Units and Measurement

Since, $\frac{x^2}{\alpha kT}$ should be dimensionless.

So, dimension of  $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$

Dimension of  $\alpha {\beta }^2$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

$z=\frac{A^2{B}^3}{C^4}$

The relative error in z is given by

$\frac{\Delta z}{z}=\frac{2\Delta A}{A}+\frac{3\Delta B}{B}+\frac{4\Delta C}{C}$

E ∝ F? A? T?
[M¹L²T? ²] ∝ [M¹L¹T? ²]? [LT? ²]? [T]?
a = 1
a + b = 2 ⇒ b = 1
-2a - 2b + c = -2
⇒ c = 2
a = 1 b = 1 c = 2
E ∝ [F] [A] [T²]

Least count = 1 mm / 100 = 0.01 mm
Diameter = main scale reading + circular scale reading
Diameter = 0 + 52 * 0.01 mm
= 0.52 mm = 0.052 cm

Least count = (1 mm) / 100 = 0.01 mm.

$m=\rho \pi r^{2}1$

$\begin{array}{rr}& \rho =\frac{\mathrm{m}}{\pi {\mathrm{r}}^{2}1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }=\frac{\mathrm{\Delta }\mathrm{m}}{\mathrm{m}}+\frac{2\mathrm{\Delta }\mathrm{r}}{\mathrm{r}}+\frac{\mathrm{\Delta }\mathrm{l}}{1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }*100=\frac{0.002}{0.4}*100+\frac{2*0.001}{0.3}*100+\frac{0.02}{5}*100\\ & =\frac{0.2}{0.4}+\frac{0.2}{0.3}+\frac{2}{5}\\ & =0.5+0.67+0.4\\ & =1.57\\ & =1.6\mathrm{\%}\end{array}$

As the factors controlling temperature and voltage supply are beyond prediction and control so the error occurred due to unpredictable fluctuations of temperature and voltage would be random errors.

The dimensional formula for energy E is [ML²T? ²].
The dimensional formula for the gravitational constant G is [M? ¹L³T? ²].
The ratio E/G has dimensions: [ML²T? ²] / [M? ¹L³T? ²] = [M²L? ¹T? ].

Plane angle and solid angle are dimensionless but have units.

$m=\rho \pi r^{2}1$

$\begin{array}{rr}& \rho =\frac{\mathrm{m}}{\pi {\mathrm{r}}^{2}1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }=\frac{\mathrm{\Delta }\mathrm{m}}{\mathrm{m}}+\frac{2\mathrm{\Delta }\mathrm{r}}{\mathrm{r}}+\frac{\mathrm{\Delta }\mathrm{l}}{1}\\ & \frac{\mathrm{\Delta }\rho }{\rho }*100=\frac{0.002}{0.4}*100+\frac{2*0.001}{0.3}*100+\frac{0.02}{5}*100\\ & =\frac{0.2}{0.4}+\frac{0.2}{0.3}+\frac{2}{5}\\ & =0.5+0.67+0.4\\ & =1.57\\ & =1.6\mathrm{\%}\end{array}$