Question 88: Units , Dimensions and Error Analysis

Question

Consider two physical quantities $$A$$ and $$B$$ related to each other as $$E=\frac{B-x^2}{A t}$$ where $$E, x$$ and $$t$$ have dimensions of energy, length and time respectively. The dimension of $$A B$$ is

Accepted Answer

$$\begin{aligned}
& {[\mathrm{B}]=\mathrm{L}^2} \
& \mathrm{~A}=\frac{\mathrm{x}^2}{\mathrm{tE}}=\frac{\mathrm{L}^2}{\mathrm{TML}^2 \mathrm{~T}^{-2}}=\frac{1}{\mathrm{MT}^{-1}} \
& {[\mathrm{~A}]=\mathrm{M}^{-1 \mathrm{~T}}} \
& {[\mathrm{AB}]=\left[\mathrm{L}^2 \mathrm{M}^{-1} \mathrm{~T}^1ight]}
\end{aligned}$$

Answer

$$L^{-2} M^1 T^0$$

Answer

$$L^2 M^{-1} T^1$$

Answer

$$L^0 M^{-1} T^1$$

Answer

$$L^{-2} M^{-1} T^1$$

Units , Dimensions and Error Analysis

Consider two physical quantities $A and B related to each other as E=\frac{B-x^2}{A t} where E, x and t have dimensions of energy, length and time respectively. The dimension of A B$ is

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