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JEE Main 2021 (16 Mar, Shift 2) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2021 (16 Mar, Shift 2)

Q23JEE Main 2021 · 16 MAR · SHIFT 2Medium

Inertia, First & Second Laws

A body of mass 2 kg moves under a force of (2i^+3j^+5k^)\left(2\hat{i} + 3\hat{j} + 5\hat{k}\right) N. It starts from rest and was at the origin initially. After 4s, its new coordinates are (8, b, 20). The value of b is _____________. (Round off to the Nearest Integer)

Numerical answer type. Work it out, then open the answer.

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Answer: 12

  1. ay=32=1.5 m/s2a_y = \frac{3}{2} = 1.5\text{ m/s}^2 (check: ax=1a_x = 1, x=12(1)(16)=8x = \frac{1}{2}(1)(16) = 8)
  2. y=12ayt2=12(1.5)(16)=12 my = \frac{1}{2}a_yt^2 = \frac{1}{2}(1.5)(16) = 12\text{ m}
  3. So b=12b = 12.
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