In an experiment, the following observation's were recorded : $L = 2.820\, m, M = 3.00 \,kg, l = 0.087 \,cm$, Diameter $D = 0.041 \,cm$ Taking $g = 9.81$ $m/{s^2}$ using the formula , $Y=\frac{{4MgL}}{{\pi {D^2}l}}$, the maximum permissible error in $Y$ is ......... $\%$
In an experiment, the following observation's were recorded : $L = 2.820\, m, M = 3.00 \,kg, l = 0.087 \,cm$, Diameter $D = 0.041 \,cm$ Taking $g = 9.81$ $m/{s^2}$ using the formula , $Y=\frac{{4MgL}}{{\pi {D^2}l}}$, the maximum permissible error in $Y$ is ......... $\%$
$7.96$
$4.56$
$6.5$
$8.42$
In an experiment, the following observation's were recorded : $L = 2.820\, m, M = 3.00 \,kg, l = 0.087 \,cm$, Diameter $D = 0.041 \,cm$ Taking $g = 9.81$ $m/{s^2}$ using the formula , $Y=\frac{{4MgL}}{{\pi {D^2}l}}$, the maximum permissible error in $Y$ is ......... $\%$
$Y = \frac{{4MgL}}{{\pi {D^2}l}}$
So maximum permissible error in
$Y = \frac{{\Delta Y}}{Y} \times 100 $ $= \left( {\frac{{\Delta M}}{M} + \frac{{\Delta g}}{g} + \frac{{\Delta L}}{L} + \frac{{2\Delta D}}{D} + \frac{{\Delta l}}{l}} \right) \times 100$
$ = \left( {\frac{1}{{300}} + \frac{1}{{981}} + \frac{1}{{2820}} + 2 \times \frac{1}{{41}} + \frac{1}{{87}}} \right) \times 100$
$ = 0.065 \times 100 = 6.5\% $
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