Units & Dimensions

JEE Main level, single correct, medium

63 of 149

Question 63 of 149

A force is represented by F=ax2+bt12F=a{x}^{2}+b{t}^{\dfrac{1}{2}}, where x=x= distance and t=t= time. The dimensions of b2a\dfrac{{b}^{2}}{a} are :

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Answer: (A) [ML3T–3][M{L}^{3}T{}^{–3}]

Given: F=ax2+bt12F=a{x}^{2}+b{t}^{\dfrac{1}{2}}
Dimension of [a]=[F][x2]=[MLT−2][L2]=[M1L−1T−2]\left[a\right]=\dfrac{\left[F\right]}{\left[{x}^{2}\right]}=\dfrac{\left[ML{T}^{-2}\right]}{\left[{L}^{2}\right]}=\left[{M}^{1}{L}^{-1}{T}^{-2}\right]
Dimension of [b]=[F][t12]=[MLT−2][T12]=[M1L1T−52]\left[b\right]=\dfrac{\left[F\right]}{\left[{t}^{\dfrac{1}{2}}\right]}=\dfrac{\left[ML{T}^{-2}\right]}{\left[{T}^{\dfrac{1}{2}}\right]}=\left[{M}^{1}{L}^{1}{T}^{-\dfrac{5}{2}}\right]
Required value, [b2a]=[M2L2T−5][M1L−1T−2]=[M1L3T−3]\left[\dfrac{{b}^{2}}{a}\right]=\dfrac{\left[{M}^{2}{L}^{2}{T}^{-5}\right]}{\left[{M}^{1}{L}^{-1}{T}^{-2}\right]}=\left[{M}^{1}{L}^{3}{T}^{-3}\right]

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