Question

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

• A. Spring constant, if $a=1,b=1$and $c=-2$
• B. Surface tension, if $a=1,b=1$and $c=-2$
• C. Force, if$a=1,b=1,c=2$
• D. Angular frequency if $a=0,b=0,c=-1$

Answer 1

The correct option is D

Angular frequency if $a=0,b=0,c=-1$

Explanation for the correct option:

Option D: Angular frequency if $a=0,b=0,c=-1$

Angular frequency if $a=0,b=0,c=-1$

Step 1: Use the angular frequency formula:

Know that the angular frequency

$\omega =\frac{2\pi }{T}$

Where,$\omega$ angular frequency, $T$ is the time.

Now the dimensional of the angular frequency,

$$\begin{gathered} =\left[M^a{L}^b{T}^c\right] \\ =\frac{\left[M^0{L}^0{T}^0\right]}{\left[T^1\right]} \\ =\left[M^0{L}^0{T}^{-1}\right] \end{gathered}$$

Therefore the angular frequency physical quantity$a=0,b=0,c=-1$.

Explanation for the incorrect option(s):

Option A: Spring constant, if $a=1,b=1$and $c=-2$

Spring constant, if $a=1,b=1$and $c=-2$

It is measured in $N/m$ and has a formula for spring constant (Newton per meter). As a result, the dimension of K is

.$$\begin{gathered} =\frac{\left[ML{T}^{-2}\right]}{\left[L\right]} \\ =\left[M{T}^{-2}\right] \end{gathered}$$

Therefore, for the physical quantity of the spring constant, the values will be $a=0,b=0,c=-2$.

Option B: Surface tension, if $a=1,b=1$and $c=-2$

Surface tension, if $a=1,b=1$and $c=-2$

The surface tension dimension is

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

Therefore, for the physical quantity of the surface tension, the values will be $a=1,b=0,c=-2$.

Option C: Force, if$a=1,b=1,c=2$

Force, if$a=1,b=1,c=2$

The dimension of the force,

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

Therefore, for the physical quantity of the force, the values will be $a=1,b=1,c=-2$.

Hence, the correct option is D.