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CBSE Class 11-science Answered

 
Asked by nitishkrnehu09 | 18 Nov, 2017, 10:11: PM
answered-by-expert Expert Answer
Dimension for frequency ν = T-1.
Dimension for Surface tension σ = MS-2
Dimension for density ρ = ML-3
 
Hence dimension for the given term is worked as follows.
 begin mathsize 12px style space open square brackets fraction numerator begin display style nu end style over denominator begin display style square root of fraction numerator sigma over denominator rho R cubed end fraction end root end style end fraction close square brackets equals space fraction numerator T to the power of negative 1 end exponent over denominator square root of begin display style fraction numerator M T to the power of negative 2 end exponent over denominator M L to the power of negative 3 end exponent L cubed end fraction end style end root end fraction equals space T to the power of negative 1 end exponent over T to the power of negative 1 end exponent space equals space d i m e n s i o n l e s s space q u a n t i t y end style
 
Let us perform the dimensional analysis for the option A
 
begin mathsize 12px style open square brackets fraction numerator k rho g R squared over denominator sigma end fraction close square brackets space equals space fraction numerator M L to the power of negative 3 end exponent L T to the power of negative 2 end exponent L squared over denominator M T to the power of negative 2 end exponent end fraction space equals space d i m e n s i o n l e s s space q u a n t i t y end style
if dimensional analysis is performed in same way to other options 
we will end up with some dimensions but not dimensionless quantity.
 
Hence option A is answer
Answered by | 20 Nov, 2017, 03:54: PM

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