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1.Units, Dimensions and Measurement
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Force $(F)$ and density $(d)$ are related as $F\, = \,\frac{\alpha }{{\beta \, + \,\sqrt d }}$ then dimension of $\alpha $ and $\beta$ are
A$M^{3/2} L^{-1/2} T^{-2}, M^{1/2} L^{-3/2} T^0$
B$M^{1/2 }L^{-3/2} T^{-2}, M^{-3/2} L^{-3/2} T^0$
C$M^{3} L^{-1} T^{-2/3}, M^{2} L^{-3} T^{2}$
D$M^{2} L^{-1/2} T^{-2}, M^{3/2} L^{-1/2} T^0$
Solution
$\alpha = \,\,\left[ {F\sqrt d } \right]\,\, = \,\,\left[ {ML{T^{ – 2}}} \right]\,\left[ {{M^{1/2}}\,{L^{ – 3/2}}} \right]\,\, = \,\,{M^{3/2}}{L^{ – 1/2}}{T^{ – 2}},\,$
$\,\left[ \beta \right]\,\, = \,\,{\left[ d \right]^{1/2}}\,\, = \,\,{\left[ {{M^1}{L^{ – 3}}\,{T^0}} \right]^{1/2}}\,\, = \,\,\left[ {{M^{1/2}}\,{L^{ – 3/2}}{T^0}} \right]$
$\,\left[ \beta \right]\,\, = \,\,{\left[ d \right]^{1/2}}\,\, = \,\,{\left[ {{M^1}{L^{ – 3}}\,{T^0}} \right]^{1/2}}\,\, = \,\,\left[ {{M^{1/2}}\,{L^{ – 3/2}}{T^0}} \right]$
Standard 11
Physics
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