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Physics · Units, Dimensions & Error Analysis

JEE Main 2025 — 23 January, Morning Shift — Question 56

. The position of a particle moving on xx-axis is given by

x(t)=Asin⁡t+Bcos⁡2t+Ct2+Dx(t)=A \sin t+B \cos ^{2} t+\mathrm{Ct}^{2}+D, where tt is time.

The dimension of ABCD\frac{A B C}{D} is-

  1. Option A:

    L

  2. Option B:

    L3 T−2\mathrm{L}^{3} \mathrm{~T}^{-2}

  3. Option C:

    L2 T−2\mathrm{L}^{2} \mathrm{~T}^{-2}

    Correct
  4. Option D:

    L2L^{2}

Answer: C

Step-by-step solution

Dimension [x(t)]=[L][x(t)]=[L]

[A]=[L][\mathrm{A}]=[\mathrm{L}]

[B]=[L][B]=[L]

[C]=[LT−2][C]=\left[\mathrm{LT}^{-2}\right]

[D]=[L][\mathrm{D}]=[\mathrm{L}]

[ABCD]=[L×L×LT−2 L]=[L2 T−2]\left[\frac{\mathrm{ABC}}{\mathrm{D}}\right]=\left[\frac{\mathrm{L} \times \mathrm{L} \times \mathrm{LT}^{-2}}{\mathrm{~L}}\right]=\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Units and Dimensions Analysis

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