Tag: measurement of length

$ H={ I }^{ 2 }RT$ (by dimensional analysis)
$\displaystyle \frac { \triangle n }{ n } =(2\frac { \triangle I }{ I } +\frac { \triangle R }{ R } +\frac { \triangle T }{ T } )$%
          $=2(2)+1+1$
          $= 6 $ %

$ [X] = [M^{-1} L^{3} T^{-2}]$

$ \Rightarrow 
Error  = (\dfrac{\triangle M}{M} +
3 \dfrac{\triangle L}{L} + 2 \dfrac{\triangle T}{T}) %  $

$ \Rightarrow 2+3\times (3) +2\times (4)$

$\Rightarrow   $ 19  %

Let $Y=\dfrac { { a }^{ 2 } }{ b{ c }^{ 3 } } $
$\displaystyle \frac { \Delta Y }{ Y } \times 100=\pm \left( 2\times \frac { \Delta a }{ a } \times 100+\frac { \Delta b }{ b } \times 100+3\times \frac { \Delta c }{ c } \times 100 \right) $
                    $=\pm \left( 2\times 1+3+3\times 2 \right)$

The percentage error in measurement of radius is given, $\dfrac{\Delta r}{r}\times 100 =0.5$%
Thus, $\dfrac{\Delta r}{r}=0.5/100=0.005$
The surface area of sphere is $S=4\pi r^2$
Take ln and differentiate, $\dfrac{\Delta S}{S}=2\dfrac{\Delta r}{r}=2\times 0.005=0.01 $
The permissible or % error in the measurement of surface area $=\dfrac{\Delta S}{S}\times 100=0.01\times 100=1$%

$K.E = \dfrac{1}{2}mv^{2}$

$ \dfrac{\triangle K.E}{K.E} =

\dfrac{\triangle m}{m} + 2 \dfrac{\triangle v}{v}$

Given, $A=\dfrac{x^p}{y^qz^r}$
Take ln and then differentiate , $\dfrac{\Delta A}{A}=\pm\left(p\dfrac{\Delta x}{x}+q\dfrac{\Delta y}{y}+r\dfrac{\Delta z}{z}\right)$
Thus the maximum percentage error in A:
$\dfrac{\Delta A}{A}\times 100=p\dfrac{\Delta x}{x}\times 100+q\dfrac{\Delta y}{y}\times 100+r\dfrac{\Delta z}{z}\times 100=(3\times 1)+(2\times 0.5)+(1\times 2)=6$ %

Here, diameter $d=44.42 $mm and $\Delta d=\pm 0.03 $mm
Thus, true value of diameter will be in between $(44.42+0.03)=44.45$ mm and$(44.42-0.03)=44.39$ mm.
The relative error $=|\dfrac{\Delta d}{d}|=\dfrac{0.03}{44.42}=6.75\times 10^{-4}$ 
The percentage error $=|\dfrac{\Delta d}{d}|\times 100=0.07 $%

$ d = ML^{-3}\rightarrow  (By \   dimensional \  analysis)$

$V=\cfrac{4}{3}\pi R^3$