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JEE Main 2026 (4 Apr, Shift 2) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2026 (4 Apr, Shift 2)

Q2JEE Main 2026 · 4 APR · SHIFT 2Medium

Inertia, First & Second Laws

At t=0t = 0, a body of mass 100 g starts moving under the influence of a force (5i^+10j^) N(5\hat{i} + 10\hat{j})\text{ N} After 2 s its position is (2xi^+5yj^) m(2x\hat{i} + 5y\hat{j})\text{ m}. The ratio x:yx : y is ______.

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Answer: (d)

  1. a⃗=5i^+10j^0.1=50i^+100j^ m/s2\vec{a} = \frac{5\hat{i} + 10\hat{j}}{0.1} = 50\hat{i} + 100\hat{j}\text{ m/s}^2
  2. From rest: r⃗=12a⃗t2=2(50i^+100j^)=100i^+200j^\vec{r} = \frac{1}{2}\vec{a}t^2 = 2(50\hat{i} + 100\hat{j}) = 100\hat{i} + 200\hat{j}
  3. 2x=100⇒x=502x = 100 \Rightarrow x = 50; 5y=200⇒y=405y = 200 \Rightarrow y = 40
  4. x:y=50:40=5:4x : y = 50 : 40 = 5 : 4. So the answer is (d).
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