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Units and Measurements question

2021 · 25 Jul · Shift 2 · Q58
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  5. /2021 · 25 Jul · Shift 2 · Q58

Units and Measurements question

2021 · 25 Jul · Shift 2 · Q58

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The force is given in terms of time t and displacement x by the equation F = A cos Bx + C sin Dt The dimensional formula of ADB{{AD} \over B}BAD​ is :
  1. A
    [M0LT−1][{M^0}L{T^{ - 1}}][M0LT−1]
  2. B
    [ML2T−3][M{L^2}{T^{ - 3}}][ML2T−3]
  3. C
    [M1L1T−2][{M^1}{L^1}{T^{ - 2}}][M1L1T−2]
  4. D
    [M2L2T−3][{M^2}{L^2}{T^{ - 3}}][M2L2T−3]
View written solutionFree

Correct answer: B

  1. Given equation

    F=Acos⁡(Bx)+Csin⁡(Dt)F = A\cos(Bx) + C\sin(Dt)F=Acos(Bx)+Csin(Dt)

    Since FFF is force, each term on the right-hand side must have the dimensions of force.

  1. Use the fact that arguments of trigonometric functions are dimensionless

    For cos⁡(Bx)\cos(Bx)cos(Bx): [Bx]=1[Bx] = 1[Bx]=1 Since [x]=L[x] = L[x]=L, we get [B]=L−1[B] = L^{-1}[B]=L−1

    For sin⁡(Dt)\sin(Dt)sin(Dt): [Dt]=1[Dt] = 1[Dt]=1 Since [t]=T[t] = T[t]=T, we get [D]=T−1[D] = T^{-1}[D]=T−1

  1. Find dimension of AAA

    In the term Acos⁡(Bx)A\cos(Bx)Acos(Bx), cosine is dimensionless, so [A]=[F]=MLT−2[A] = [F] = MLT^{-2}[A]=[F]=MLT−2

  1. Find dimension of ADB\dfrac{AD}{B}BAD​

    [ADB]=[A][D][B]\left[\frac{AD}{B}\right] = \frac{[A][D]}{[B]}[BAD​]=[B][A][D]​

    Substitute the dimensions: [ADB]=(MLT−2)(T−1)L−1\left[\frac{AD}{B}\right] = \frac{(MLT^{-2})(T^{-1})}{L^{-1}}[BAD​]=L−1(MLT−2)(T−1)​

    =(MLT−3)(L)= (MLT^{-3})(L)=(MLT−3)(L)

    =ML2T−3= ML^2T^{-3}=ML2T−3

  1. Match with options

    [ADB]=[M1L2T−3]\left[\frac{AD}{B}\right] = [M^1L^2T^{-3}][BAD​]=[M1L2T−3]

    This corresponds to Option B.

  1. Comparison with stored answer

    Stored correct answer: B

    Derived answer: B

    Hence, the derived answer agrees with the stored answer.

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