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Units and Measurements question

2025 · 23 Jan · Shift 1 · Q55
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Units and Measurements question

2025 · 23 Jan · Shift 1 · Q55

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The position of a particle moving on xxx-axis is given by x(t)=Asin⁡t+Bcos⁡2t+Ct2+Dx(t)=A \sin t+B \cos ^2 t+C t^2+Dx(t)=Asint+Bcos2t+Ct2+D, where ttt is time. The dimension of ABCD\frac{A B C}{D}DABC​ is
  1. A
    L\mathrm{L}L
  2. B
    L3 T−2\mathrm{L}^3 \mathrm{~T}^{-2}L3 T−2
  3. C
    L2 T−2\mathrm{L}^2 \mathrm{~T}^{-2}L2 T−2
  4. D
    L2\mathrm{L}^2L2
View written solutionFree

Correct answer: C

  1. Since x(t)x(t)x(t) represents position on the xxx-axis, its dimension is length: [x]=L[x]=L[x]=L

  2. In the expression x(t)=Asin⁡t+Bcos⁡2t+Ct2+D,x(t)=A\sin t + B\cos^2 t + Ct^2 + D,x(t)=Asint+Bcos2t+Ct2+D, each term must have the same dimensions as xxx, i.e. LLL.

  3. Now analyze each term:

  • For Asin⁡tA\sin tAsint: Since sin⁡t\sin tsint is dimensionless, [A]=L[A]=L[A]=L

  • For Bcos⁡2tB\cos^2 tBcos2t: Since cos⁡2t\cos^2 tcos2t is dimensionless, [B]=L[B]=L[B]=L

  • For Ct2Ct^2Ct2: [C][t2]=L[C][t^2]=L[C][t2]=L [C]T2=L[C]T^2=L[C]T2=L [C]=LT−2[C]=LT^{-2}[C]=LT−2

  • For DDD: [D]=L[D]=L[D]=L

  1. Therefore, [ABCD]=[A][B][C][D]\left[\frac{ABC}{D}\right]=\frac{[A][B][C]}{[D]}[DABC​]=[D][A][B][C]​ Substitute the dimensions: [ABCD]=(L)(L)(LT−2)L\left[\frac{ABC}{D}\right]=\frac{(L)(L)(LT^{-2})}{L}[DABC​]=L(L)(L)(LT−2)​ =L2T−2=L^2T^{-2}=L2T−2

  2. Hence the correct option is: L2T−2\boxed{L^2T^{-2}}L2T−2​ which is option C.

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