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Find the dimensions of (i) force (ii) surface tension and (iii) momentum in terms of frequency, velocity and density as fundamental units. |
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Answer» SOLUTION :Frequency `n=T^(-1),T=1/n` Velocity `v=LT^(-1),L=vT=v/n` Density `d=ML^(-3),M=dL^(3)=(dv^(3))/(n^(3))` (i) Force `=MLT^(-2)=(dv^(3))/(n^(3))xxv/nxxn^(2)=(dv^(4))/(n^(2))` (ii) SURFACE tension `=MT^(-2)=(dv^(3))/(n^(3))xxn^(2)=(dv^(3))/n` (iii) Momentum `=MLT^(-1)=(dv^(3))/(n^(3))xxv/nxxn=(dv^(4))/(n^(3))` |
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