Question

The expression for a physical quantity is $\sqrt{\frac{T}{\mu }}$ where T = Force and $\mu =\frac{Mass}{Length}$What is the physical quantity

• A. mass
• B. speed
• C. time
• D. distance

Answer 1

The correct option is B

speed

Step 1: Given data

An expression of a physical quantity is given,

$\sqrt{\frac{T}{\mu }}$

Where T is given as force, $T=ma$

$\mu$ is the value of mass per unit length, $\mu =\frac{m}{L}$

Substituting the value of T and $\mu$ we get,

$=\sqrt{\frac{mass\times acceleration}{{\displaystyle \raisebox{1ex}{$mass$}\!\left/ \!\raisebox{-1ex}{$length$}\right.}}}$-------------(1)

Step 2: Calculation of the dimensional formula

The dimensional formula for mass is $\left[M\right]$

The dimensional formula for acceleration is, $a=\frac{m}{s^2}=\frac{\left[L\right]}{{\left[T\right]}^2}=\left[L{T}^{-2}\right]$

The dimensional formula for length is,$L=\left[L\right]$

Substituting these dimensional formula in equation (1) we get,

$=\sqrt{\frac{\left[M\right][L{T}^{-2}]}{{\displaystyle \raisebox{1ex}{$\left[M\right]$}\!\left/ \!\raisebox{-1ex}{$\left[L\right]$}\right.}}}$

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

So, the dimensional representation of a given quantity is$\left[L\right]\left[T^{-1}\right]$ it is the dimensional formula of speed

So, $\sqrt{\frac{T}{\mu }}$is the expression for the physical quantity velocity or speed

Hence,

Option (b) is correct