Question

The dimensional formula of physical quantity is $\left[M^a{L}^b{T}^c\right]$. Then that physical quantity is

• A. surface tension if a = 1, b = 1, c = -2
• B. force if a = 1, b = 1, c = 2
• C. angular frequency if a = 0, b = 0, c = -1
• D. spring constant if a = 1, b = -1, c = -2

Answer 1

The correct option is C angular frequency if a = 0, b = 0, c = -1

1) Surface Tension $\left[S\right]=\frac{Force\left[F\right]}{Length\left[L\right]}$

So, $S=\frac{\left[M^1{L}^1{T}^{-2}\right]}{\left[L^1\right]}$

$\therefore$ Surface Tension $=S=\left[M^1{L}^0{T}^{-2}\right]$

2) Force $\left[F\right]=$ mass (m) $\times$ acceleration (a)

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

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

3) Angular frequency $=w=\frac{red\left[d\varphi \right]}{\sec \left[dt\right]}$

$w=\left[M^0{L}^0{T}^{-1}\right]$

4) Spring constant $\left[K\right]=\frac{Forceapplied\left[F\right]}{Extensionofspring\left[x\right]}$

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

$K=\left[M^1{T}^{-2}\right]$

$K=\left[M^1{L}^0{T}^{-2}\right]$

$\therefore$ Option C is correct.