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 least count of a stop watch is $0.2\, second$. The time of $20\, oscillations$ of a pendulum is measured to be $25\, second$. The percentage error in the measurement of time will be ........ $\%$

The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left(\frac{x}{100}\right) \% .$ If the relative errors in measuring the mass and the diameter are $6.0 \%$ and $1.5 \%$ respectively, the value of $x$ is

$Assertion$: In the measurement of physical quantities direct and indirect methods are used.

$Reason$ : The accuracy and precision of measuring instruments along with errors in measurements should be taken into account, while expressing the result.

Write a note on combination of error.