcase is

  • (a)15000 J
  • (b)6 J
  • (c)350 J
  • (d)Zero

  Answer(d) Zero — the force he exerts on the suitcase is vertically upward while the displacement is horizontal, so θ = 90°, cos 90° = 0 and W = F s cos θ = 0. 15000 J is the designed trap: it is 300 × 50, the product you get by multiplying blindly and ignoring direction.

• practice — not a real PYQ

  An electric pump lifts 200 kg of water through a vertical height of 10 m in 20 seconds. Taking g = 10 m s⁻², the power of the pump is

  • (a)100 W
  • (b)1000 W
  • (c)2000 W
  • (d)20000 W

  Answer(b) 1000 W — the work done is mgh = 200 × 10 × 10 = 20000 J, and power is work divided by time, 20000 ÷ 20 = 1000 W, i.e. 1 kW. 20000 W is the trap: 20000 J is the work, not the power, and the '20 seconds' in the stem exists precisely to force the extra division.