If the length of a cylinder is l=(4.00 pm 0.0 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\,{l}\,\, = \,\,(4.00\,\, \pm \,\,0.01\,\,)\,cm$

$\frac{{\Delta {l}}}{{l}}\, = \,\frac{{0.01}}{{4.0}}\, = \,0.0025$

$r\,\, = \,\,(0.250\,\, \pm \,\

Vedclass · Accepted Answer

$1.21$

Vedclass · Answer

$\,{l}\,\, = \,\,(4.00\,\, \pm \,\,0.01\,\,)\,cm$

$\frac{{\Delta {l}}}{{l}}\, = \,\frac{{0.01}}{{4.0}}\, = \,0.0025$

$r\,\, = \,\,(0.250\,\, \pm \,\,0.001\,)\,\,cm\,\,$

$\frac{{\Delta r}}{r}\, = \,\frac{{0.001}}{{0.250}}\, = \,0.004$

$m\,\, = \,\,(6.25\,\, \pm \,\,0.01)\,\,g\,\,\,\,$

$\frac{{\Delta m}}{m} = \frac{{0.01}}{{6.25}}\, = \,0.0016$

density = mass/volume $ = \,\,\frac{m}{{\pi {r^2}{l}}}\,$

$ \frac{{\Delta ho }}{ho }\, = \,\frac{{\Delta m}}{m}\, + \,\frac{{2\Delta r}}{r} = \frac{{\Delta {l}}}{{l}}$

$\frac{{\Delta ho }}{ho } 	imes \,100\,\% $ $ = \left( {\frac{{\Delta m}}{m} 	imes \,100} ight)\%  + $ $2\left( {\frac{{\Delta r}}{r} 	imes 100} ight)\%  + \left( {\frac{{\Delta {l}}}{{l}} 	imes \,100} ight)\% $

$ = \,(0.0025\, 	imes \,100)\,\% \,\, + \,\,2(0.004\, 	imes \,100)\% \,\, + \,\,(0.0016\, 	imes 100)\,\% $

$\, = \,0.25\% \, + \,0.8\%  + \,0.16\,\%  = \,1.21\% $

If the length of a cylinder is $l=(4.00 \pm 0.01) cm$, radius $r =(0.250 \pm 0.001) \;cm$ and mass $m =6.25 \pm 0.01\; g$. Calculate the percentage error in determination of density.

Similar Questions

The mass and volume of a body are found to be $(5.00 ± 0.05)\,kg$ and $(1.00 ± 0.05)\,m^3$ respectively. Then the maximum possible percentage error in its density is .......... $\%$

The mass of the body is $10.000\,g$ and its volume is $10.00\,cm^3$. If the measured values are expressed upto the correct significant figures, the maximum error in the measurement of density is