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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2019 · 8 April · Evening Shift · Physics · Units and Dimensions

If Surface tension ( S ) , Moment of Inertia ( I ) and Planck's...

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:

Options

  1. AS1/2I1/2h0{S}^{1 / 2}{I}^{1 / 2}{h}^{0}
  2. BS1/2I3/2h−1{S}^{1 / 2}{I}^{3 / 2}{h}^{- 1}
  3. CS3/2I1/2h0{S}^{3 / 2}{I}^{1 / 2}{h}^{0}
  4. DS1/2I1/2h−1{S}^{1 / 2}{I}^{1 / 2}{h}^{- 1}

Correct answer

A. S1/2I1/2h0{S}^{1 / 2}{I}^{1 / 2}{h}^{0}

Step-by-step solution

P∝SaIbhcP ∝ {S}^{a}{I}^{b}{h}^{c}
[MLT−1]=[MT−2]a[ML2][ ML{T}^{- 1} ] = {\left[M{T}^{- 2}\right]}^{a}{\left[M{L}^{2}\right]}^{b}{\left[M{L}^{2}{T}^{- 1}\right]}^{C}<br/>[MLT−1]=[Ma+b+cL2b+2cT−2a−c]<br/>\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=12,b=12 and c=0a = \dfrac{1}{2}, b = \dfrac{1}{2} \text{ and } c = 0
P∝S12I12h0P ∝ {S}^{\dfrac{1}{2}}{I}^{\dfrac{1}{2}}{h}^{0}
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