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JEE Main 2025 (28 Jan, Shift 2) · Power

Power question from JEE Main 2025 (28 Jan, Shift 2)

Q7JEE Main 2025 · 28 JAN · SHIFT 2Medium

Power

A body of mass 4 kg is placed on a plane at a point PP having coordinates (3,4) m(3, 4)\text{ m}. Under the action of forceF⃗=(2i^+3j^) N\vec{F} = (2\hat{i} + 3\hat{j})\text{ N}, it moves to a new point Q having coordinates (6,10) m(6, 10)\text{ m} in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of:

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Answer: (d)

  1. Displacement s⃗=(6−3)i^+(10−4)j^=3i^+6j^\vec{s} = (6-3)\hat{i} + (10-4)\hat{j} = 3\hat{i} + 6\hat{j} m.
  2. W=F⃗⋅s⃗=6+18=24 JW = \vec{F} \cdot \vec{s} = 6 + 18 = 24\text{ J}, so Pavg=244=6 WP_{avg} = \dfrac{24}{4} = 6\text{ W}.
  3. a⃗=F⃗m=0.5i^+0.75j^\vec{a} = \dfrac{\vec{F}}{m} = 0.5\hat{i} + 0.75\hat{j}; from rest v⃗=a⃗t=2i^+3j^\vec{v} = \vec{a}t = 2\hat{i} + 3\hat{j} m/s at 4 s.
  4. Pinst=F⃗⋅v⃗=4+9=13 WP_{inst} = \vec{F} \cdot \vec{v} = 4 + 9 = 13\text{ W}
  5. So the answer is (d) 6:136:13.
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