Question

Which of the following is dimensionally correct? ($\rho$=density,$\eta$=coefficient of viscosity,$P$=pressure,$S$=surface tension,$r$=radius,$g$=gravitational constant)

• A. $h=\frac{2Scos\theta }{\rho \times rg}$
• B. $v=\frac{P}{\rho }$
• C. $V=\frac{P{r}^{4}t}{\eta }$
• D. none of the above

Answer 1

The correct option is A $h=\frac{2Scos\theta }{\rho \times rg}$

Dimensional formula of the following are given as-

$S=M{T}^{-2}$, $g=L{T}^{-2}$, $\rho =M{L}^{-3}$, $r=L$, $h=L$, $v=L{T}^{-1}$, $P=M{L}^{-1}{T}^{-2}$, $t=T$, $\eta =M{L}^{-1}{T}^{-1}$

Equating LHS and RHS of the following options :

(A) : LHS = $L$

RHS $=\frac{\left[M{T}^{-2}\right]}{\left[M{L}^{-3}\right]\left[L\right]\left[L{T}^{-2}\right]}=\left[L\right]$

$⟹$ LHS = RHS

(B) : LHS = $L{T}^{-1}$

RHS $=\frac{\left[M{L}^{-1}{T}^{-2}\right]}{M{L}^{-3}}=\left[L^2{T}^{-2}\right]$

$⟹$ LHS $\ne$ RHS

(C) : LHS = $L{T}^{-1}$

RHS $=\frac{\left[M{L}^{-1}{T}^{-2}\right]\left[L^4\right]\left[T\right]}{\left[M{L}^{-1}{T}^{-1}\right]}=\left[L^4\right]$

$⟹$ LHS $\ne$ RHS

Thus option A is correct.