A packet contains silver powder of mass 20.23 | vedclass

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

Question

(B) Given:Initial mass $m_1 = 20.23 \,g \pm 0.01 \,g$Mass removed $m_2 = 5.75 \,g \pm 0.01 \,g$Let the remaining mass be $m = m_1 - m_2$.The value of

Vedclass · Accepted Answer

$14.48 \,g \pm 0.02 \,g$

Vedclass · Answer

(B) Given:Initial mass $m_1 = 20.23 \,g \pm 0.01 \,g$Mass removed $m_2 = 5.75 \,g \pm 0.01 \,g$Let the remaining mass be $m = m_1 - m_2$.The value of the remaining mass is $20.23 - 5.75 = 14.48 \,g$.According to the rules of error propagation,when two quantities are subtracted,their absolute errors are added.Therefore,the absolute error in the remaining mass is $\Delta m = \Delta m_1 + \Delta m_2 = 0.01 \,g + 0.01 \,g = 0.02 \,g$.Thus,the mass of the powder left back is $(14.48 \pm 0.02) \,g$.

$A$ packet contains silver powder of mass $20.23 \,g \pm 0.01 \,g$. Some of the powder of mass $5.75 \,g \pm 0.01 \,g$ is taken out from it. The mass of the powder left back is ................

Similar Questions

Error in volume of a sphere is $6\%$. Error in its radius will be .......... $\%$

The potential difference across the ends of a conductor is $(30 \pm 0.3) \ V$ and the current through the conductor is $(5 \pm 0.1) \ A$. The error in the determination of the resistance of the conductor is: (in $\%$)

The maximum error in the measurement of mass and length is $4 \%$ and $3 \%$ respectively. The error in the measurement of density of a cube will be (in $\%$)

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

$A$ wire has a mass $0.3 \pm 0.003 \, g$,radius $0.5 \pm 0.005 \, mm$ and length $6 \pm 0.06 \, cm$. The maximum percentage error in the measurement of its density is .......... $\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module