A packet contains silver powder of mass 20.23 | vedclass

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(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 ................

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