Question 106: Units , Dimensions and Error Analysis

Question

A force is represented by $F=a x^2+b t^{\frac{1}{2}}$
where $x=$ distance and $t=$ time. The dimensions of $b^2 / a$ are:

Accepted Answer

To determine the dimensions of $\frac{b^2}{a}$, let's start by identifying the dimensions of each term in the equation $F=a x^2+b t^{\frac{1}{2}}$, where $F$ represents force, $x$ represents distance, and $t$ represents time.
The dimension of force ($F$) is given by [MLT
-2
], where $M$ is mass, $L$ is length, and $T$ is time.
The term $ax^2$ has the same dimension as force, so:
$[a] = \frac{[F]}{[x]^2} = \frac{MLT^{-2}}{L^2} = M L^{-1} T^{-2}$
The term $bt^{\frac{1}{2}}$ also has the same dimension as force, which gives:
$[b] = \frac{[F]}{[t]^{\frac{1}{2}}} = \frac{MLT^{-2}}{T^{\frac{1}{2}}} = M L T^{-\frac{5}{2}}$
Now, to find $\frac{b^2}{a}$, we substitute the dimensions of $b$ and $a$:
$\left[\frac{b^2}{a}\right] = \frac{\left(M^2 L^2 T^{-5}\right)}{\left(M L^{-1} T^{-2}\right)} = [M^1 L^3 T^{-3}]$
Therefore, the dimensions of $\frac{b^2}{a}$ are $\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]$, which corresponds to mass times length cubed per time cubed.

Answer

$\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]$

Answer

$\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]$

Answer

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

Answer

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

Units , Dimensions and Error Analysis

A force is represented by F=a x^2+b t^{\frac{1}{2}} where x= distance and t= time. The dimensions of b^2 / a are:

Choose the correct answer: