Units & Dimensions

JEE Main level, single correct, medium

105 of 149

Question 105 of 149

The force is given in terms of time tt and displacement xx by the equation F=Acos⁡Bx+Csin⁡DtF=A\cos Bx+C\sin Dt
The dimensional formula of ADB\dfrac{AD}{B} is:

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Answer: (B) [ML2T−3]\left[{ML}^{2}{T}^{-3}\right]

[A]=[MLT−2][A]=\left[{MLT}^{-2}\right]
[B]=[L−1][B]=\left[{L}^{-1}\right]
[D]=[T−1][D]=\left[{T}^{-1}\right]
[ADB]=[MLT−2][T−1][L−1]\left[\dfrac{AD}{B}\right]=\dfrac{\left[{MLT}^{-2}\right]\left[{T}^{-1}\right]}{\left[{L}^{-1}\right]}
[ADB]=[ML2T−3]\left[\dfrac{AD}{B}\right]=\left[{ML}^{2}{T}^{-3}\right]

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