Question 78: 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: