Units and Measurement

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

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

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

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

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

Let 'x' he the value of one division of main scale

              $x=\frac{1}{20}cm=0.05cm$  

              Let y be value of one division on venire scale given

              10 y = 9 x

              $y=\frac{9x}{10}$  

              Least count = $x-\frac{9x}{10}=\frac{x}{10}=\frac{0.05}{10}$  

              = 0.005 cm

              = 5 * 10^-2 mm

Least  count of Vernier = 0.1mm

Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

a = dimensionless of b

a = MLT^-2

Least  count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

Pressure * time $=\frac{Ft}{A}=\left[ML{T}^{-2}\right]\left[T\right]\left[L^{-2}\right]=\left[M{L}^{-1}{T}^{-1}\right]$  

 Coefficient of viscosity, $\eta =\left[M{L}^{-1}{T}^{-1}\right]$  

  $F_{viscous}=-\eta A\frac{\Delta v}{\Delta y}$  

(1)      2.3056 -> 4 decimals

10.138 -> 3 decimals

-7.4671 -> 4 decimals

_________________

4.9765

Answer should have 3 decimals.

(2)  2.38 * 1.0 = 2.38 -> answer should have 2 significant digits

(3)   $\frac{8.05}{3.1}=2.59\approx 2.6$  answer should have 2SD

(4)  1.11 - 0.1 = 1.01 $\approx$  1.0 but is an intermediate step so we keep 1 digit extra. 1.01 * 9.0 = 9.09 $\approx$  9.1 -> both 1.01 and 9.0 have 2SD

Least  count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

Using F = MA = $m\frac{V}{T}$ $\frac{}{}$

m = FTV¹

Zero error = 0.0 + 4 * 0.01 = 0.04 cm
Reading = 2.5a + 7 * 0.01 = 2.57 cm
Correct reading = 2.57 cm - 0.04 cm = 2.53 cm