Units and Measurement

Solid angle $d\Omega =\frac{dA}{r^2}$ has dimensions $\left[M^0 L^0 T^0\right]$

Strain  $=\frac{\Delta l}{l}$ has dimensions $\left[M^0 L^0 T^0\right]$

Angle measured in radians is also dimensionless  $\left[M^0 L^0 T^0\right]$

$\theta =\frac{l}{r}$

$T=2\pi \sqrt[]{\frac{\mathcal{l}}{g}}$

$g=4{\pi }^2\frac{\mathcal{l}}{T^2}$

$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta \mathcal{l}}{\mathcal{l}}+\frac{2\Delta T}{T}$

Student 1 will get least error because it has more number of oscillation

Vernier calipers are instruments for making accurate linear measurements. They are commonly used to measure the diameter of a sphere or the depth of a hole. Its precision comes from a sliding Vernier scale that runs along a fixed main scale. 

There are two types of errors in measurements. One is a systematic error, which comes from a fault in the measuring device. The other is a random error in measurement, which is caused by a sudden change in the environment.  

Let Mass α  $\left[t^a{\upsilon }^b{\mathcal{l}}^c\right]\Rightarrow \left[M^1{L}^0{T}^0\right]\equiv T^a*{\left[L{T}^{-1}\right]}^b*{\left[M{L}^2{T}^{-1}\right]}^c$

=> a – b – c = 0    c = 1, and b + 2c = 0 Þ b = -2c = -2

=> a = b + c = 1 – 2 = -1

From question :

9MSD = 10VSD

LC = 1MSD – 1VSD = 1MSD - $\frac{9}{10}MSD=0.1MSD=0.1*1mm=0.1mm$

Reading = 10mm + 8 * 0.1 mm = 10.8 mm

Final reading = 10.8 mm – 0.04 mm = 10.4 mm = 104 * 10^-2 cm

So radius = 52 * 10^-2 cm

Using F = MA = $m\frac{V}{T}$  

=> m = FTV^-1

(a) Torque ® ML²T^-2 ↑ (iii)

(b) Impulse ® MlT^-1 ↑ (i)

(c) Tension ↑ MLT^-2 ↑ (iv)

(d) Surface Tension ↑  $\frac{ML{T}^{-2}}{L}=M{T}^{-2}$ (ii)

$\to \left[W\right]=\left[\tau \theta \right]=\left[M{L}^2{T}^{-2}\right]\to correct$

$\to \left[h\right]=\left[L\right]=\left[\frac{2scos\theta }{\rho rg}\right]\to correct$

$\left[v\right]=\left[L^3\right]\left[\frac{\pi p{a}^4}{8\eta L}\right]=\left[L^3{T}^{-1}\right]\to$ Incorrect

$\to J=\epsilon \frac{\partial E}{\partial t}\to$ correct

$\left[A\stackrel{\downarrow }{L}{}^{-2}\right]\left[A\stackrel{\downarrow }{L}{}^{-2}\right]$

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

$\left[\rho \right]=M{L}^{-3}=F{L}^{-1}{T}^2{L}^{-3}=F{L}^{-4}{T}^2$