Units & Dimensions

JEE Main level, single correct, medium

123 of 149

Question 123 of 149

If Surface tension(S)(S), Moment of Inertia (I)(I) and Planck's constant (h),(h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be:

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Answer: (A) S1/2I1/2h0{S}^{1/2}{I}^{1/2}{h}^{0}

P∝SaIbhcP\propto {S}^{a}{I}^{b}{h}^{c}
[MLT−1]=[MT−2]a[ML2]b[ML2T−1]C[ML{T}^{-1}]={\left[M{T}^{-2}\right]}^{a}{\left[M{L}^{2}\right]}^{b}{\left[M{L}^{2}{T}^{-1}\right]}^{C}
[MLT−1]=[Ma+b+cL2b+2cT−2a−c]\left[ML{T}^{-1}\right]=[{M}^{a+b+c}{L}^{2b+2c}{T}^{-2a-c}]
From comparing the power on both side
a+b+c=1a+b+c=1
2b+2c=12b+2c=1
−2a−c=−1-2a-c=-1
From the above three equations we will get
a=\dfrac{1}{2},b=\dfrac{1}{2} &c=0
P∝S12I12h0P\propto {S}^{\dfrac{1}{2}}{I}^{\dfrac{1}{2}}{h}^{0}

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