The density rho of a uniform cylinder is dete | vedclass

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

Question

(B) The density $\rho$ of a cylinder is given by the formula $\rho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 l} = \frac{4m}{\pi d^2 l}$.To find the relativ

Vedclass · Accepted Answer

$0.82$

Vedclass · Answer

(B) The density $ho$ of a cylinder is given by the formula $ho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 l} = \frac{4m}{\pi d^2 l}$.To find the relative error, we use the formula $\frac{\Delta ho}{ho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$.Substituting the given values:$\frac{\Delta ho}{ho} = \frac{0.02}{97.42} + 2 	imes \left( \frac{0.02}{20.20} ight) + \frac{0.05}{8.35}$.Calculating each term:$\frac{0.02}{97.42} \approx 0.000205$,$2 	imes \left( \frac{0.02}{20.20} ight) \approx 0.001980$,$\frac{0.05}{8.35} \approx 0.005988$.Adding these values: $\frac{\Delta ho}{ho} \approx 0.000205 + 0.001980 + 0.005988 = 0.008173$.The percentage error is $\frac{\Delta ho}{ho} 	imes 100 \% \approx 0.008173 	imes 100 \% = 0.8173 \% \approx 0.82 \%$.

Similar Questions

$A$ force $F$ is applied on a square plate of side $L$. If the percentage error in determining $F$ is $3 \%$ and that in $L$ is $2 \%$,then the percentage error in determining the pressure is (in $\%$)

Two resistors of resistances $R_1 = (100 \pm 3) \,\Omega$ and $R_2 = (200 \pm 4) \,\Omega$ are connected in series. The maximum absolute error and percentage error in the equivalent resistance of the series combination are:

$A$ wire has mass $(0.3 \pm 0.003) \text{ g}$, radius $(0.5 \pm 0.005) \text{ cm}$, and length $(6 \pm 0.06) \text{ cm}$. The maximum percentage error in the measurement of density is (in $\%$)

What is called as relative error? Define fractional error.

Vedclass Test Series

Exam Paper Generator

Online Exam Module