6 questions · timed · auto-graded
Explanation:
Dimensionless quantities may have units.
$\text{n}\propto\text{size of u}$
$\text{n}\propto\text{u}^2$
$\text{n}\propto\sqrt{\text{u}}$
$\text{n}\propto\frac{1}{\text{u}}$
Explanation:
The larger the unit used to express the physical quantity, the lesser will be the numerical value.
Example: 1kg of sugar can be expressed as 1000g or 10000mg of sugar.
Here, g (gram) is the larger quantity as compared to mg (milligram), but the numerical value used with gram is lesser than the numerical value used with milligram.
Explanation:
We define length and time separately as it is not possible to define velocity without using these quantities. This means that one fundamental quantity depends on the other. So, these quantities cannot be listed as fundamental quantities in any system of units.
Explanation:
A unitless quantity never has a non-zero dimension.
Explantion:
[ax] = [x2]
⇒ [a] = [x] ...(1)
Dimension of LHS = Dimension of RHS
$\Rightarrow\Big[\frac{\text{dx}}{\sqrt{\text{x}}^2}\Big]=\big[\text{a}^{\text{n}}\big]$
$\Rightarrow\Big[\frac{\text{L}}{\text{L}}\Big]=\big[\text{a}^{\text{n}}\big] \ ...(2)$
$\Rightarrow\big[\text{L}^{0}\big]=[\text{a}^{\text{n}}]$
$\text{n}=0$
Explantion:
$\text{[Work done] = [ML}^2\text{ T}^{-2}]$
$\text{[Linear momentum] = [MLT}^{-1}]$
$\text{[Pressure] = [ML}^{-1}\text{T}^{-2}]$
$\text{[Energy per unit volume] = [ML}^{-1}\text{T}^{-2}]$
From the above, we can see that pressure and energy per unit volume have the same dimension, i.e., $\text{ML}^{-1}\text{T}^{-2}.$