Which of the following equations is dimensionally incorrect?\ $t=$ time, $h=$ height, $s=$ surface

Question

Which of the following equations is dimensionally incorrect?\ $t=$ time, $h=$ height, $s=$ surface tension, $	heta=$ angle,$ho=$ density $, a, r=$ radius,$g=$ acceleration due to gravity, $v=$ volume, $p=$ pressure, $W=$ work done, $	au=$ torque, $\varepsilon=$ permittivity, $E=$ electric field, $J=$ current density, $L=$ length\

Professor PC Thomas Classes · Answer

$ \begin{aligned} & (\mathrm{b}): t=[\mathrm{T}], h=[\mathrm{L}], s=\left[\mathrm{MT}^{-2}ight], 	heta=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0ight] \ & ho=\left[\mathrm{ML}^{-3}ight], a, r=[\mathrm{L}], g=\left[\mathrm{LT}^{-2}ight], v=\left[\mathrm{L}^3ight] \ & p=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight], W=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight], 	au=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight] \ & \varepsilon=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2ight], E=\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}ight] \ & J=\left[\mathrm{AL}^{-2}ight], L=[\mathrm{L}], \eta=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}ight] \end{aligned} $ \ (a) $W=	au 	heta ;\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]$ Since the dimensions of both sides of the equation is same, the equation is dimensionally correct. \ (b) $v=\frac{\pi p a^4}{8 \eta L} ;\left[\mathrm{L}^3ight]=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight]\left[\mathrm{L}^4ight]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}ight][\mathrm{L}]} 
eq\left[\mathrm{L}^3 \mathrm{~T}^{-1}ight]$ So the equation is dimensionally incorrect. \ (c) $h=\frac{2 s \cos 	heta}{r ho g} ;[\mathrm{L}]=\frac{\left[\mathrm{MT}^{-2}ight]}{[\mathrm{L}]\left[\mathrm{ML}^{-3}ight]\left[\mathrm{LT}^{-2}ight]}=[\mathrm{L}]$ So the equation is dimensionally correct. \ (d) $J=\varepsilon \frac{\partial E}{\partial t} ;\left[\mathrm{AL}^{-2}ight]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2ight] 	imes \frac{\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}ight]}{[\mathrm{T}]}=\left[\mathrm{AL}^{-2}ight]$ So, the equation is dimensionally correct.