A pair of Oxen exerts a force of 140 newton while ploughing the field. The field ploughed is 15 meter long. The work done in ploughing the length of the field is :

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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.

Why the other options are wrong — (a) 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) 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) 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.

Concept: '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.

State PCS 2025 Q126 — Science & Technology PYQ with Answer

State PCS 2025 Q126 — Science & Technology PYQ with Answer